Math trigonometry ...... due tomorrow....... (2023)

MargaretL.LialJohnHornsbyDavidI.SchneiderCallieDaniels-Trigonometry-Pearson2016.pdf

TrigonometryL I A L | H O R N S B Y | S C H N E I D E R | D A N I E L S

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Boston Columbus Indianapolis New York San FranciscoAmsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto

Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

Margaret L. LialAmerican River College

John HornsbyUniversity of New Orleans

David I. SchneiderUniversity of Maryland

Callie J. DanielsSt. Charles Community College

TrigonometryELEVENTH EDITION

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To Butch, Peggy, Natalie, and Alexis—and in memory of MarkE.J.H.

To Coach Lonnie Myers—thank you for your leadership on and off the court.

C.J.D.

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Contents

Preface xi

Resources for Success xvi

1 Trigonometric Functions 11.1 Angles 2Basic Terminology ■ Degree Measure ■ Standard Position ■ Coterminal Angles

1.2 Angle Relationships and Similar Triangles 10Geometric Properties ■ Triangles

Chapter 1 Quiz (Sections 1.1–1.2) 21

1.3 Trigonometric Functions 22The Pythagorean Theorem and the Distance Formula ■ Trigonometric Functions ■ Quadrantal Angles

1.4 Using the Definitions of the Trigonometric Functions 30Reciprocal Identities ■ Signs and Ranges of Function Values ■ Pythagorean Identities ■ Quotient Identities

Test Prep 39 ■ Review Exercises 42 ■ Test 45

2 Acute Angles and Right Triangles 472.1 Trigonometric Functions of Acute Angles 48Right-Triangle-Based Definitions of the Trigonometric Functions ■Cofunctions ■ How Function Values Change as Angles Change ■ Trigonometric Function Values of Special Angles

2.2 Trigonometric Functions of Non-Acute Angles 56Reference Angles ■ Special Angles as Reference Angles ■ Determination of Angle Measures with Special Reference Angles

2.3 Approximations of Trigonometric Function Values 64Calculator Approximations of Trigonometric Function Values ■ Calculator Approximations of Angle Measures ■ An Application

Chapter 2 Quiz ( Sections 2.1–2.3) 71

2.4 Solutions and Applications of Right Triangles 72Historical Background ■ Significant Digits ■ Solving Triangles ■ Angles of Elevation or Depression

2.5 Further Applications of Right Triangles 82Historical Background ■ Bearing ■ Further Applications

Test Prep 91 ■ Review Exercises 93 ■ Test 97

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vi CONTENTS

3 Radian Measure and the Unit Circle 993.1 Radian Measure 100Radian Measure ■ Conversions between Degrees and Radians ■ Trigonometric Function Values of Angles in Radians

3.2 Applications of Radian Measure 106Arc Length on a Circle ■ Area of a Sector of a Circle

3.3 The Unit Circle and Circular Functions 116Circular Functions ■ Values of the Circular Functions ■ Determining a Number with a Given Circular Function Value ■ Applications of Circular Functions ■ Function Values as Lengths of Line Segments

Chapter 3 Quiz (Sections 3.1–3.3) 126

3.4 Linear and Angular Speed 127Linear Speed ■ Angular Speed

Test Prep 133 ■ Review Exercises 135 ■ Test 138

4 Graphs of the Circular Functions 1394.1 Graphs of the Sine and Cosine Functions 140Periodic Functions ■ Graph of the Sine Function ■ Graph of the Cosine Function ■ Techniques for Graphing, Amplitude, and Period ■ Connecting Graphs with Equations ■ A Trigonometric Model

4.2 Translations of the Graphs of the Sine and Cosine Functions 153

Horizontal Translations ■ Vertical Translations ■ Combinations of Translations ■ A Trigonometric Model

Chapter 4 Quiz ( Sections 4.1 –4.2) 164

4.3 Graphs of the Tangent and Cotangent Functions 164Graph of the Tangent Function ■ Graph of the Cotangent Function ■ Techniques for Graphing ■ Connecting Graphs with Equations

4.4 Graphs of the Secant and Cosecant Functions 173Graph of the Secant Function ■ Graph of the Cosecant Function ■ Techniques for Graphing ■ Connecting Graphs with Equations ■ Addition of Ordinates

Summary Exercises on Graphing Circular Functions 181

4.5 Harmonic Motion 181Simple Harmonic Motion ■ Damped Oscillatory Motion

Test Prep 187 ■ Review Exercises 189 ■ Test 193

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viiCONTENTS

5 Trigonometric Identities 1955.1 Fundamental Identities 196Fundamental Identities ■ Uses of the Fundamental Identities

5.2 Verifying Trigonometric Identities 202Strategies ■ Verifying Identities by Working with One Side ■ Verifying Identities by Working with Both Sides

5.3 Sum and Difference Identities for Cosine 211Difference Identity for Cosine ■ Sum Identity for Cosine ■ Cofunction Identities ■ Applications of the Sum and Difference Identities ■ Verifying an Identity

5.4 Sum and Difference Identities for Sine and Tangent 221Sum and Difference Identities for Sine ■ Sum and Difference Identities for Tangent ■ Applications of the Sum and Difference Identities ■ Verifying an Identity

Chapter 5 Quiz (Sections 5.1–5.4) 230

5.5 Double-Angle Identities 230Double-Angle Identities ■ An Application ■ Product-to-Sum and Sum-to-Product Identities

5.6 Half-Angle Identities 238Half-Angle Identities ■ Applications of the Half-Angle Identities ■Verifying an Identity

Summary Exercises on Verifying Trigonometric Identities 245

Test Prep 246 ■ Review Exercises 248 ■ Test 250

6 Inverse Circular Functions and Trigonometric Equations 2516.1 Inverse Circular Functions 252Inverse Functions ■ Inverse Sine Function ■ Inverse Cosine Function ■ Inverse Tangent Function ■ Other Inverse Circular Functions ■ Inverse Function Values

6.2 Trigonometric Equations I 268Linear Methods ■ Zero-Factor Property Method ■ Quadratic Methods ■ Trigonometric Identity Substitutions ■ An Application

6.3 Trigonometric Equations II 275Equations with Half-Angles ■ Equations with Multiple Angles ■ An Application

Chapter 6 Quiz ( Sections 6.1–6.3) 282

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viii

7 Applications of Trigonometry and Vectors 2957.1 Oblique Triangles and the Law of Sines 296Congruency and Oblique Triangles ■ Derivation of the Law of Sines ■ Solutions of SAA and ASA Triangles (Case 1) ■ Area of a Triangle

7.2 The Ambiguous Case of the Law of Sines 306Description of the Ambiguous Case ■ Solutions of SSA Triangles (Case 2) ■ Analyzing Data for Possible Number of Triangles

7.3 The Law of Cosines 312Derivation of the Law of Cosines ■ Solutions of SAS and SSS Triangles (Cases 3 and 4) ■ Heron’s Formula for the Area of a Triangle ■ Derivation of Heron’s Formula

Chapter 7 Quiz ( Sections 7.1–7.3) 325

7.4 Geometrically Defined Vectors and Applications 326Basic Terminology ■ The Equilibrant ■ Incline Applications ■ Navigation Applications

7.5 Algebraically Defined Vectors and the Dot Product 336Algebraic Interpretation of Vectors ■ Operations with Vectors ■ The Dot Product and the Angle between Vectors

Summary Exercises on Applications of Trigonometry and Vectors 344

Test Prep 346 ■ Review Exercises 349 ■ Test 353

6.4 Equations Involving Inverse Trigonometric Functions 282Solution for x in Terms of y Using Inverse Functions ■ Solution of Inverse Trigonometric Equations

Test Prep 289 ■ Review Exercises 291 ■ Test 293

CONTENTS

8 Complex Numbers, Polar Equations, and Parametric Equations 3558.1 Complex Numbers 356Basic Concepts of Complex Numbers ■ Complex Solutions of Quadratic Equations (Part 1) ■ Operations on Complex Numbers ■ Complex Solutions of Quadratic Equations (Part 2) ■ Powers of i

8.2 Trigonometric (Polar) Form of Complex Numbers 366The Complex Plane and Vector Representation ■ Trigonometric (Polar) Form ■ Converting between Rectangular and Trigonometric (Polar) Forms ■ An Application of Complex Numbers to Fractals

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ixCONTENTS

8.3 The Product and Quotient Theorems 372Products of Complex Numbers in Trigonometric Form ■ Quotients of Complex Numbers in Trigonometric Form

8.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers 378

Powers of Complex Numbers (De Moivre’s Theorem) ■ Roots of Complex Numbers

Chapter 8 Quiz (Sections 8.1–8.4) 385

8.5 Polar Equations and Graphs 385Polar Coordinate System ■ Graphs of Polar Equations ■ Conversion from Polar to Rectangular Equations ■ Classification of Polar Equations

8.6 Parametric Equations, Graphs, and Applications 398Basic Concepts ■ Parametric Graphs and Their Rectangular Equivalents ■ The Cycloid ■ Applications of Parametric Equations

Test Prep 406 ■ Review Exercises 409 ■ Test 412

Appendices 413Appendix A Equations and Inequalities 413

Basic Terminology of Equations ■ Linear Equations ■ Quadratic Equations ■ Inequalities ■ Linear Inequalities and Interval Notation ■ Three-Part Inequalities

Appendix B Graphs of Equations 422The Rectangular Coordinate System ■ Equations in Two Variables ■ Circles

Appendix C Functions 428Relations and Functions ■ Domain and Range ■ Determining Whether Relations Are Functions ■ Function Notation ■ Increasing, Decreasing, and Constant Functions

Appendix D Graphing Techniques 438Stretching and Shrinking ■ Reflecting ■ Symmetry ■ Translations

Answers to Selected Exercises A-1Photo Credits C-1Index I-1

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Preface

WELCOME TO THE 11TH EDITIONIn the eleventh edition of Trigonometry, we continue our ongoing commitment to providing the best possible text to help instructors teach and students succeed. In this edition, we have remained true to the pedagogical style of the past while staying focused on the needs of today’s students. Support for all classroom types (traditional, hybrid, and online) may be found in this classic text and its supplements backed by the power of Pearson’s MyMathLab.

In this edition, we have drawn upon the extensive teaching experience of the Lial team, with special consideration given to reviewer suggestions. General updates include enhanced readability with improved layout of examples, better use of color in displays, and language written with students in mind. All calculator screenshots have been updated and now provide color displays to enhance students’ conceptual understanding. Each homework section now begins with a group of Concept Preview exercises, assignable in MyMathLab, which may be used to ensure students’ understanding of vocabulary and basic concepts prior to beginning the regular homework exercises.

Further enhancements include numerous current data examples and exercises that have been updated to reflect current information. Additional real-life exercises have been included to pique student interest; answers to writing exercises have been provided; better consistency has been achieved between the directions that introduce examples and those that introduce the corresponding exercises; and better guidance for rounding of answers has been provided in the exercise sets.

The Lial team believes this to be our best Trigonometry edition yet, and we sin-cerely hope that you enjoy using it as much as we have enjoyed writing it. Additional textbooks in this series are as follows:

College Algebra, Twelfth EditionCollege Algebra & Trigonometry, Sixth EditionPrecalculus, Sixth Edition

HIGHLIGHTS OF NEW CONTENT ■ Discussion of the Pythagorean theorem and the distance formula has been

moved from an appendix to Chapter 1.

■ In Chapter 2, the two sections devoted to applications of right triangles now begin with short historical vignettes, to provide motivation and illustrate how trigonometry developed as a tool for astronomers.

■ The example solutions of applications of angular speed in Chapter 3 have been rewritten to illustrate the use of unit fractions.

■ In Chapter 4, we have included new applications of periodic functions. They involve modeling monthly temperatures of regions in the southern hemisphere and fractional part of the moon illuminated for each day of a particular month. The example of addition of ordinates in Section 4.4 has been rewritten, and a new example of analysis of damped oscillatory motion has been included in Section 4.5.

■ Chapter 5 now presents a derivation of the product-to-sum identity for the product sin A cos B.

■ In Chapter 6, we include several new screens of periodic function graphs that differ in appearance from typical ones. They pertain to the music phenomena of pressure of a plucked spring, beats, and upper harmonics.

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xii PREFACE

■ The two sections in Chapter 7 on vectors have been reorganized but still cover the same material as in the previous edition. Section 7.4 now introduces geometrically defined vectors and applications, and Section 7.5 follows with algebraically defined vectors and the dot product.

■ In Chapter 8, the examples in Section 8.1 have been reordered for a better flow with respect to solving quadratic equations with complex solutions.

■ For visual learners, numbered Figure and Example references within the text are set using the same typeface as the figure number itself and bold print for the example. This makes it easier for the students to identify and connect them. We also have increased our use of a “drop down” style, when appropri-ate, to distinguish between simplifying expressions and solving equations, and we have added many more explanatory side comments. Guided Visual-izations, with accompanying exercises and explorations, are now available and assignable in MyMathLab.

■ Trigonometry is widely recognized for the quality of its exercises. In the eleventh edition, nearly 500 are new or modified, and many present updated real-life data. Furthermore, the MyMathLab course has expanded coverage of all exercise types appearing in the exercise sets, as well as the mid-chapter Quizzes and Summary Exercises.

FEATURES OF THIS TEXTSUPPORT FOR LEARNING CONCEPTSWe provide a variety of features to support students’ learning of the essential topics of trigonometry. Explanations that are written in understandable terms, figures and graphs that illustrate examples and concepts, graphing technology that supports and enhances algebraic manipulations, and real-life applications that enrich the topics with meaning all provide opportunities for students to deepen their understanding of mathematics. These features help students make mathematical connections and expand their own knowledge base.

■ Examples Numbered examples that illustrate the techniques for working exercises are found in every section. We use traditional explanations, side comments, and pointers to describe the steps taken—and to warn students about common pitfalls. Some examples provide additional graphing calcula-tor solutions, although these can be omitted if desired.

■ Now Try Exercises Following each numbered example, the student is directed to try a corresponding odd-numbered exercise (or exercises). This feature allows for quick feedback to determine whether the student has understood the principles illustrated in the example.

■ Real-Life Applications We have included hundreds of real-life applica-tions, many with data updated from the previous edition. They come from fields such as sports, biology, astronomy, geology, music, and environmental studies.

■ Function Boxes Special function boxes offer a comprehensive, visual introduction to each type of trigonometric function and also serve as an excellent resource for reference and review. Each function box includes a table of values, traditional and calculator-generated graphs, the domain, the range, and other special information about the function. These boxes are assignable in MyMathLab.

■ Figures and Photos Today’s students are more visually oriented than ever before, and we have updated the figures and photos in this edition to

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promote visual appeal. Guided Visualizations with accompanying exercises and explorations are now available and assignable in MyMathLab.

■ Use of Graphing Technology We have integrated the use of graphing calculators where appropriate, although this technology is completely op-tional and can be omitted without loss of continuity. We continue to stress that graphing calculators support understanding but that students must first master the underlying mathematical concepts. Exercises that require the use of a graphing calculator are marked with the icon .

■ Cautions and Notes Text that is marked CAUTION warns students of common errors, and NOTE comments point out explanations that should receive particular attention.

■ Looking Ahead to Calculus These margin notes offer glimpses of how the topics currently being studied are used in calculus.

SUPPORT FOR PRACTICING CONCEPTSThis text offers a wide variety of exercises to help students master trigonometry. The extensive exercise sets provide ample opportunity for practice, and the exercise problems generally increase in difficulty so that students at every level of under-standing are challenged. The variety of exercise types promotes understanding of the concepts and reduces the need for rote memorization.

■ NEW Concept Preview Each exercise set now begins with a group of CONCEPT PREVIEW exercises designed to promote understanding of vo-cabulary and basic concepts of each section. These new exercises are assign-able in MyMathLab and will provide support especially for hybrid, online, and flipped courses.

■ Exercise Sets In addition to traditional drill exercises, this text includes writing exercises, optional graphing calculator problems , and multiple-choice, matching, true/false, and completion exercises. Concept Check exer-cises focus on conceptual thinking. Connecting Graphs with Equations exercises challenge students to write equations that correspond to given graphs.

■ Relating Concepts Exercises Appearing at the end of selected exer-cise sets, these groups of exercises are designed so that students who work them in numerical order will follow a line of reasoning that leads to an un-derstanding of how various topics and concepts are related. All answers to these exercises appear in the student answer section, and these exercises are assignable in MyMathLab.

■ Complete Solutions to Selected Exercises Complete solutions to all exercises marked are available in the eText. These are often exercises that extend the skills and concepts presented in the numbered examples.

SUPPORT FOR REVIEW AND TEST PREPAmple opportunities for review are found within the chapters and at the ends of chapters. Quizzes that are interspersed within chapters provide a quick assessment of students’ understanding of the material presented up to that point in the chapter. Chapter “Test Preps” provide comprehensive study aids to help students prepare for tests.

■ Quizzes Students can periodically check their progress with in-chapter quizzes that appear in all chapters. All answers, with corresponding section references, appear in the student answer section. These quizzes are assign-able in MyMathLab.

xiiiPREFACE

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xiv

■ Summary Exercises These sets of in-chapter exercises give students the all-important opportunity to work mixed review exercises, requiring them to synthesize concepts and select appropriate solution methods.

■ End-of-Chapter Test Prep Following the final numbered section in each chapter, the Test Prep provides a list of Key Terms, a list of New Symbols (if applicable), and a two-column Quick Review that includes a section-by-section summary of concepts and examples. This feature con-cludes with a comprehensive set of Review Exercises and a Chapter Test. The Test Prep, Review Exercises, and Chapter Test are assignable in MyMathLab. Additional Cumulative Review homework assignments are available in MyMathLab, following every chapter.

PREFACE

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MyMathLab®Get the most out of

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MyMathLab® Online Course for Trigonometry by Lial, Hornsby, Schneider, and Daniels

to give students the practice they need to develop a conceptual understanding of

classroom formats (traditional, hybrid, and online).

Concept Preview Exercises

Exercise sets now begin with a group of Concept Preview Exer-

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Learning Catalytics is a “bring your own device” system of prebuilt

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MyNotes and MyClassroomExamplesMyNotes provide a note-taking structure for students to use while they read the text or watch the MyMathLab videos. MyClassroom Examples offer structure for notes taken during lecture and are for use with the Classroom Examples found in

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Resources for SuccessStudent SupplementsStudent’s Solutions ManualBy Beverly Fusfield

Provides detailed solutions to all odd-numbered text exercises

ISBN: 0-13-431021-7 & 978-0-13-431021-3

Video Lectures with Optional Captioning

Feature Quick Reviews and Example Solutions:Quick Reviews cover key definitions and procedures from each section.Example Solutions walk students through the detailed solution process for every example in the textbook.

Ideal for distance learning or supplemental instruction at home or on campus

Include optional text captioning Available in MyMathLab®

MyNotes Available in MyMathLab and offer structure for

students as they watch videos or read the text Include textbook examples along with ample space

for students to write solutions and notes Include key concepts along with prompts for

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Customizable so that instructors can add their own examples or remove examples that are not covered in their courses

MyClassroomExamples Available in MyMathLab and offer structure for

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opportunities to keep students engaged Customizable so that instructors can add their

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Instructor SupplementsAnnotated Instructor’s Edition

Provides answers in the margins to almost all text exercises, as well as helpful Teaching Tips and Classroom Examples

Includes sample homework assignments indicated by exercise numbers underlined in blue within each end-of-section exercise set

Sample homework exercises assignable in MyMathLab

ISBN: 0-13-421764-0 & 978-0-13-421764-2

Online Instructor’s Solutions ManualBy Beverly Fusfield

Provides complete solutions to all text exercises Available in MyMathLab or downloadable from

Pearson Education’s online catalog

Online Instructor’s Testing ManualBy David Atwood

Includes diagnostic pretests, chapter tests, final exams, and additional test items, grouped by section, with answers provided

Available in MyMathLab or downloadable from Pearson Education’s online catalog

TestGen® Enables instructors to build, edit, print, and administer

tests Features a computerized bank of questions developed

to cover all text objectives Available in MyMathLab or downloadable from

Pearson Education’s online catalog

Online PowerPoint Presentation and Classroom Example PowerPoints

Written and designed specifically for this text Include figures and examples from the text Provide Classroom Example PowerPoints that include

full worked-out solutions to all Classroom Examples Available in MyMathLab or downloadable from

Pearson Education’s online catalog

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xviii ACKNOWLEDGMENTS

ACKNOWLEDGMENTSWe wish to thank the following individuals who provided valuable input into this edition of the text.

John E. Daniels – Central Michigan UniversityMary Hill – College of DuPageRene Lumampao – Austin Community CollegeRandy Nichols – Delta CollegePatty Schovanec – Texas Tech UniversityDeanna M. Welsch – Illinois Central College

Our sincere thanks to those individuals at Pearson Education who have sup-ported us throughout this revision: Anne Kelly, Christine O’Brien, Joe Vetere, and Danielle Simbajon. Terry McGinnis continues to provide behind-the-scenes guid-ance for both content and production. We have come to rely on her expertise during all phases of the revision process. Marilyn Dwyer of Cenveo® Publishing Services, with the assistance of Carol Merrigan, provided excellent production work. Special thanks go out to Paul Lorczak and Perian Herring for their excellent accuracy-checking. We thank Lucie Haskins, who provided an accurate index, and Jack Hornsby, who provided assistance in creating calculator screens, researching data updates, and proofreading.

As an author team, we are committed to providing the best possible college algebra course to help instructors teach and students succeed. As we continue to work toward this goal, we welcome any comments or suggestions you might send, via e-mail, to [emailprotected]

Margaret L. LialJohn HornsbyDavid I. SchneiderCallie J. Daniels

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mailto:[emailprotected]

A sequence of similar triangles, a topic covered in this introductory chapter, can be used to approximate the spiral of the chambered nautilus.

Angles

Angle Relationships and Similar Triangles

Chapter 1 Quiz

Trigonometric Functions

Using the Definitions of the Trigonometric Functions

1.1

1.2

1.3

1.4

Trigonometric Functions1

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2 CHAPTER 1 Trigonometric Functions

Degree Measure The most common unit for measuring angles is the degree. Degree measure was developed by the Babylonians 4000 yr ago. To use degree measure, we assign 360 degrees to a complete rotation of a ray.* In Figure 4, notice that the terminal side of the angle corresponds to its initial side when it makes a complete rotation.

One degree, written 1°, represents 1

360 of a complete rotation.

Therefore, 90° represents 90360 =14 of a complete rotation, and 180° represents

180360 =

12 of a complete rotation.

An angle measuring between 0° and 90° is an acute angle. An angle mea-suring exactly 90° is a right angle. The symbol m is often used at the vertex of a right angle to denote the 90° measure. An angle measuring more than 90° but less than 180° is an obtuse angle, and an angle of exactly 180° is a straight angle.

1.1 Angles

Basic Terminology Two distinct points A and B determine a line called line AB. The portion of the line between A and B, including points A and B them-selves, is line segment AB, or simply segment AB. The portion of line AB that starts at A and continues through B, and on past B, is the ray AB. Point A is the endpoint of the ray. See Figure 1.

In trigonometry, an angle consists of two rays in a plane with a common endpoint, or two line segments with a common endpoint. These two rays (or segments) are the sides of the angle, and the common endpoint is the vertex of the angle. Associated with an angle is its measure, generated by a rotation about the vertex. See Figure 2. This measure is determined by rotating a ray starting at one side of the angle, the initial side, to the position of the other side, the terminal side. A counterclockwise rotation generates a positive measure, and a clockwise rotation generates a negative measure. The rotation can consist of more than one complete revolution.

Figure 3 shows two angles, one positive and one negative.

■ Basic Terminology■ Degree Measure■ Standard Position■ Coterminal Angles

A BLine AB

Segment AB

A B

Ray AB

A B

Figure 1

Terminal side

Vertex A

Angle A

Initial side

Figure 2Positive angle Negative angle

Figure 3

An angle can be named by using the name of its vertex. For example, the angle on the right in Figure 3 can be named angle C. Alternatively, an angle can be named using three letters, with the vertex letter in the middle. Thus, the angle on the right also could be named angle ACB or angle BCA.

*The Babylonians were the first to subdivide the circumference of a circle into 360 parts. There are various theories about why the number 360 was chosen. One is that it is approximately the number of days in a year, and it has many divisors, which makes it convenient to work with in computations.

A complete rotation of a raygives an angle whose measure

is 360°. of a complete

rotation gives an angle whosemeasure is 1°.

3601

Figure 4

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3 1.1 Angles

In Figure 5, we use the Greek letter U (theta)* to name each angle. The table in the margin lists the upper- and lowercase Greek letters, which are often used in trigonometry.

* In addition to u (theta), other Greek letters such as a (alpha) and b (beta) are used to name angles.

Acute angle0° < u < 90°

Right angleu = 90°

Obtuse angle

90° < u < 180°

Straight angleu = 180°

Figure 5

If the sum of the measures of two positive angles is 90°, the angles are comple-mentary and the angles are complements of each other. Two positive angles with measures whose sum is 180° are supplementary, and the angles are supplements.

The Greek Letters

Α a alphaΒ b betaΓ g gamma∆ d deltaΕ e epsilonΖ z zetaΗ h etaϴ u thetaΙ i iotaΚ k kappaΛ l lambdaΜ m muΝ n nuΞ j xiΟ o omicronΠ p piΡ r rhoΣ s sigmaΤ t tauΥ y upsilonΦ f phiΧ x chiΨ c psiΩ v omega

EXAMPLE 1 Finding the Complement and the Supplement of an Angle

Find the measure of (a) the complement and (b) the supplement of an angle measuring 40°.

SOLUTION

(a) To find the measure of its complement, subtract the measure of the angle from 90°.

90° - 40° = 50° Complement of 40°(b) To find the measure of its supplement, subtract the measure of the angle

from 180°.180° - 40° = 140° Supplement of 40°

■✔ Now Try Exercise 11.

EXAMPLE 2 Finding Measures of Complementary and Supplementary Angles

Find the measure of each marked angle in Figure 6.

SOLUTION

(a) Because the two angles in Figure 6(a) form a right angle, they are comple-mentary angles.

6x + 3x = 90 Complementary angles sum to 90°. 9x = 90 Combine like terms.

x = 10 Divide by 9.Don’t stop here.

Be sure to determine the measure of each angle by substituting 10 for x in 6x and 3x. The two angles have measures of 61102 = 60° and 31102 = 30°.

(b) The angles in Figure 6(b) are supplementary, so their sum must be 180°.

4x + 6x = 180 Supplementary angles sum to 180°. 10x = 180 Combine like terms.

x = 18 Divide by 10.The angle measures are 4x = 41182 = 72° and 6x = 61182 = 108°.

■✔ Now Try Exercises 23 and 25.

(6x)°(3x)°

(a)

(4x)° (6x)°

(b)

Figure 6

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4 CHAPTER 1 Trigonometric Functions

The measure of angle A in Figure 7 is 35°. This measure is often expressed by saying that m 1angle A 2 is 35°, where m1angle A2 is read “the measure of angle A.” The symbolism m1angle A2 = 35° is abbreviated as A = 35°.

Traditionally, portions of a degree have been measured with minutes and seconds. One minute, written 1′, is 160 of a degree.

1′ =160° or 60′ = 1°

One second, 1″, is 160 of a minute.

1″ =160′=

13600

° or 60″ = 1′ and 3600″ = 1°

The measure 12° 42′ 38″ represents 12 degrees, 42 minutes, 38 seconds.

A = 35°x

Figure 7

EXAMPLE 4 Converting between Angle Measures

(a) Convert 74° 08′ 14″ to decimal degrees to the nearest thousandth.

(b) Convert 34.817° to degrees, minutes, and seconds to the nearest second.

SOLUTION

(a) 74° 08′ 14″

= 74° + 860°+ 14

3600° 08′ # 1°60′ = 860° and 14″ # 1°3600″ = 143600°

≈ 74° + 0.1333° + 0.0039° Divide to express the fractions as decimals. ≈ 74.137° Add and round to the nearest thousandth.

An alternative way to measure angles involves decimal degrees. For example,

12.4238° represents 12 4238

10,000°.

EXAMPLE 3 Calculating with Degrees, Minutes, and Seconds

Perform each calculation.

(a) 51° 29′ + 32° 46′ (b) 90° - 73° 12′

SOLUTION

(a) 51° 29′ + 32° 46′ 83° 75′

Add degrees and minutes separately.

The sum 83° 75′ can be rewritten as follows.

83° 75′

= 83° + 1° 15′ 75′ = 60′ + 15′ = 1° 15′ = 84° 15′ Add.

(b) 90° 89° 60′ Write 90° as 89° 60′. - 73° 12′ can be written - 73° 12′ 16° 48′

■✔ Now Try Exercises 41 and 45.

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5 1.1 Angles

(b) 34.817°

= 34° + 0.817° Write as a sum. = 34° + 0.817160′2 0.817° # 60′1° = 0.817160′2 = 34° + 49.02′ Multiply. = 34° + 49′ + 0.02′ Write 49.02′ as a sum. = 34° + 49′ + 0.02160″2 0.02′ # 60″1′ = 0.02160″2 = 34° + 49′ + 1.2″ Multiply. ≈ 34° 49′ 01″ Approximate to the nearest second.

■✔ Now Try Exercises 61 and 71.

This screen shows how the TI-84 Plus performs the conversions in Example 4. The ▶DMS option is found in the ANGLE Menu.

Standard Position An angle is in standard position if its vertex is at the origin and its initial side lies on the positive x-axis. The angles in Figures 8(a) and 8(b) are in standard position. An angle in standard position is said to lie in the quadrant in which its terminal side lies. An acute angle is in quadrant I (Figure 8(a)) and an obtuse angle is in quadrant II (Figure 8(b)). Figure 8(c) shows ranges of angle measures for each quadrant when 0° 6 u 6 360°.

Terminal side

Vertex Initial side

Q I

50°

Q II

160°

0°360°180°

90°

270°

Q II90° < u < 180°

Q I0° < u < 90°

Q III180° < u < 270°

Q IV270° < u < 360°

(a) (b) (c)

Figure 8

Angles in standard position

Coterminal Angles A complete rotation of a ray results in an angle meas-uring 360°. By continuing the rotation, angles of measure larger than 360° can be produced. The angles in Figure 9 with measures 60° and 420° have the same initial side and the same terminal side, but different amounts of rotation. Such angles are coterminal angles. Their measures differ by a multiple of 360°. As shown in Figure 10, angles with measures 110° and 830° are coterminal.

60°420°

Coterminalangles

Figure 9

110°830°

Coterminalangles

Figure 10

Quadrantal Angles

Angles in standard position whose terminal sides lie on the x-axis or y-axis, such as angles with measures 90°, 180°, 270°, and so on, are quadrantal angles.

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6 CHAPTER 1 Trigonometric Functions

EXAMPLE 5 Finding Measures of Coterminal Angles

Find the angle of least positive measure that is coterminal with each angle.

(a) 908° (b) -75° (c) -800°

SOLUTION

(a) Subtract 360° as many times as needed to obtain an angle with measure greater than 0° but less than 360°.

908° - 2 # 360° = 188° Multiply 2 # 360°. Then subtract. An angle of 188° is coterminal with an angle of 908°. See Figure 11.

188°908°

Figure 11

285°–75°

Figure 12

(b) Add 360° to the given negative angle measure to obtain the angle of least positive measure. See Figure 12.

-75° + 360° = 285°

3 # 360° = 1080°. Add 1080° to -800° to obtain

-800° + 1080° = 280°.■✔ Now Try Exercises 81, 91, and 95.

Sometimes it is necessary to find an expression that will generate all angles coterminal with a given angle. For example, we can obtain any angle coterminal with 60° by adding an integer multiple of 360° to 60°. Let n represent any inte-ger. Then the following expression represents all such coterminal angles.

60° + n # 360° Angles coterminal with 60°The table below shows a few possibilities.

Examples of Angles Coterminal with 60°

Value of n Angle Coterminal with 60° 2 60° + 2 # 360° = 780° 1 60° + 1 # 360° = 420° 0 60° + 0 # 360° = 60° (the angle itself)-1 60° + 1-12 # 360° = -300°-2 60° + 1-22 # 360° = -660°

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7 1.1 Angles

This table shows some examples of coterminal quadrantal angles.

Examples of Coterminal Quadrantal Angles

Quadrantal Angle U Coterminal with U

0° {360°, {720° 90° -630°, -270°, 450°180° -180°, 540°, 900°270° -450°, -90°, 630°

EXAMPLE 6 Analyzing Revolutions of a Disk Drive

A constant angular velocity disk drive spins a disk at a constant speed. Suppose a disk makes 480 revolutions per min. Through how many degrees will a point on the edge of the disk move in 2 sec?

SOLUTION The disk revolves 480 times in 1 min, or 48060 times = 8 times per sec

(because 60 sec = 1 min). In 2 sec, the disk will revolve 2 # 8 = 16 times. Each revolution is 360°, so in 2 sec a point on the edge of the disk will revolve

16 # 360° = 5760°.A unit analysis expression can also be used.

480 rev1 min

* 1 min60 sec

* 360°1 rev

* 2 sec = 5760° Divide out common units.■✔ Now Try Exercise 123.

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. One degree, written 1°, represents of a complete rotation.

2. If the measure of an angle is x°, its complement can be expressed as - x°.

3. If the measure of an angle is x°, its supplement can be expressed as - x°.

4. The measure of an angle that is its own complement is .

5. The measure of an angle that is its own supplement is .

6. One minute, written 1′, is of a degree.

7. One second, written 1″, is of a minute.

8. 12° 30′ written in decimal degrees is .

9. 55.25° written in degrees and minutes is .

10. If n represents any integer, then an expression representing all angles coterminal with 45° is 45° + .

1.1 Exercises

Find the measure of (a) the complement and (b) the supplement of an angle with the given measure. See Examples 1 and 3.

11. 30° 12. 60° 13. 45° 14. 90°

15. 54° 16. 10° 17. 1° 18. 89°

19. 14° 20′ 20. 39° 50′ 21. 20° 10′ 30″ 22. 50° 40′ 50″

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8 CHAPTER 1 Trigonometric Functions

Find the measure of each marked angle. See Example 2.

23.

(7x)° (11x)°

24.(20x + 10)° (3x + 9)°

25.

(2x)°

(4x)°

26.

(5x + 5)°

(3x + 5)°

27.(–4x)°

(–14x)°

28.

(9x)°

29. supplementary angles with measures 10x + 7 and 7x + 3 degrees

30. supplementary angles with measures 6x - 4 and 8x - 12 degrees

31. complementary angles with measures 9x + 6 and 3x degrees

32. complementary angles with measures 3x - 5 and 6x - 40 degrees

Find the measure of the smaller angle formed by the hands of a clock at the following times.

33. 34.

35. 3:15 36. 9:45 37. 8:20 38. 6:10

Perform each calculation. See Example 3.

39. 62° 18′ + 21° 41′ 40. 75° 15′ + 83° 32′ 41. 97° 42′ + 81° 37′

42. 110° 25′ + 32° 55′ 43. 47° 29′ - 71° 18′ 44. 47° 23′ - 73° 48′

45. 90° - 51° 28′ 46. 90° - 17° 13′ 47. 180° - 119° 26′

48. 180° - 124° 51′ 49. 90° - 72° 58′ 11″ 50. 90° - 36° 18′ 47″

51. 26° 20′ + 18° 17′ - 14° 10′ 52. 55° 30′ + 12° 44′ - 8° 15′

Convert each angle measure to decimal degrees. If applicable, round to the nearest thou-sandth of a degree. See Example 4(a).

53. 35° 30′ 54. 82° 30′ 55. 112° 15′ 56. 133° 45′

57. -60° 12′ 58. -70° 48′ 59. 20° 54′ 36″ 60. 38° 42′ 18″

61. 91° 35′ 54″ 62. 34° 51′ 35″ 63. 274° 18′ 59″ 64. 165° 51′ 09″

Convert each angle measure to degrees, minutes, and seconds. If applicable, round to the nearest second. See Example 4(b).

65. 39.25° 66. 46.75° 67. 126.76° 68. 174.255°

69. -18.515° 70. -25.485° 71. 31.4296° 72. 59.0854°

73. 89.9004° 74. 102.3771° 75. 178.5994° 76. 122.6853°

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9 1.1 Angles

Find the angle of least positive measure (not equal to the given measure) that is coterminal with each angle. See Example 5.

77. 32° 78. 86° 79. 26° 30′ 80. 58° 40′

81. -40° 82. -98° 83. -125° 30′ 84. -203° 20′

85. 361° 86. 541° 87. -361° 88. -541°

89. 539° 90. 699° 91. 850° 92. 1000°

93. 5280° 94. 8440° 95. -5280° 96. -8440°

Give two positive and two negative angles that are coterminal with the given quadrantal angle.

97. 90° 98. 180° 99. 0° 100. 270°

Write an expression that generates all angles coterminal with each angle. Let n represent any integer.

101. 30° 102. 45° 103. 135° 104. 225°

105. -90° 106. -180° 107. 0° 108. 360°

109. Why do the answers to Exercises 107 and 108 give the same set of angles?

110. Concept Check Which two of the following are not coterminal with r°?

A. 360° + r° B. r° - 360° C. 360° - r° D. r° + 180°

Concept Check Sketch each angle in standard position. Draw an arrow representing the correct amount of rotation. Find the measure of two other angles, one positive and one negative, that are coterminal with the given angle. Give the quadrant of each angle, if applicable.

111. 75° 112. 89° 113. 174° 114. 234°

115. 300° 116. 512° 117. -61° 118. -159°

119. 90° 120. 180° 121. -90° 122. -180°

Solve each problem. See Example 6.

123. Revolutions of a Turntable A turntable in a shop makes 45 revolutions per min. How many revolutions does it make per second?

124. Revolutions of a Windmill A windmill makes 90 revolutions per min. How many revolutions does it make per second?

125. Rotating Tire A tire is rotating 600 times per min. Through how many degrees does a

point on the edge of the tire move in 12 sec?

126. Rotating Airplane Propeller An airplane propeller rotates 1000 times per min. Find the number of degrees that a point on the edge of the propeller will rotate in 2 sec.

127. Rotating Pulley A pulley rotates through 75° in 1 min. How many rotations does the pulley make in 1 hr?

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10 CHAPTER 1 Trigonometric Functions

128. Surveying One student in a surveying class measures an angle as 74.25°, while another student measures the same angle as 74° 20′. Find the difference between these measurements, both to the nearest minute and to the nearest hundredth of a degree.

129. Viewing Field of a Telescope As a consequence of Earth’s rotation, celestial objects such as the moon and the stars appear to move across the sky, rising in the east and setting in the west. As a result, if a telescope on Earth remains stationary while viewing a celestial object, the object will slowly move outside the viewing field of the telescope. For this reason, a motor is often attached to telescopes so that the telescope rotates at the same rate as Earth. Determine how long it should take the motor to turn the telescope through an angle of 1 min in a direction per-pendicular to Earth’s axis.

130. Angle Measure of a Star on the American Flag Determine the measure of the angle in each point of the five-pointed star appearing on the American flag. (Hint: Inscribe the star in a circle, and use the following theorem from geometry: An angle whose vertex lies on the circumference of a circle is equal to half the central angle that cuts off the same arc. See the figure.)

74.25°

1.2 Angle Relationships and Similar Triangles

Geometric Properties In Figure 13, we extended the sides of angle NMP to form another angle, RMQ. The pair of angles NMP and RMQ are vertical angles. Another pair of vertical angles, NMQ and PMR, are also formed. Vertical angles have the following important property.

■ Geometric Properties■ Triangles

Q R

N P

Vertical angles

Figure 13

Vertical Angles

Vertical angles have equal measures.

Parallel lines are lines that lie in the same plane and do not intersect. Figure 14 shows parallel lines m and n. When a line q intersects two parallel lines, q is called a transversal. In Figure 14, the transversal intersecting the parallel lines forms eight angles, indicated by numbers.

1 23 4

5 67 8

Transversal

Parallel lines

Figure 14

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111.2 Angle Relationships and Similar Triangles

We learn in geometry that the degree measures of angles 1 through 8 in Figure 14 possess some special properties. The following table gives the names of these angles and rules about their measures.

Angle Pairs of Parallel Lines Intersected by a Transversal

Name Sketch Rule

Alternate interior angles

(also 3 and 6)

Angle measures are equal.

Alternate exterior angles1

8 (also 2 and 7)

Angle measures are equal.

Interior angles on the same side of a transversal

(also 3 and 5)

Angle measures add to 180°.

Corresponding angles

(also 1 and 5, 3 and 7, 4 and 8)

Angle measures are equal.

(3x + 2)°

(5x – 40)°

1 23

Figure 15

EXAMPLE 1 Finding Angle Measures

Find the measures of angles 1, 2, 3, and 4 in Figure 15, given that lines m and n are parallel.

SOLUTION Angles 1 and 4 are alternate exterior angles, so they are equal.

3x + 2 = 5x - 40 Alternate exterior angles have equal measures. 42 = 2x Subtract 3x and add 40. 21 = x Divide by 2.

Angle 1 has measure

3x + 2 = 3 # 21 + 2 Substitute 21 for x. = 65°. Multiply, and then add.

Angle 4 has measure

5x - 40 = 5 # 21 - 40 Substitute 21 for x. = 65°. Multiply, and then

subtract.

Angle 2 is the supplement of a 65° angle, so it has measure

180° - 65° = 115°.

Angle 3 is a vertical angle to angle 1, so its measure is also 65°. (There are other ways to determine these measures.)

■✔ Now Try Exercises 11 and 19.

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12 CHAPTER 1 Trigonometric Functions

Triangles An important property of triangles, first proved by Greek geom-eters, deals with the sum of the measures of the angles of any triangle.

Angle Sum of a Triangle

The sum of the measures of the angles of any triangle is 180°.

A rather convincing argument for the truth of this statement uses any size trian-gle cut from a piece of paper. Tear each corner from the triangle, as suggested in Figure 16(a). We should be able to rearrange the pieces so that the three angles form a straight angle, which has measure 180°, as shown in Figure 16(b).

1 3

(a)

(b)

Figure 16

EXAMPLE 2 Applying the Angle Sum of a Triangle Property

The measures of two of the angles of a triangle are 48° and 61°. See Figure 17. Find the measure of the third angle, x.

SOLUTION 48° + 61° + x = 180° The sum of the angles is 180°. 109° + x = 180° Add. x = 71° Subtract 109°.

The third angle of the triangle measures 71°.■✔ Now Try Exercises 13 and 23.

48° 61°

Figure 17

Types of Triangles

All acute One right angle One obtuse angle

Angles

Acute triangle Right triangle Obtuse triangle

All sides equal Two sides equal No sides equal

Sides

Equilateral triangle Isosceles triangle Scalene triangle

Similar triangles are triangles of exactly the same shape but not neces-sarily the same size. Figure 18 on the next page shows three pairs of similar triangles. The two triangles in Figure 18(c) have not only the same shape but also the same size. Triangles that are both the same size and the same shape are congruent triangles. If two triangles are congruent, then it is possible to pick one of them up and place it on top of the other so that they coincide.

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131.2 Angle Relationships and Similar Triangles

(a) (b) (c)

Figure 18

Conditions for Similar Triangles

Triangle ABC is similar to triangle DEF if the following conditions hold.

1. Corresponding angles have the same measure.

2. Corresponding sides are proportional. (That is, the ratios of their corre-sponding sides are equal.)

The triangular supports for a child’s swing set are congruent (and thus simi-lar) triangles, machine-produced with exactly the same dimensions each time. These supports are just one example of similar triangles. The supports of a long bridge, all the same shape but increasing in size toward the center of the bridge, are examples of similar (but not congruent) figures. See the photo.

Consider the correspondence between triangles ABC and DEF in Figure 19.

Angle A corresponds to angle D.

Angle B corresponds to angle E.

Angle C corresponds to angle F.

Side AB corresponds to side DE.

Side BC corresponds to side EF.

Side AC corresponds to side DF.

The small arcs found at the angles in Figure 19 denote the corresponding angles in the triangles.

D F

Figure 19

EXAMPLE 3 Finding Angle Measures in Similar Triangles

In Figure 20, triangles ABC and NMP are similar. Find all unknown angle measures.

SOLUTION First, we find the measure of angle M using the angle sum prop-erty of a triangle.

104° + 45° + M = 180° The sum of the angles is 180°. 149° + M = 180° Add.

M = 31° Subtract 149°.

The triangles are similar, so corresponding angles have the same measure. Because C corresponds to P and P measures 104°, angle C also measures 104°. Angles B and M correspond, so B measures 31°.

■✔ Now Try Exercise 49.

Figure 20

45°

C B

N 45°

104°

If two triangles are congruent, then they must be similar. However, two similar triangles need not be congruent.

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14 CHAPTER 1 Trigonometric Functions

EXAMPLE 4 Finding Side Lengths in Similar Triangles

Given that triangle ABC and triangle DFE in Figure 21 are similar, find the lengths of the unknown sides of triangle DFE.

SOLUTION Similar triangles have corre-sponding sides in proportion. Use this fact to find the unknown side lengths in triangle DFE.

Side DF of triangle DFE corresponds to side AB of triangle ABC, and sides DE and AC correspond. This leads to the following proportion.

816

= DF24

Recall this property of proportions from algebra.

If ab=

, then ad = bc.

We use this property to solve the equation for DF.

816

= DF24

Corresponding sides are proportional.

8 # 24 = 16 # DF If ab = cd , then ad = bc. 192 = 16 # DF Multiply. 12 = DF Divide by 16.

Side DF has length 12.Side EF corresponds to CB. This leads to another proportion.

816

= EF32

Corresponding sides are proportional.

8 # 32 = 16 # EF If ab = cd , then ad = bc. 16 = EF Solve for EF.

Side EF has length 16. ■✔ Now Try Exercise 55.

➡➡

24 FD

Figure 21

EXAMPLE 5 Finding the Height of a Flagpole

Workers must measure the height of a building flagpole. They find that at the instant when the shadow of the building is 18 m long, the shadow of the flagpole is 27 m long. The building is 10 m high. Find the height of the flagpole.

SOLUTION Figure 22 shows the information given in the problem. The two triangles are similar, so corresponding sides are in proportion.

MN10

= 2718

Corresponding sides are proportional.

MN10

= 32

Write 2718 in lowest terms.

MN # 2 = 10 # 3 Property of proportions MN = 15 Solve for MN.

The flagpole is 15 m high. ■✔ Now Try Exercise 59.

27 M

Figure 22

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151.2 Angle Relationships and Similar Triangles

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. The sum of the measures of the angles of any triangle is .

2. An isosceles right triangle has one angle and equal sides.

3. An equilateral triangle has equal sides.

4. If two triangles are similar, then their corresponding are proportional and their corresponding have equal measure.

1.2 Exercises

CONCEPT PREVIEW In each figure, find the measures of the numbered angles, given that lines m and n are parallel.

131° 12 3

4 56 7

120°10

31 2

4 55°5

8 67

CONCEPT PREVIEW Name the corresponding angles and the corresponding sides of each pair of similar triangles.

P R

Q 8.

A C

B Q

9. (EA is parallel to CD.) 10. (HK is parallel to EF.)

H K

Find the measure of each marked angle. In Exercises 19–22, m and n are parallel. See Examples 1 and 2.

11.

(5x – 129)° (2x – 21)°

12.

(11x – 37)° (7x + 27)°

13.

(210 – 3x)°

(x + 20)°

x°

14.

(10x – 20)°

(x + 15)°

(x + 5)°

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16 CHAPTER 1 Trigonometric Functions

15. (2x – 120)°

(x – 30)°

x + 15 °1_2( )

16. (2x + 16)°

(5x – 50)°

(3x – 6)°

17. (6x + 3)°

(9x + 12)°(4x – 3)°

18. (–5x)°

(7 – 12x)°(–8x + 3)°

19.

(2x – 5)°(x + 22)°

20.

(2x + 61)°

(6x – 51)°

21.

(4x – 56)°

(x + 1)°

22.

(10x + 11)°

(15x – 54)°

The measures of two angles of a triangle are given. Find the measure of the third angle. See Example 2.

23. 37°, 52° 24. 29°, 104° 25. 147° 12′, 30° 19′

26. 136° 50′, 41° 38′ 27. 74.2°, 80.4° 28. 29.6°, 49.7°

29. 51° 20′ 14″, 106° 10′ 12″ 30. 17° 41′ 13″, 96° 12′ 10″

31. Concept Check Can a triangle have angles of measures 85° and 100°?

32. Concept Check Can a triangle have two obtuse angles?

Concept Check Classify each triangle as acute, right, or obtuse. Also classify each as equilateral, isosceles, or scalene. See the discussion following Example 2.

33. 34.

120°

35.

60° 60°

60°

36.

37. 90°

38.

130°

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171.2 Angle Relationships and Similar Triangles

39.

40.

60°

41.

96°

42.

r60° 60°

60° 43.

50° 50°

44.

45. Angle Sum of a Triangle Use this figure to dis-cuss why the measures of the angles of a triangle must add up to the same sum as the measure of a straight angle.

46. Carpentry Technique The following tech-nique is used by carpenters to draw a 60° angle with a straightedge and a compass. Why does this technique work? (Source: Hamilton, J. E. and M. S. Hamilton, Math to Build On, Con-struction Trades Press.)

“Draw a straight line segment, and mark a point near the midpoint. Now place the tip on the marked point, and draw a semicircle. Without changing the setting of the compass, place the tip at the right intersection of the line and the semicircle, and then mark a small arc across the semicircle. Finally, draw a line segment from the marked point on the original segment to the point where the arc crosses the semicircle. This will form a 60° angle with the original segment.”

90°

1 2

2 514

QRm and n are parallel.

60°

Point markedon line

Tip placedhere

Find all unknown angle measures in each pair of similar triangles. See Example 3.

47.

C B

42°

Q 48.

49.

C30°

M30°

106°

50.

78°

46°

74°

28°W

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18 CHAPTER 1 Trigonometric Functions

51. X

38°

P 38°

52.

64°

20°

Q R

V U

Find the unknown side lengths in each pair of similar triangles. See Example 4.

53.

54. a

2010

55.

12 12a

56.

57.

58. 21

12m

Solve each problem. See Example 5.

59. Height of a Tree A tree casts a shadow 45 m long. At the same time, the shadow cast by a vertical 2-m stick is 3 m long. Find the height of the tree.

60. Height of a Lookout Tower A forest fire lookout tower casts a shadow 180 ft long at the same time that the shadow of a 9-ft truck is 15 ft long. Find the height of the tower.

61. Lengths of Sides of a Triangle On a photograph of a triangular piece of land, the lengths of the three sides are 4 cm, 5 cm, and 7 cm, respectively. The shortest side of the actual piece of land is 400 m long. Find the lengths of the other two sides.

62. Height of a Lighthouse The Biloxi lighthouse in the figure casts a shadow 28 m long at 7 a.m. At the same time, the shadow of the lighthouse keeper, who is 1.75 m tall, is 3.5 m long. How tall is the lighthouse?

28 m

3.5 m

NOT TO SCALE

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191.2 Angle Relationships and Similar Triangles

63. Height of a Building A house is 15 ft tall. Its shadow is 40 ft long at the same time that the shadow of a nearby building is 300 ft long. Find the height of the building.

64. Height of a Carving of Lincoln Assume that Lincoln was 6 13 ft tall and his head 34 ft long. Knowing that the carved head of Lincoln at Mt. Rushmore is 60 ft tall, find how tall his entire body would be if it were carved into the mountain.

60 ft

In each figure, there are two similar triangles. Find the unknown measurement. Give approximations to the nearest tenth.

65.

x50

120100

66.

40160

67. c

10 90

100

68. m

5 75

Solve each problem.

69. Solar Eclipse on Earth The sun has a diameter of about 865,000 mi with a maxi-mum distance from Earth’s surface of about 94,500,000 mi. The moon has a smaller diam-eter of 2159 mi. For a total solar eclipse to occur, the moon must pass between Earth and the sun. The moon must also be close enough to Earth for the moon’s umbra (shadow) to reach the surface of Earth. (Source: Karttunen, H., P. Kröger, H. Oja, M. Putannen, and K. Donners, Editors, Funda-mental Astronomy, Fourth Edition, Springer-Verlag.)

(a) Calculate the maximum distance, to the nearest thousand miles, that the moon can be from Earth and still have a total solar eclipse occur. (Hint: Use similar triangles.)

(b) The closest approach of the moon to Earth’s surface was 225,745 mi and the far-thest was 251,978 mi. (Source: World Almanac and Book of Facts.) Can a total solar eclipse occur every time the moon is between Earth and the sun?

70. Solar Eclipse on Neptune (Refer to Exercise 69.) The sun’s distance from Neptune is approximately 2,800,000,000 mi (2.8 billion mi). The largest moon of Neptune is Triton, with a diameter of approximately 1680 mi. (Source: World Almanac and Book of Facts.)

(a) Calculate the maximum distance, to the nearest thousand miles, that Triton can be from Neptune for a total eclipse of the sun to occur on Neptune. (Hint: Use similar triangles.)

(b) Triton is approximately 220,000 mi from Neptune. Is it possible for Triton to cause a total eclipse on Neptune?

EarthMoon

Umbra

Sun

NOT TO SCALE

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20 CHAPTER 1 Trigonometric Functions

2°

10°20°

71. Solar Eclipse on Mars (Refer to Exercise 69.) The sun’s distance from the surface of Mars is approximately 142,000,000 mi. One of Mars’ two moons, Phobos, has a maximum diameter of 17.4 mi. (Source: World Almanac and Book of Facts.)

(a) Calculate the maximum distance, to the nearest hundred miles, that the moon Phobos can be from Mars for a total eclipse of the sun to occur on Mars.

(b) Phobos is approximately 5800 mi from Mars. Is it possible for Phobos to cause a total eclipse on Mars?

72. Solar Eclipse on Jupiter (Refer to Exercise 69.) The sun’s distance from the sur-face of Jupiter is approximately 484,000,000 mi. One of Jupiter’s moons, Gany-mede, has a diameter of 3270 mi. (Source: World Almanac and Book of Facts.)

(a) Calculate the maximum distance, to the nearest thousand miles, that the moon Ganymede can be from Jupiter for a total eclipse of the sun to occur on Jupiter.

(b) Ganymede is approximately 665,000 mi from Jupiter. Is it possible for Ganymede to cause a total eclipse on Jupiter?

73. Sizes and Distances in the Sky Astronomers use degrees, minutes, and seconds to measure sizes and distances in the sky along an arc from the horizon to the zenith point directly overhead. An adult observer on Earth can judge distances in the sky using his or her hand at arm’s length. An outstretched hand will be about 20 arc degrees wide from the tip of the thumb to the tip of the little finger. A clenched fist at arm’s length measures about 10 arc degrees, and a thumb corresponds to about 2 arc degrees. (Source: Levy, D. H., Skywatching, The Nature Company.)

(a) The apparent size of the moon is about 31 arc minutes. Approximately what part of your thumb would cover the moon?

(b) If an outstretched hand plus a fist cover the distance between two bright stars, about how far apart in arc degrees are the stars?

74. Estimates of Heights There is a relatively simple way to make a reasonable esti-mate of a vertical height.

Step 1 Hold a 1-ft ruler vertically at arm’s length and approach the object to be measured.

Step 2 Stop when one end of the ruler lines up with the top of the object and the other end with its base.

Step 3 Now pace off the distance to the object, taking normal strides. The number of paces will be the approximate height of the object in feet.

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21CHAPTER 1 Quiz

Reasons

(a) CG =CG1

= AGAD

(b)AGAD

= EGBD

(c)EGEF

= EGBD

= EG1

(d) CG ft = EG paces(e) What is the height of the tree in feet?

1. Find the measure of (a) the complement and (b) the supplement of an angle measur-ing 19°.

Find the measure of each unknown angle.

(3x + 5)° (5x + 15)°

(5x – 1)°

(2x)°

4. (3x + 3)°

(4x – 8)°

(13x + 45)°

5. (–14x + 18)°

(–6x + 2)°

m and n are parallel.

Solve each problem.

6. Perform each conversion.

(a) 77° 12′ 09″ to decimal degrees (b) 22.0250° to degrees, minutes, seconds

7. Find the angle of least positive measure (not equal to the given measure) that is coterminal with each angle.

(a) 410° (b) -60° (c) 890° (d) 57°

Chapter 1 Quiz (Sections 1.1—1.2)

1 pace

1 ftH

E F G

Furnish the reasons in parts (a)–(d), which refer to the figure. (Assume that the length of one pace is EF.) Then answer the question in part (e).

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22 CHAPTER 1 Trigonometric Functions

8. Rotating Flywheel A flywheel rotates 300 times per min. Through how many degrees does a point on the edge of the flywheel move in 1 sec?

9. Length of a Shadow If a vertical antenna 45 ft tall casts a shadow 15 ft long, how long would the shadow of a 30-ft pole be at the same time and place?

10. Find the values of x and y.

(a)

(b) 40°

40°

82°82°

(10x + 8)°

1.3 Trigonometric Functions

The Pythagorean Theorem and the Distance Formula The distance between any two points in a plane can be found by using a formula derived from the Pythagorean theorem.

■ The Pythagorean Theorem and the Distance Formula

■ Trigonometric Functions

■ Quadrantal Angles Pythagorean Theorem

In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

a2 + b2 = c2Leg b

HypotenusecLeg a

To find the distance between two points P1x1, y12 and R1x2, y22, draw a line segment connecting the points, as shown in Figure 23. Complete a right trian-gle by drawing a line through 1x1, y12 parallel to the x-axis and a line through 1x2, y22 parallel to the y-axis. The ordered pair at the right angle is 1x2, y12.

The horizontal side of the right triangle in Figure 23 has length x2 - x1, while the vertical side has length y2 - y1. If d represents the distance between the two original points, then by the Pythagorean theorem,

d2 = 1x2 - x122 + 1 y2 - y122.Solving for d, we obtain the distance formula.

R(x2, y2)

(x2, y1)

dy2 – y1

x2 – x1P(x1, y1)

Figure 23

Distance Formula

Suppose that P1x1, y12 and R1x2, y22 are two points in a coordinate plane. The distance between P and R, written d1P, R2, is given by the following formula.

d 1P, R 2 = !1x2 − x1 2 2 + 1 y2 − y1 2 2That is, the distance between two points in a coordinate plane is the square root of the sum of the square of the difference between their x-coordinates and the square of the difference between their y-coordinates.

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231.3 Trigonometric Functions

Trigonometric Functions To define the six trigonometric functions, we start with an angle u in standard position and choose any point P having coordi-nates 1x, y2 on the terminal side of angle u. (The point P must not be the vertex of the angle.) See Figure 24. A perpendicular from P to the x-axis at point Q determines a right triangle, having vertices at O, P, and Q. We find the distance r from P1x, y2 to the origin, 10, 02, using the distance formula.

d1O, P2 = 21x2 - x122 + 1y2 - y122 Distance formula r = 21x - 022 + 1y - 022 Substitute 1x, y2 for 1x2, y22 and 10, 02 for 1x1, y12. r = 2x2 + y2 Subtract.

Notice that r + 0 because this is the undirected distance.The six trigonometric functions of angle u are

sine, cosine, tangent, cotangent, secant, and cosecant,

abbreviated sin, cos, tan, cot, sec, and csc.

P(x, y)

Figure 24

Trigonometric Functions

Let 1x, y2 be a point other than the origin on the terminal side of an angle u in standard position. The distance from the point to the origin is

r = 2x2 + y2. The six trigonometric functions of u are defined as follows.sin U =

yr cos U =

tan U =yx 1x 3 0 2

csc U =ry 1 y 3 0 2 sec U = r

x 1x 3 0 2 cot U = x

y 1 y 3 0 2

EXAMPLE 1 Finding Function Values of an Angle

The terminal side of an angle u in standard position passes through the point 18, 152. Find the values of the six trigonometric functions of angle u.

SOLUTION Figure 25 shows angle u and the triangle formed by dropping a perpendicular from the point 18, 152 to the x-axis. The point 18, 152 is 8 units to the right of the y-axis and 15 units above the x-axis, so

x = 8 and y = 15. Now use r = 2x2 + y2 .r = 282 + 152 = 264 + 225 = 2289 = 17

We can now use these values for x, y, and r to find the values of the six trigono-metric functions of angle u.

sin u =yr

= 1517

cos u = xr

= 817

tan u =yx

= 158

csc u = ry

= 1715

sec u = rx

= 178

cot u = xy

= 815

■✔ Now Try Exercise 13.

0 8

17 15

(8, 15)

x = 8y = 15r = 17

Figure 25

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24 CHAPTER 1 Trigonometric Functions

EXAMPLE 2 Finding Function Values of an Angle

The terminal side of an angle u in standard position passes through the point 1-3, -42. Find the values of the six trigonometric functions of angle u.

SOLUTION As shown in Figure 26, x = -3 and y = -4.

r = 21-322 + 1-422 r = 2x2 + y2 r = 225 Simplify the radicand. r = 5 r 7 0

Now we use the definitions of the trigonometric functions.

sin u = -45

= - 45

cos u = -35

= - 35

tan u = -4-3 =43

csc u = 5-4 = - 54

sec u = 5-3 = - 53

cot u = -3-4 =34

■✔ Now Try Exercise 17.

–3

–4

x = –3y = –4r = 5

(–3, –4)

Figure 26

We can find the six trigonometric functions using any point other than the origin on the terminal side of an angle. To see why any point can be used, refer to Figure 27, which shows an angle u and two distinct points on its terminal side. Point P has coordinates 1x, y2, and point P′ (read “P-prime”) has coordi-nates 1x′, y′2. Let r be the length of the hypotenuse of triangle OPQ, and let r ′ be the length of the hypotenuse of triangle OP′Q′. Because corresponding sides of similar triangles are proportional, we have

=y′r′

. Corresponding sides are proportional.

Thus sin u = yr is the same no matter which point is used to find it. A similar result holds for the other five trigonometric functions.

We can also find the trigonometric function values of an angle if we know the equation of the line coinciding with the terminal ray. Recall from algebra that the graph of the equation

Ax + By = 0 Linear equation in two variables

is a line that passes through the origin 10, 02. If we restrict x to have only nonpos-itive or only nonnegative values, we obtain as the graph a ray with endpoint at the origin. For example, the graph of x + 2y = 0, x Ú 0, shown in Figure 28, is a ray that can serve as the terminal side of an angle u in standard position. By choosing a point on the ray, we can find the trigonometric function values of the angle.

y(x !, y !)

OP = rOP! = r!

O Q Q!

(x, y)

Figure 27

x + 2y = 0, x ≥ 0

Figure 28

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251.3 Trigonometric Functions

Recall that when the equation of a line is written in the form

y = mx + b, Slope-intercept form

the coefficient m of x gives the slope of the line. In Example 3, the equation x + 2y = 0 can be written as y = - 12 x, so the slope of this line is -

12 . Notice

that tan u = - 12 .

In general, it is true that m = tan U.

0(2, –1)

x + 2y = 0, x ≥ 0

x = 2y = –1r =

Figure 29

EXAMPLE 3 Finding Function Values of an Angle

Find the six trigonometric function values of an angle u in standard position, if the terminal side of u is defined by x + 2y = 0, x Ú 0.

SOLUTION The angle is shown in Figure 29. We can use any point except 10, 02 on the terminal side of u to find the trigonometric function values. We choose x = 2 and find the corresponding y-value.

x + 2y = 0, x Ú 0 2 + 2y = 0 Let x = 2.

2y = -2 Subtract 2. y = -1 Divide by 2.

The point 12, -12 lies on the terminal side, and so r = 222 + 1-122 = 25. Now we use the definitions of the trigonometric functions.

sin u =yr

= -125 = -125 # 2525 = - 255 cos u = x

r= 225 = 225 # 2525 = 2255

tan u =yx

= -12

= - 12

csc u = ry

= 25-1 = - 25 sec u = rx = 252 cot u = xy = 2-1 = -2Multiply by 2525 , a form of 1, to rationalize the denominators.

■✔ Now Try Exercise 51.

Quadrantal Angles If the terminal side of an angle in standard position lies along the y-axis, any point on this terminal side has x-coordinate 0. Similarly, an angle with terminal side on the x-axis has y-coordinate 0 for any point on the terminal side. Because the values of x and y appear in the denominators of some trigonometric functions, and because a fraction is undefined if its denominator is 0, some trigonometric function values of quadrantal angles (i.e., those with terminal side on an axis) are undefined.

When determining trigonometric function values of quadrantal angles, Figure 30 can help find the ratios. Because any point on the terminal side can be used, it is convenient to choose the point one unit from the origin, with r = 1. (Later we will extend this idea to the unit circle.)

0(–1, 0)

(1, 0)

(0, 1)

(0, –1)

x = 0y = 1r = 1

x = 1y = 0r = 1

x = –1y = 0r = 1

x = 0y = –1r = 1

Figure 30

NOTE The trigonometric function values we found in Examples 1–3 are exact. If we were to use a calculator to approximate these values, the decimal results would not be acceptable if exact values were required.

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26 CHAPTER 1 Trigonometric Functions

To find the function values of a quadrantal angle, determine the position of the terminal side, choose the one of these four points that lies on this terminal side, and then use the definitions involving x, y, and r.

A calculator in degree mode returns the correct values for sin 90° and cos 90°. The screen shows an ERROR message for tan 90°, because 90° is not in the domain of the tangent function.

EXAMPLE 4 Finding Function Values of Quadrantal Angles

Find the values of the six trigonometric functions for each angle.

(a) an angle of 90°

(b) an angle u in standard position with terminal side passing through 1-3, 02SOLUTION

(a) Figure 31 shows that the terminal side passes through 10, 12. So x = 0, y = 1, and r = 1. Thus, we have the following.

sin 90° = 11

= 1 cos 90° = 01

= 0 tan 90° = 10

Undefined

csc 90° = 11

= 1 sec 90° = 10

Undefined cot 90° = 01

= 0

0(–1, 0)

(1, 0)

(0, 1)

(0, –1)

x = 0y = 1r = 1

x = 1y = 0r = 1

x = –1y = 0r = 1

x = 0y = –1r = 1

Figure 30 (repeated)

(0, 1)

90°

Figure 31

0(–3, 0)

Figure 32

(b) Figure 32 shows the angle. Here, x = -3, y = 0, and r = 3, so the trigono-metric functions have the following values.

sin u = 03

= 0 cos u = -33

= -1 tan u = 0-3 = 0

csc u = 30

Undefined sec u = 3-3 = -1 cot u =-30

Undefined

Verify that these values can also be found using the point 1-1, 02.■✔ Now Try Exercises 23, 67, 69, and 71.

The conditions under which the trigonometric function values of quadrantal angles are undefined are summarized here.

Conditions for Undefined Function Values

Identify the terminal side of a quadrantal angle.

If the terminal side of the quadrantal angle lies along the y-axis, then the tangent and secant functions are undefined.

If the terminal side of the quadrantal angle lies along the x-axis, then the cotangent and cosecant functions are undefined.

The function values of some commonly used quadrantal angles, 0°, 90°, 180°, 270°, and 360°, are summarized in the table on the next page. They can be determined when needed by using Figure 30 and the method of Example 4(a).

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271.3 Trigonometric Functions

Function Values of Quadrantal Angles

U sin U cos U tan U cot U sec U csc U

0° 0 1 0 Undefined 1 Undefined 90° 1 0 Undefined 0 Undefined 1180° 0 -1 0 Undefined -1 Undefined270° -1 0 Undefined 0 Undefined -1360° 0 1 0 Undefined 1 Undefined

The values given in this table can be found with a calculator that has trigo-nometric function keys. Make sure the calculator is set to degree mode.

TI-84 Plus

Figure 33

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. The Pythagorean theorem for right triangles states that the sum of the squares of the lengths of the legs is equal to the square of the .

2. In the definitions of the sine, cosine, secant, and cosecant functions, r is interpreted geometrically as the distance from a given point 1x, y2 on the terminal side of an angle u in standard position to the .

3. For any nonquadrantal angle u, sin u and csc u will have the sign. (same/opposite) 4. If cot u is undefined, then tan u = .

5. If the terminal side of an angle u lies in quadrant III, then the values of tan u and cot u are , and all other trigonometric function values are

(positive/negative) . (positive/negative)

6. If a quadrantal angle u is coterminal with 0° or 180°, then the trigonometric func-tions and are undefined.

1.3 Exercises

CONCEPT PREVIEW The terminal side of an angle u in standard position passes through the point 1-3, -32. Use the figure to find the following values. Rationalize denominators when applicable.

7. r 8. sin u

9. cos u 10. tan u

–3

(–3, –3)

For other quadrantal angles such as -90°, -270°, and 450°, first determine the coterminal angle that lies between 0° and 360°, and then refer to the table entries for that particular angle. For example, the function values of a -90° angle would correspond to those of a 270° angle.

CAUTION One of the most common errors involving calculators in trigonometry occurs when the calculator is set for radian measure, rather than degree measure. Be sure to set your calculator to degree mode. See Figure 33.

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28 CHAPTER 1 Trigonometric Functions

Sketch an angle u in standard position such that u has the least positive measure, and the given point is on the terminal side of u. Then find the values of the six trigonometric func-tions for each angle. Rationalize denominators when applicable. See Examples 1, 2, and 4.

11. 15, -122 12. 1-12, -52 13. 13, 42 14. 1-4, -3215. 1-8, 152 16. 115, -82 17. 1-7, -242 18. 1-24, -7219. 10, 22 20. 10, 52 21. 1-4, 02 22. 1-5, 0223. 10, -42 24. 10, -32 25. A1, 23 B 26. A -1, 23 B27. A22, 22 B 28. A -22, -22 B 29. A -223, -2 B 30. A -223, 2 BConcept Check Suppose that the point 1x, y2 is in the indicated quadrant. Determine whether the given ratio is positive or negative. Recall that r = 2x2 + y2. (Hint: Drawing a sketch may help.)

31. II, xr 32. III,

yr 33. IV,

34. IV, xy

35. II, yr

36. III, xr 37. IV,

xr 38. IV,

yr 39. II,

40. II, yx

41. III, yx

42. III, xy

43. III, rx

44. III, ry

45. I, xy

46. I, yx

47. I, yr 48. I,

xr 49. I,

50. I, ry

An equation of the terminal side of an angle u in standard position is given with a restriction on x. Sketch the least positive such angle u, and find the values of the six trigonometric functions of u. See Example 3.

51. 2x + y = 0, x Ú 0 52. 3x + 5y = 0, x Ú 0 53. -6x - y = 0, x … 0

54. -5x - 3y = 0, x … 0 55. -4x + 7y = 0, x … 0 56. 6x - 5y = 0, x Ú 0

57. x + y = 0, x Ú 0 58. x - y = 0, x Ú 0 59. -23x + y = 0, x … 060. 23x + y = 0, x … 0 61. x = 0, y Ú 0 62. y = 0, x … 0Find the indicated function value. If it is undefined, say so. See Example 4.

63. cos 90° 64. sin 90° 65. tan 180° 66. cot 90°

67. sec 180° 68. csc 270° 69. sin1-270°2 70. cos1-90°2 71. cot 540° 72. tan 450° 73. csc1-450°2 74. sec1-540°2 75. sin 1800° 76. cos 1800° 77. csc 1800°

78. cot 1800° 79. sec 1800° 80. tan 1800°

81. cos1-900°2 82. sin1-900°2 83. tan1-900°2 84. How can the answer to Exercise 83 be given once the answers to Exercises 81 and

82 have been determined?

Use trigonometric function values of quadrantal angles to evaluate each expression. An expression such as cot2 90° means 1cot 90°22, which is equal to 02 = 0. 85. cos 90° + 3 sin 270° 86. tan 0° - 6 sin 90°

87. 3 sec 180° - 5 tan 360° 88. 4 csc 270° + 3 cos 180°

89. tan 360° + 4 sin 180° + 5 cos2 180° 90. 5 sin2 90° + 2 cos2 270° - tan 360°

91. sin2 180° + cos2 180° 92. sin2 360° + cos2 360°

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291.3 Trigonometric Functions

93. sec2 180° - 3 sin2 360° + cos 180° 94. 2 sec 0° + 4 cot2 90° + cos 360°

95. -2 sin4 0° + 3 tan2 0° 96. -3 sin4 90° + 4 cos3 180°

97. sin21-90°2 + cos21-90°2 98. cos21-180°2 + sin21-180°2If n is an integer, n # 180° represents an integer multiple of 180°, 12n + 12 # 90° repre-sents an odd integer multiple of 90°, and so on. Determine whether each expression is equal to 0, 1, or -1, or is undefined.

99. cos312n + 12 # 90°4 100. sin3n # 180°4 101. tan3n # 180°4102. sin3270° + n # 360°4 103. tan312n + 12 # 90°4 104. cot3n # 180°4 105. cot312n + 12 # 90°4 106. cos3n # 360°4 107. sec312n + 12 # 90°4 108. csc3n # 180°4Concept Check In later chapters we will study trigonometric functions of angles other than quadrantal angles, such as 15°, 30°, 60°, 75°, and so on. To prepare for some impor-tant concepts, provide conjectures in each exercise. Use a calculator set to degree mode.

109. The angles 15° and 75° are complementary. Determine sin 15° and cos 75°. Make a conjecture about the sines and cosines of complementary angles, and test this hypothesis with other pairs of complementary angles.

110. The angles 25° and 65° are complementary. Determine tan 25° and cot 65°. Make a conjecture about the tangents and cotangents of complementary angles, and test this hypothesis with other pairs of complementary angles.

111. Determine sin 10° and sin1-10°2. Make a conjecture about the sine of an angle and the sine of its negative, and test this hypothesis with other angles.

112. Determine cos 20° and cos1-20°2. Make a conjecture about the cosine of an angle and the cosine of its negative, and test this hypothesis with other angles.

Set a TI graphing calculator to parametric and degree modes. Use the window values shown in the first screen, and enter the equations shown in the second screen. The corre-sponding graph in the third screen is a circle of radius 1. Trace to move a short distance around the circle. In the third screen, the point on the circle corresponds to the angle T = 25°. Because r = 1, cos 25° is X = 0.90630779 and sin 25° is Y = 0.42261826.

−1.8

−1.2

1.2

1.8

Use this information to answer each question.

113. Use the right- and left-arrow keys to move to the point corresponding to 20° 1T = 202. Approximate cos 20° and sin 20° to the nearest thousandth.114. For what angle T, 0° … T … 90°, is cos T ≈ 0.766?

115. For what angle T, 0° … T … 90°, is sin T ≈ 0.574?

116. For what angle T, 0° … T … 90° does cos T equal sin T?

117. As T increases from 0° to 90°, does the cosine increase or decrease? What about the sine?

118. As T increases from 90° to 180°, does the cosine increase or decrease? What about the sine?

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30 CHAPTER 1 Trigonometric Functions

1.4 Using the Definitions of the Trigonometric Functions

Identities are equations that are true for all values of the variables for which all expressions are defined.1x + y22 = x2 + 2xy + y2 21x + 32 = 2x + 6 Identities

■ Reciprocal Identities■ Signs and Ranges of

Function Values■ Pythagorean

Identities■ Quotient Identities

Reciprocal Identities Recall the definition of a reciprocal.

The reciprocal of a nonzero number x is 1x

Examples: The reciprocal of 2 is 12 , and the reciprocal of 8

11 is 118 . There is no

reciprocal for 0 because 10 is undefined.

The definitions of the trigonometric functions in the previous section were written so that functions in the same column were reciprocals of each other. Because sin u = yr and csc u =

ry ,

sin u = 1csc u

and csc u = 1sin u

, provided sin u ≠ 0.

Also, cos u and sec u are reciprocals, as are tan u and cot u. The reciprocal identities hold for any angle u that does not lead to a 0 denominator.

The screen in Figure 34(a) shows that csc 90° = 1 and sec1-180°2 = -1 using appropriate reciprocal identities. The third entry uses the reciprocal func-tion key x-1 to evaluate sec1-180°2. Figure 34(b) shows that attempting to find sec 90° by entering 1cos 90° produces an ERROR message, indicating that the reciprocal is undefined. See Figure 34(c). ■

Reciprocal Identities

For all angles u for which both functions are defined, the following identi-ties hold.

sin U =1

csc U cos U =

1sec U

tan U =1

cot U

csc U =1

sin U sec U =

1cos U

cot U =1

tan U(a)

(b)

(c)

Figure 34

This is the inverse cosine function, which will be discussed later in the text.

This is the reciprocal function, which correctly evaluates sec1-180°2, as seen in Figure 34(a).1cos1-180°22-1 = 1cos1-180°2 = sec1-180°2

CAUTION Be sure not to use the inverse trigonometric function keys to find reciprocal function values. For example, consider the following.

cos-11-180°2 ≠ 1cos1-180°22-1

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311.4 Using the Definitions of the Trigonometric Functions

The reciprocal identities can be written in different forms. For example,

sin U =1

csc U is equivalent to csc U =

1sin U

and 1sin U 2 1csc U 2 = 1.EXAMPLE 1 Using the Reciprocal Identities

Find each function value.

(a) cos u, given that sec u = 53 (b) sin u, given that csc u = - 212

SOLUTION

(a) We use the fact that cos u is the reciprocal of sec u.

cos u = 1sec u

= 153

= 1 , 53

= 1 # 35

= 35

Simplify the complex fraction.

(b) sin u = 1csc u

sin u is the reciprocal of csc u.

= 1

- 2122 Substitute csc u = - 2122 . = - 2212 Simplify the complex fraction as in part (a). = - 2

223 212 = 24 # 3 = 223 = - 123 Divide out the common factor 2. = - 123 # 2323 Rationalize the denominator. = - 23

3 Multiply.

■✔ Now Try Exercises 11 and 19.

Signs and Ranges of Function Values In the definitions of the trigo-nometric functions, r is the distance from the origin to the point 1x, y2. This distance is undirected, so r 7 0. If we choose a point 1x, y2 in quadrant I, then both x and y will be positive, and the values of all six functions will be positive.

A point 1x, y2 in quadrant II satisfies x 6 0 and y 7 0. This makes the values of sine and cosecant positive for quadrant II angles, while the other four functions take on negative values. Similar results can be obtained for the other quadrants.

This important information is summarized here.

Signs of Trigonometric Function Values

U in Quadrant sin U cos U tan U cot U sec U csc U

I + + + + + +

II + − − − − +

III − − + + − −

IV − + − − + −

IISine and cosecant

positive

x < 0, y > 0, r > 0

IAll functions

positive

x > 0, y > 0, r > 0

IIITangent and cotangent

positive

x < 0, y < 0, r > 0

IVCosine and secant

positive

x > 0, y < 0, r > 0x

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32 CHAPTER 1 Trigonometric Functions

EXAMPLE 2 Determining Signs of Functions of Nonquadrantal Angles

Determine the signs of the trigonometric functions of an angle in standard posi-tion with the given measure.

(a) 87° (b) 300° (c) -200°

SOLUTION

(a) An angle of 87° is in the first quadrant, with x, y, and r all positive, so all of its trigonometric function values are positive.

(b) A 300° angle is in quadrant IV, so the cosine and secant are positive, while the sine, cosecant, tangent, and cotangent are negative.

■✔ Now Try Exercises 27, 29, and 33.

(a)

0 x

(r, 0)

(0, r)

(r, 0)

(0, –r)

(b)

Figure 35

Figure 35(a) shows an angle u as it increases in measure from near 0° toward 90°. In each case, the value of r is the same. As the measure of the angle increases, y increases but never exceeds r, so y … r. Dividing both sides by the positive number r gives

yr … 1.

In a similar way, angles in quadrant IV as in Figure 35(b) suggest that

-1 …yr

so -1 … yr … 1

and −1 " sin U " 1. yr = sin u for any angle u.Similarly, −1 " cos U " 1.

The tangent of an angle is defined as yx . It is possible that x 6 y, x = y, or

x 7 y. Thus, yx can take any value, so tan U can be any real number, as can cot U.

EXAMPLE 3 Identifying the Quadrant of an Angle

Identify the quadrant (or possible quadrants) of an angle u that satisfies the given conditions.

(a) sin u 7 0, tan u 6 0 (b) cos u 6 0, sec u 6 0

SOLUTION

(a) Because sin u 7 0 in quadrants I and II and tan u 6 0 in quadrants II and IV, both conditions are met only in quadrant II.

(b) The cosine and secant functions are both negative in quadrants II and III, so in this case u could be in either of these two quadrants.

■✔ Now Try Exercises 43 and 49.

NOTE Because numbers that are reciprocals always have the same sign, the sign of a function value automatically determines the sign of the recipro-cal function value.

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331.4 Using the Definitions of the Trigonometric Functions

The functions sec u and csc u are reciprocals of the functions cos u and sin u, respectively, making

sec U " −1 or sec U # 1 and csc U " −1 or csc U # 1.

In summary, the ranges of the trigonometric functions are as follows.

Ranges of Trigonometric Functions

Trigonometric Function of U

Range (Set-Builder Notation)

Range (Interval Notation)

sin u, cos u 5 y 0 E y E … 16 3-1, 14tan u, cot u 5 y 0 y is a real number6 1-∞, ∞2sec u, csc u 5 y 0 E y E Ú 16 1-∞, -14 ´ 31, ∞2

EXAMPLE 4 Determining Whether a Value Is in the Range of a Trigonometric Function

Determine whether each statement is possible or impossible.

(a) sin u = 2.5 (b) tan u = 110.47 (c) sec u = 0.6

SOLUTION

(a) For any value of u, we know that -1 … sin u … 1. Because 2.5 7 1, it is impossible to find a value of u that satisfies sin u = 2.5.

(b) The tangent function can take on any real number value. Thus, tan u = 110.47 is possible.

■✔ Now Try Exercises 53, 57, and 59.

The six trigonometric functions are defined in terms of x, y, and r, where the Pythagorean theorem shows that r2 = x2 + y2 and r 7 0. With these relation-ships, knowing the value of only one function and the quadrant in which the angle lies makes it possible to find the values of the other trigonometric functions.

EXAMPLE 5 Finding All Function Values Given One Value and the Quadrant

Suppose that angle u is in quadrant II and sin u = 23 . Find the values of the five remaining trigonometric functions.

SOLUTION Choose any point on the terminal side of angle u. For simplicity, since sin u = yr , choose the point with r = 3.

sin u = 23

Given value

= 23

Substitute yr for sin u.

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34 CHAPTER 1 Trigonometric Functions

Remember both roots.

–2

2x = –

5y = 2r = 3

(–Ë

5, 2)

–Ë

Figure 36

These have rationalized

denominators.

Because u is in quadrant II, x must be negative. Choose x = -25 so that the point A -25, 2 B is on the terminal side of u. See Figure 36.

cos u = xr

= -253

= - 253

sec u = rx

= 3-25 = - 325 # 2525 = - 3255

tan u =yx

= 2-25 = - 225 # 2525 = - 2255

cot u = xy

= -252

= - 252

csc u = ry

= 32 ■✔ Now Try Exercise 75.

Pythagorean Identities We now derive three new identities.

x2 + y2 = r2 Pythagorean theorem

r2+

r2= r

r2 Divide by r2.

a xrb 2 + a y

rb 2 = 1 Power rule for exponents; ambm = Aab Bm

1cos u22 + 1sin u22 = 1 cos u = xr , sin u = yr sin2 U + cos2 U = 1 Apply exponents; commutative property

1cos u22 and cos2 u are equivalent forms.

Starting again with x2 + y2 = r2 and dividing through by x2 gives the following.

x2+

x2= r

x2 Divide by x2.

1 + a yxb 2 = a r

xb 2 Power rule for exponents

1 + 1tan u22 = 1sec u22 tan u = yx , sec u = rx tan2 U + 1 = sec2 U Apply exponents; commutative property

Because yr =

23 and r = 3, it follows that y = 2. We must find the value of x.

x2 + y2 = r2 Pythagorean theorem x2 + 22 = 32 Substitute. x2 + 4 = 9 Apply exponents.

x2 = 5 Subtract 4.

x = 25 or x = -25 Square root property: If x2 = k, then x = 2k or x = -2k.

Similarly, dividing through by y2 leads to another identity.

1 + cot2 U = csc2 U

These three identities are the Pythagorean identities because the original equa-tion that led to them, x2 + y2 = r2, comes from the Pythagorean theorem.

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351.4 Using the Definitions of the Trigonometric Functions

Pythagorean Identities

For all angles u for which the function values are defined, the following identities hold.

sin2 U + cos2 U = 1 tan2 U + 1 = sec2 U 1 + cot2 U = csc2 U

We give only one form of each identity. However, algebraic transforma-tions produce equivalent forms. For example, by subtracting sin2 u from both sides of sin2 u + cos2 u = 1, we obtain an equivalent identity.

cos2 u = 1 - sin2 u Alternative form

It is important to be able to transform these identities quickly and also to rec-ognize their equivalent forms.

Quotient Identities Consider the quotient of the functions sin u and cos u, for cos u ≠ 0.

sin ucos u

=yrxr

=yr

, xr

=yr# r

= tan u

Similarly, cos usin u = cot u, for sin u ≠ 0. Thus, we have the quotient identities.

LOOKING AHEAD TO CALCULUSThe reciprocal, Pythagorean, and

quotient identities are used in calculus

to find derivatives and integrals of

trigonometric functions. A standard

technique of integration called

trigonometric substitution relies on the Pythagorean identities.

Quotient Identities

For all angles u for which the denominators are not zero, the following identities hold.

sin Ucos U

= tan U cos Usin U

= cot U

EXAMPLE 6 Using Identities to Find Function Values

Find sin u and tan u, given that cos u = - 234 and sin u 7 0.SOLUTION Start with the Pythagorean identity that includes cos u.

sin2 u + cos2 u = 1 Pythagorean identity

sin2 u + ¢ - 234

≤2 = 1 Replace cos u with - 234 . sin2 u + 3

16= 1 Square - 234 .

sin2 u = 1316

Subtract 316 .

sin u = {2134

Take square roots.

sin u = 2134

Choose the positive square root because sin u is positive.

Choose the correct sign here.

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36 CHAPTER 1 Trigonometric Functions

To find tan u, use the values of cos u and sin u and the quotient identity for tan u.

tan u = sin ucos u

=213

- 234 = 2134 a - 423 b = - 21323 = - 21323 # 2323 = - 2393 Rationalize the denominator.

■✔ Now Try Exercise 79.

EXAMPLE 7 Using Identities to Find Function Values

Find sin u and cos u, given that tan u = 43 and u is in quadrant III.

SOLUTION Because u is in quadrant III, sin u and cos u will both be negative.

It is tempting to say that since tan u = sin ucos u and tan u =43 , then sin u = -4 and

cos u = -3. This is incorrect, however—both sin u and cos u must be in the interval 3-1, 14.

We use the Pythagorean identity tan2 u + 1 = sec2 u to find sec u, and then the reciprocal identity cos u = 1sec u to find cos u.

tan2 u + 1 = sec2 u Pythagorean identity

a 43b 2 + 1 = sec2 u tan u = 43 169

+ 1 = sec2 u Square 43 .

259

= sec2 u Add.

- 53

= sec u Choose the negative square root because sec u is negative when u is in quadrant III.

- 35

= cos u Secant and cosine are reciprocals.

Be careful to choose the correct

sign here.

Now we use this value of cos u to find sin u.

sin2 u = 1 - cos2 u Pythagorean identity (alternative form)

sin2 u = 1 - a - 35b 2 cos u = - 35

sin2 u = 1 - 925

Square - 35 .

sin2 u = 1625

Subtract.

sin u = - 45

Choose the negative square root.

Again, be careful.

■✔ Now Try Exercise 77.

CAUTION In exercises like Examples 5 and 6, be careful to choose the correct sign when square roots are taken. Refer as needed to the diagrams preceding Example 2 that summarize the signs of the functions.

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371.4 Using the Definitions of the Trigonometric Functions

NOTE Example 7 can also be worked by sketching u in standard position in quadrant III, finding r to be 5, and then using the definitions of sin u and cos u in terms of x, y, and r. See Figure 37.

When using this method, be sure to choose the correct signs for x and y as determined by the quadrant in which the terminal side of u lies. This is analogous to choosing the correct signs after apply-ing the Pythagorean identities.

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. Given cos u = 1sec u , two equivalent forms of this identity are sec u =1

and

cos u # = 1. 2. Given tan u = 1cot u , two equivalent forms of this identity are cot u =

and

tan u # = 1. 3. For an angle u measuring 105°, the trigonometric functions and

are positive, and the remaining trigonometric functions are negative.

4. If sin u 7 0 and tan u 7 0, then u is in quadrant .

1.4 Exercises

CONCEPT PREVIEW Determine whether each statement is possible or impossible.

5. sin u = 12

, csc u = 2 6. tan u = 2, cot u = -2 7. sin u 7 0, csc u 6 0

8. cos u = 1.5 9. cot u = -1.5 10. sin2 u + cos2 u = 2

Use the appropriate reciprocal identity to find each function value. Rationalize denomi-nators when applicable. See Example 1.

11. sec u, given that cos u = 23 12. sec u, given that cos u =58

13. csc u, given that sin u = - 37 14. csc u, given that sin u = - 8

15. cot u, given that tan u = 5 16. cot u, given that tan u = 18

17. cos u, given that sec u = - 52 18. cos u, given that sec u = - 117

19. sin u, given that csc u = 282 20. sin u, given that csc u = 224321. tan u, given that cot u = -2.5 22. tan u, given that cot u = -0.01

23. sin u, given that csc u = 1.25 24. cos u, given that sec u = 8

25. Concept Check What is wrong with the following item that appears on a trigonom-etry test?

“Find sec u, given that cos u = 32

.”

26. Concept Check What is wrong with the statement tan 90° = 1cot 90° ?

x = –3y = –4r = 5

(–3, –4)

–4

–3

–45

Figure 37

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38 CHAPTER 1 Trigonometric Functions

Identify the quadrant (or possible quadrants) of an angle u that satisfies the given condi-tions. See Example 3.

39. sin u 7 0, csc u 7 0 40. cos u 7 0, sec u 7 0 41. cos u 7 0, sin u 7 0

42. sin u 7 0, tan u 7 0 43. tan u 6 0, cos u 6 0 44. cos u 6 0, sin u 6 0

45. sec u 7 0, csc u 7 0 46. csc u 7 0, cot u 7 0 47. sec u 6 0, csc u 6 0

48. cot u 6 0, sec u 6 0 49. sin u 6 0, csc u 6 0 50. tan u 6 0, cot u 6 0

51. Why are the answers to Exercises 41 and 45 the same?

52. Why is there no angle u that satisfies tan u 7 0, cot u 6 0?

Determine whether each statement is possible or impossible. See Example 4.

53. sin u = 2 54. sin u = 3 55. cos u = -0.96

56. cos u = -0.56 57. tan u = 0.93 58. cot u = 0.93

59. sec u = -0.3 60. sec u = -0.9 61. csc u = 100

62. csc u = -100 63. cot u = -4 64. cot u = -6

Use identities to solve each of the following. Rationalize denominators when applicable. See Examples 5–7.

65. Find cos u, given that sin u = 35 and u is in quadrant II.

66. Find sin u, given that cos u = 45 and u is in quadrant IV.

67. Find csc u, given that cot u = - 12 and u is in quadrant IV.

68. Find sec u, given that tan u = 273 and u is in quadrant III.69. Find tan u, given that sin u = 12 and u is in quadrant II.

70. Find cot u, given that csc u = -2 and u is in quadrant III.

71. Find cot u, given that csc u = -1.45 and u is in quadrant III.

72. Find tan u, given that sin u = 0.6 and u is in quadrant II.

Give all six trigonometric function values for each angle u. Rationalize denominators when applicable. See Examples 5–7.

73. tan u = - 158 , and u is in quadrant II 74. cos u = - 35 , and u is in quadrant III

75. sin u = 257 , and u is in quadrant I 76. tan u = 23, and u is in quadrant III77. cot u = 238 , and u is in quadrant I 78. csc u = 2, and u is in quadrant II79. sin u = 226 , and cos u 6 0 80. cos u = 258 , and tan u 6 081. sec u = -4, and sin u 7 0 82. csc u = -3, and cos u 7 0

83. sin u = 1 84. cos u = 1

Determine the signs of the trigonometric functions of an angle in standard position with the given measure. See Example 2.

27. 74° 28. 84° 29. 218° 30. 195°

31. 178° 32. 125° 33. -80° 34. -15°

35. 855° 36. 1005° 37. -345° 38. -640°

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39CHAPTER 1 Test Prep

Work each problem.

85. Derive the identity 1 + cot2 u = csc2 u by dividing x2 + y2 = r2 by y2.

86. Derive the quotient identity cos usin u = cot u.

87. Concept Check True or false: For all angles u, sin u + cos u = 1. If the statement is false, give an example showing why.

88. Concept Check True or false: Since cot u = cos usin u , if cot u =12 with u in quadrant I,

then cos u = 1 and sin u = 2. If the statement is false, explain why.

Concept Check Suppose that -90° 6 u 6 90°. Find the sign of each function value.

97. cos u

2 98. sec

2 99. sec1u + 180°2 100. cos1u + 180°2

101. sec1-u2 102. cos1-u2 103. cos1u - 180°2 104. sec1u - 180°2

109. Concept Check The screen below was obtained with the calculator in degree mode. Use it to justify that an angle of 14,879° is a quad-rant II angle.

110. Concept Check The screen below was obtained with the calculator in degree mode. In which quadrant does a 1294° angle lie?

Concept Check Find a solution for each equation.

105. tan13u - 4°2 = 1cot15u - 8°2 106. cos16u + 5°2 = 1sec14u + 15°2

107. sin14u + 2°2 csc13u + 5°2 = 1 108. sec12u + 6°2 cos15u + 3°2 = 1

Concept Check Suppose that 90° 6 u 6 180°. Find the sign of each function value.

89. sin 2u 90. csc 2u 91. tan u

2 92. cot

93. cot1u + 180°2 94. tan1u + 180°2 95. cos1-u2 96. sec1-u2

Chapter 1 Test Prep

Key Terms

1.1 line line segment

(or segment) ray endpoint of a ray angle side of an anglevertex of an angleinitial side terminal side positive angle

negative angle degree acute angle right angle obtuse angle straight angle complementary angles

(complements) supplementary angles

(supplements) minute

second angle in standard

position quadrantal angle coterminal angles

1.2 vertical angles parallel lines transversal similar triangles congruent triangles

1.3 sine (sin) cosine (cos) tangent (tan) cotangent (cot) secant (sec) cosecant (csc) degree mode

1.4 reciprocal

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40 CHAPTER 1 Trigonometric Functions

New Symbols

right angle symbol (for a right triangle)U Greek letter theta

° degree′ minute″ second

Quick ReviewConcepts Examples

70° and 90° - 70° = 20° are complementary.

70° and 180° - 70° = 110° are supplementary.

Angles

Types of AnglesTwo positive angles with a sum of 90° are complementary angles.

Two positive angles with a sum of 180° are supplementary angles.

1 degree = 60 minutes 11° = 60′ 21 minute = 60 seconds 11′ = 60″ 2

Coterminal angles have measures that differ by a multiple of 360°. Their terminal sides coincide when in standard position.

1.1

30′ # 1°60′ = 3060° and 45″ # 1°3600″ = 453600°.

= 15.5125° Decimal degrees

15° 30′ 45″

= 15° + 3060

°+ 45

3600

The acute angle u in the figure is in standard position. If u measures 46°, find the measure of a positive and a negative coterminal angle.

46° + 360° = 406°

46° - 360° = -314°

Angle Relationships and Similar Triangles1.2

Vertical angles have equal measures.

When a transversal intersects two parallel lines, the follow-ing angles formed have equal measure:

alternate interior angles,

alternate exterior angles, and

corresponding angles.

Interior angles on the same side of a transversal are supplementary.

Angle Sum of a TriangleThe sum of the measures of the angles of any triangle is 180°.

Find the measures of angles 1, 2, 3, and 4.

12 (12x – 24)°

3 (4x + 12)°4

m and n are parallel lines.

16x - 12 = 180

x = 12

Angle 2 has measure 12 # 12 - 24 = 120°.Angle 3 has measure 4 # 12 + 12 = 60°.Angle 1 is a vertical angle to angle 2, so its measure is 120°.

Angle 4 corresponds to angle 2, so its measure is 120°.

The measures of two angles of a triangle are 42° 20′ and 35° 10′. Find the measure of the third angle, x.

42° 20′ + 35° 10′ + x = 180° The sum of the angles is 180°. 77° 30′ + x = 180°

x = 102° 30′

12x - 24 + 4x + 12 = 180 Interior angles on the same side of a transversal are supplementary.

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41CHAPTER 1 Test Prep

Find the unknown side length.

x20

1550

20= 50 + 15

Corresponding sides of similar triangles are proportional.

50x = 1300 Property of proportions

x = 26 Divide by 50.

Concepts Examples

Similar triangles have corresponding angles with the same measures and corresponding sides proportional.

Congruent triangles are the same size and the same shape.

Using the Definitions of the Trigonometric Functions1.4

If cot u = - 23 , find tan u.

tan u = 1cot u

= 1- 23

= - 32

Find sin u and tan u, given that cos u = 235 and sin u 6 0. sin2 u + cos2 u = 1 Pythagorean identity

sin2 u + ¢235

≤2 = 1 Replace cos u with 235 . sin2 u + 3

25= 1 Square 235 .

sin2 u = 2225

Subtract 325 .

sin u = - 2225

Choose the negative root.

Reciprocal Identities

sin U =1

csc U cos U =

1sec U

tan U =1

cot U

csc U =1

sin U sec U =

1cos U

cot U =1

tan U

Pythagorean Identities

sin2 U + cos2 U = 1 tan2 U + 1 = sec2 U

1 + cot2 U = csc2 U

Trigonometric Functions

Trigonometric FunctionsLet 1x, y2 be a point other than the origin on the terminal side of an angle u in standard position. The distance from the point to the origin is

r = 2x2 + y2. The six trigonometric functions of u are defined as follows.

1.3

sin U =yr cos U =

tan U =yx

1x 3 0 2csc U =

1 y 3 0 2 sec U = rx

1x 3 0 2 cot U = xy

1 y 3 0 2

If the point 1-2, 32 is on the terminal side of an angle u in standard position, find the values of the six trigono-metric functions of u.

Here x = -2 and y = 3, so

r = 21-222 + 32 = 24 + 9 = 213. sin u = 3213

13 cos u = - 2213

13 tan u = - 3

csc u = 2133

sec u = - 2132

cot u = - 23

See the summary table of trigonometric function values for quadrantal angles in this section.

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42 CHAPTER 1 Trigonometric Functions

Concepts Examples

Quotient Identities

sin Ucos U

= tan U cos Usin U

= cot U

To find tan u, use the values of sin u and cos u from the preceding page and the quotient identity tan u = sin ucos u .

tan u = sin ucos u

=- 222523

= - 22223 # 2323 = - 2663Simplify the complex fraction, and rationalize the denominator.

Signs of the Trigonometric Functions

IISine and cosecant

positive

x < 0, y > 0, r > 0

IAll functions

positive

x > 0, y > 0, r > 0

IIITangent and cotangent

positive

x < 0, y < 0, r > 0

IVCosine and secant

positive

x > 0, y < 0, r > 0x

Identify the quadrant(s) of any angle u that satisfies sin u 6 0, tan u 7 0.

Because sin u 6 0 in quadrants III and IV, and tan u 7 0 in quadrants I and III, both conditions are met only in quadrant III.

1. Give the measures of the complement and the supplement of an angle measuring 35°.

Find the angle of least positive measure that is coterminal with each angle.

2. -51° 3. -174° 4. 792°

Work each problem.

5. Rotating Propeller The propeller of a speedboat rotates 650 times per min. Through how many degrees does a point on the edge of the propeller rotate in 2.4 sec?

6. Rotating Pulley A pulley is rotating 320 times per min. Through how many degrees does a point on the edge of the pulley move in 23 sec?

Convert decimal degrees to degrees, minutes, seconds, and convert degrees, minutes, seconds to decimal degrees. If applicable, round to the nearest second or the nearest thousandth of a degree.

7. 119° 08′ 03″ 8. 47° 25′ 11″ 9. 275.1005° 10. -61.5034°

Find the measure of each marked angle.

11.

(5x + 5)°

(4x)° (4x – 20)°

12.

(9x + 4)°

(12x – 14)°

Chapter 1 Review Exercises

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43CHAPTER 1 Review Exercises

13.

(4x + 5)°

(6x – 45)°

m and n are parallel.

14.

60°30°

x°y°

Solve each problem.

15. Length of a Road A camera is located on a satellite with its lens positioned at C in the figure. Length PC represents the dis-tance from the lens to the film PQ, and BA represents a straight road on the ground. Use the measurements given in the figure to find the length of the road. (Source: Kastner, B., Space Mathematics, NASA.)

16. Express u in terms of a and b.

Not to scale

30 km

150 mm

1.25 mm

B A

P Q

" #

Find all unknown angle measures in each pair of similar triangles.

17.

82°

Q S

R12°

86°

18.

X Y

41°V U

T32°

Find the unknown side lengths in each pair of similar triangles.

19.

16 16 7p

q 20.

n m

3040

In each figure, there are two similar triangles. Find the unknown measurement.

21.

22.

711

23. Length of a Shadow If a tree 20 ft tall casts a shadow 8 ft long, how long would the shadow of a 30-ft tree be at the same time and place?

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44 CHAPTER 1 Trigonometric Functions

Find the six trigonometric function values for each angle. Rationalize denominators when applicable.

24.

(–3, –3)

25.

(1, –√3)θ

26.

0(–2, 0)

Find the values of the six trigonometric functions for an angle in standard position having each given point on its terminal side. Rationalize denominators when applicable.

27. 13, -42 28. 19, -22 29. 1-8, 152 30. 11, -52 31. A623, -6 B 32. A -222, 222 BAn equation of the terminal side of an angle u in standard position is given with a restriction on x. Sketch the least positive such angle u, and find the values of the six trigonometric functions of u.

33. 5x - 3y = 0, x Ú 0 34. y = -5x, x … 0 35. 12x + 5y = 0, x Ú 0

Give all six trigonometric function values for each angle u. Rationalize denominators when applicable.

39. cos u = - 58 , and u is in quadrant III 40. sin u =235 , and cos u 6 0

41. sec u = -25, and u is in quadrant II 42. tan u = 2, and u is in quadrant III43. sec u = 54 , and u is in quadrant IV 44. sin u = -

25 , and u is in quadrant III

45. Decide whether each statement is possible or impossible.

(a) sec u = - 23 (b) tan u = 1.4 (c) cos u = 5

46. Concept Check If, for some particular angle u, sin u 6 0 and cos u 7 0, in what quadrant must u lie? What is the sign of tan u?

Complete the table with the appropriate function values of the given quadrantal angles. If the value is undefined, say so.

U sin U cos U tan U cot U sec U csc U

36. 180°

37. -90°

38. Concept Check If the terminal side of a quadrantal angle lies along the y-axis, which of its trigonometric functions are undefined?

Solve each problem.

47. Swimmer in Distress A lifeguard located 20 yd from the water spots a swimmer in distress. The swimmer is 30 yd from shore and 100 yd east of the lifeguard. Suppose the lifeguard runs and then swims to the swimmer in a direct line, as shown in the figure. How far east from his original position will he enter the water? (Hint: Find the value of x in the sketch.)

20 yd

30 yd

x100 – x

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45CHAPTER 1 Test

48. Angle through Which the Celestial North Pole Moves At present, the north star Polaris is located very near the celestial north pole. However, because Earth is inclined 23.5°, the moon’s gravi-tational pull on Earth is uneven. As a result, Earth slowly precesses (moves in) like a spinning top, and the direction of the celestial north pole traces out a circular path once every 26,000 yr. See the figure. For example, in approximately A.D. 14,000 the star Vega—not the star Polaris—will be located at the celestial north pole. As viewed from the center C of this circular path, calculate the angle (to the nearest second) through which the celestial north pole moves each year. (Source: Zeilik, M., S. Gregory, and E. Smith, Introduc-tory Astronomy and Astrophysics, Second Edition, Saunders College Publishers.)

49. Depth of a Crater on the Moon The depths of unknown craters on the moon can be approximated by comparing the lengths of their shadows to the shadows of nearby craters with known depths. The crater Aristillus is 11,000 ft deep, and its shadow was measured as 1.5 mm on a photograph. Its companion crater, Autolycus, had a shadow of 1.3 mm on the same photograph. Use similar triangles to determine the depth of the crater Autolycus to the nearest hundred feet. (Source: Webb, T., Celestial Objects for Common Telescopes, Dover Publications.)

50. Height of a Lunar Peak The lunar mountain peak Huygens has a height of 21,000 ft. The shadow of Huygens on a photograph was 2.8 mm, while the nearby mountain Bradley had a shadow of 1.8 mm on the same photograph. Calculate the height of Bradley. (Source: Webb, T., Celestial Objects for Common Telescopes, Dover Publications.)

23.5° S

1. Give the measures of the complement and the supplement of an angle measuring 67°.

Find the measure of each marked angle.

(7x + 19)° (2x – 1)°

(–8x + 30)°

(–3x + 5)°

4. (4x – 30)°

(5x – 70)°

Chapter 1 Test

5. 6. 7.

(8x + 14)°

(10x – 10)°m

m and n are parallel.

(32 – 2x)°

(20x + 10)°

(2x + 18)°

(8x)°

(12x + 40)°

(12x)°

Perform each conversion.

8. 74° 18′ 36″ to decimal degrees 9. 45.2025° to degrees, minutes, seconds

Solve each problem.

10. Find the angle of least positive measure that is coterminal with each angle.

(a) 390° (b) -80° (c) 810°

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46 CHAPTER 1 Trigonometric Functions

11. Rotating Tire A tire rotates 450 times per min. Through how many degrees does a point on the edge of the tire move in 1 sec?

12. Length of a Shadow If a vertical pole 30 ft tall casts a shadow 8 ft long, how long would the shadow of a 40-ft pole be at the same time and place?

13. Find the unknown side lengths in this pair of similar triangles.

Sketch an angle u in standard position such that u has the least positive measure, and the given point is on the terminal side of u. Then find the values of the six trigonometric functions for the angle. If any of these are undefined, say so.

14. 12, -72 15. 10, -22

18. If the terminal side of a quadrantal angle lies along the negative x-axis, which two of its trigonometric function values are undefined?

19. Identify the possible quadrant(s) in which u must lie under the given conditions.

(a) cos u 7 0, tan u 7 0 (b) sin u 6 0, csc u 6 0 (c) cot u 7 0, cos u 6 0

20. Decide whether each statement is possible or impossible.

(a) sin u = 1.5 (b) sec u = 4 (c) tan u = 10,000

21. Find the value of sec u if cos u = - 712 .

22. Find the five remaining trigonometric function values of u if sin u = 37 and u is in quadrant II.

Work each problem.

16. Draw a sketch of an angle in standard position having the line with the equation 3x - 4y = 0, x … 0, as its terminal side. Indicate the angle of least positive measure u, and find the values of the six trigonometric functions of u.

17. Complete the table with the appropriate function values of the given quadrantal angles. If the value is undefined, say so.

U sin U cos U tan U cot U sec U csc U

90° -360°

630°

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Trigonometry is used in safe roadway design to provide sufficient visibility around curves as well as a smooth-flowing, comfortable ride.

Trigonometric Functions of Acute Angles

Trigonometric Functions of Non-Acute Angles

Approximations of Trigonometric Function Values

Chapter 2 Quiz

Solutions and Applications of Right Triangles

Further Applications of Right Triangles

2.1

2.2

2.3

2.4

2.5

Acute Angles and Right Triangles2

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48 CHAPTER 2 Acute Angles and Right Triangles

Right-Triangle-Based Definitions of the Trigonometric FunctionsAngles in standard position can be used to define the trigonometric functions. There is also another way to approach them: as ratios of the lengths of the sides of right triangles.

Figure 1 shows an acute angle A in standard position. The definitions of the trigonometric function values of angle A require x, y, and r. As drawn in Figure 1, x and y are the lengths of the two legs of the right triangle ABC, and r is the length of the hypotenuse.

The side of length y is the side opposite angle A, and the side of length x is the side adjacent to angle A. We use the lengths of these sides to replace x and y in the definitions of the trigonometric functions, and the length of the hypot-enuse to replace r, to obtain the following right-triangle-based definitions. In the definitions, we use the standard abbreviations for the sine, cosine, tangent, cosecant, secant, and cotangent functions.

2.1 Trigonometric Functions of Acute Angles■ Right-Triangle-Based

Definitions of the Trigonometric Functions

■ Cofunctions■ How Function Values

Change as Angles Change

■ Trigonometric Function Values of Special Angles

Right-Triangle-Based Definitions of Trigonometric Functions

Let A represent any acute angle in standard position.

sin A =yr=

side opposite Ahypotenuse

csc A =ry=

hypotenuseside opposite A

cos A =xr=

side adjacent to Ahypotenuse

sec A =rx=

hypotenuseside adjacent to A

tan A =yx=

side opposite Aside adjacent to A

cot A =xy=

side adjacent to Aside opposite A

NOTE We will sometimes shorten wording like “side opposite A” to just “side opposite” when the meaning is obvious.

EXAMPLE 1 Finding Trigonometric Function Values of an Acute Angle

Find the sine, cosine, and tangent values for angles A and B in the right triangle in Figure 2.

SOLUTION The length of the side opposite angle A is 7, the length of the side adjacent to angle A is 24, and the length of the hypotenuse is 25.

sin A =side oppositehypotenuse

= 725

cos A =side adjacenthypotenuse

= 2425

tan A =side oppositeside adjacent

= 724

The length of the side opposite angle B is 24, and the length of the side adjacent to angle B is 7.

sin B = 2425

cos B = 725

tan B = 247

Use the right-triangle-based definitions of the trigonometric functions.

■✔ Now Try Exercise 7.

Figure 1

(x, y)

Figure 2

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492.1 Trigonometric Functions of Acute Angles

The cofunction identities state the following.

Cofunction values of complementary angles are equal.

Cofunction Identities

For any acute angle A, the following hold.

sin A = cos 190° − A 2 sec A = csc 190° − A 2 tan A = cot 190° − A 2 cos A = sin 190° − A 2 csc A = sec 190° − A 2 cot A = tan 190° − A 2

EXAMPLE 2 Writing Functions in Terms of Cofunctions

Write each function in terms of its cofunction.

(a) cos 52° (b) tan 71° (c) sec 24°

SOLUTION

(a) Cofunctions

cos 52° = sin190° - 52°2 = sin 38° cos A = sin190° - A2 Complementary angles

(b) tan 71° = cot190° - 71°2 = cot 19° (c) sec 24° = csc 66°■✔ Now Try Exercises 25 and 27.

Figure 3

Whenever we use A, B, and C to name angles in a right triangle, C will be the right angle.

NOTE The cosecant, secant, and cotangent ratios are reciprocals of the sine, cosine, and tangent values, respectively, so in Example 1 we have

csc A = 257

sec A = 2524

cot A = 247

csc B = 2524

sec B = 257 and cot B = 7

24 .

Cofunctions Figure 3 shows a right triangle with acute angles A and B and a right angle at C. The length of the side opposite angle A is a, and the length of the side opposite angle B is b. The length of the hypotenuse is c. By the pre-ceding definitions, sin A = ac . Also, cos B =

ac . Thus, we have the following.

sin A =ac= cos B

Similarly, tan A =ab= cot B and sec A =

cb= csc B.

In any right triangle, the sum of the two acute angles is 90°, so they are complementary. In Figure 3, A and B are thus complementary, and we have established that sin A = cos B. This can also be written as follows.

sin A = cos190° - A2 B = 90° - AThis is an example of a more general relationship between cofunction pairs.

sine, cosine

tangent, cotangent

secant, cosecant (+)+*

Cofunction pairs

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50 CHAPTER 2 Acute Angles and Right Triangles

How Function Values Change as Angles Change Figure 4 shows three right triangles. From left to right, the length of each hypotenuse is the same, but angle A increases in measure. As angle A increases in measure from 0° to 90°, the length of the side opposite angle A also increases.

EXAMPLE 3 Solving Equations Using Cofunction Identities

Find one solution for each equation. Assume all angles involved are acute angles.

(a) cos1u + 4°2 = sin13u + 2°2 (b) tan12u - 18°2 = cot1u + 18°2SOLUTION

(a) Sine and cosine are cofunctions, so cos1u + 4°2 = sin13u + 2°2 is true if the sum of the angles is 90°.

1u + 4°2 + 13u + 2°2 = 90° Complementary angles 4u + 6° = 90° Combine like terms.

4u = 84° Subtract 6° from each side. u = 21° Divide by 4.

(b) Tangent and cotangent are cofunctions.

12u - 18°2 + 1u + 18°2 = 90° Complementary angles 3u = 90° Combine like terms. u = 30° Divide by 3.

■✔ Now Try Exercises 31 and 33.

In the ratio

sin A =side oppositehypotenuse

=yr

as angle A increases, the numerator of this fraction also increases, while the denominator is fixed. Therefore, sin A increases as A increases from 0° to 90°.

As angle A increases from 0° to 90°, the length of the side adjacent to A decreases. Because r is fixed, the ratio xr decreases. This ratio gives cos A, showing that the values of cosine decrease as the angle measure changes from 0° to 90°. Finally, increasing A from 0° to 90° causes y to increase and x to decrease, making the values of

yx = tan A increase.

A similar discussion shows that as A increases from 0° to 90°, the values of sec A increase, while the values of cot A and csc A decrease.

r y

sin A =

r y

Figure 4

As A increases, y increases. Because r is fixed, sin A increases.

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512.1 Trigonometric Functions of Acute Angles

Trigonometric Function Values of Special Angles Certain special angles, such as 30°, 45°, and 60°, occur so often in trigonometry and in more advanced mathematics that they deserve special study. We start with an equilateral triangle, a triangle with all sides of equal length. Each angle of such a triangle measures 60°. Although the results we will obtain are independent of the length, for convenience we choose the length of each side to be 2 units. See Figure 5(a).

Bisecting one angle of this equilateral triangle leads to two right triangles, each of which has angles of 30°, 60°, and 90°, as shown in Figure 5(b). An angle bisector of an equilateral triangle also bisects the opposite side. Thus the shorter leg has length 1. Let x represent the length of the longer leg.

22 = 12 + x2 Pythagorean theorem 4 = 1 + x2 Apply the exponents. 3 = x2 Subtract 1 from each side.

23 = x Square root property; choose the positive root.Figure 6 summarizes our results using a 30°9 60° right triangle. As shown in the figure, the side opposite the 30° angle has length 1. For the 30° angle,

hypotenuse = 2, side opposite = 1, side adjacent = 23 .Now we use the definitions of the trigonometric functions.

sin 30° =side oppositehypotenuse

= 12

cos 30° =side adjacenthypotenuse

= 232

tan 30° =side oppositeside adjacent

= 123 = 123 # 2323 = 233 csc 30° = 2

1= 2

sec 30° = 223 = 223 # 2323 = 2233 cot 30° = 23

1= 23

Rationalize the denominators.

EXAMPLE 4 Comparing Function Values of Acute Angles

Determine whether each statement is true or false.

(a) sin 21° 7 sin 18° (b) sec 56° … sec 49°SOLUTION

(a) In the interval from 0° to 90°, as the angle increases, so does the sine of the angle. This makes sin 21° 7 sin 18° a true statement.

(b) For fixed r, increasing an angle from 0° to 90° causes x to decrease. Therefore, sec u = rx increases. The statement sec 56° … sec 49° is false.

■✔ Now Try Exercises 41 and 47.

Figure 6

60°

30°

√3

60°

60°2

2 2

Equilateral triangle

(b)

Figure 5

60°

30°

60°1

2 2

30°

190°90°

x x

30°– 60° right triangle

(a)

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52 CHAPTER 2 Acute Angles and Right Triangles

We find the values of the trigonometric functions for 45° by starting with a 45°9 45° right triangle, as shown in Figure 7. This triangle is isosceles. For simplicity, we choose the lengths of the equal sides to be 1 unit. (As before, the results are independent of the length of the equal sides.) If r represents the length of the hypotenuse, then we can find its value using the Pythagorean theorem.

12 + 12 = r 2 Pythagorean theorem 2 = r 2 Simplify.

22 = r Choose the positive root.Now we use the measures indicated on the 45°9 45° right triangle in Figure 7.

sin 45° = 122 = 222 cos 45° = 122 = 222 tan 45° = 11 = 1 csc 45° = 22

1= 22 sec 45° = 22

1= 22 cot 45° = 1

1= 1

Function values for 30°, 45°, and 60° are summarized in the table that follows.

EXAMPLE 5 Finding Trigonometric Function Values for 60°

Find the six trigonometric function values for a 60° angle.

SOLUTION Refer to Figure 6 to find the following ratios.

sin 60° = 232

cos 60° = 12

tan 60° = 231

= 23 csc 60° = 223 = 2233 sec 60° = 21 = 2 cot 60° = 123 = 233

■✔ Now Try Exercises 49, 51, and 53.

NOTE The results in Example 5 can also be found using the fact that cofunction values of the complementary angles 60° and 30° are equal.

Function Values of Special Angles

U sin U cos U tan U cot U sec U csc U

30° 12

!32

!33

!3 2 !33

45° !22

!22

1 1 !2 !260° !3

12 !3 !33 2 2 !33

NOTE You will be able to reproduce this table quickly if you learn the values of sin 30°, sin 45°, and sin 60°. Then you can complete the rest of the table using the reciprocal, cofunction, and quotient identities.

Figure 7

45°

r = √21

45°– 45° right triangle

Figure 6 (repeated)

60°

30°

√3

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532.1 Trigonometric Functions of Acute Angles

A calculator can find trigonometric function values at the touch of a key. So why do we spend so much time finding values for special angles? We do this because a calculator gives only approximate values in most cases instead of exact values. A scientific calculator gives the following approximation for tan 30°.

tan 30° ≈ 0.57735027 ≈ means “is approximately equal to.”

Earlier, however, we found the exact value.

tan 30° = 233

Exact value

Figure 8 shows mode display options for the TI-84 Plus. Figure 9 displays the output when evaluating the tangent, sine, and cosine of 30°. (The calculator should be in degree mode to enter angle measure in degrees.)

2.1 Exercises

CONCEPT PREVIEW Match each trigonometric function in Column I with its value in Column II. Choices may be used once, more than once, or not at all.

1. sin 30° 2. cos 45°

3. tan 45° 4. sec 60°

5. csc 60° 6. cot 30°

A. 23 B. 1 C. 12

D. 232

E. 223

3 F.

233

G. 2 H. 222

I. 22Find exact values or expressions for sin A, cos A, and tan A. See Example 1.

A20

10. A

y k

Figure 8 Figure 9 ■

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54 CHAPTER 2 Acute Angles and Right Triangles

Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C.

A C

Use the Pythagorean theorem to find the unknown side length. Then find exact values of the six trigonometric functions for angle B. Rationalize denominators when applicable. See Example 1.

11. a = 5, b = 12 12. a = 3, b = 4 13. a = 6, c = 7

14. b = 7, c = 12 15. a = 3, c = 10 16. b = 8, c = 11

17. a = 1, c = 2 18. a = 22, c = 2 19. b = 2, c = 520. Concept Check Give the six cofunction identities.

Write each function in terms of its cofunction. Assume that all angles labeled u are acute angles. See Example 2.

21. cos 30° 22. sin 45° 23. csc 60°

24. cot 73° 25. sec 39° 26. tan 25.4°

27. sin 38.7° 28. cos1u + 20°2 29. sec1u + 15°230. Concept Check With a calculator, evaluate sin190° - u2 and cos u for various

values of u. (Check values greater than 90° and less than 0°.) Comment on the results.

Find one solution for each equation. Assume that all angles involved are acute angles. See Example 3.

31. tan a = cot1a + 10°2 32. cos u = sin12u - 30°233. sin12u + 10°2 = cos13u - 20°2 34. sec1b + 10°2 = csc12b + 20°235. tan13B + 4°2 = cot15B - 10°2 36. cot15u + 2°2 = tan12u + 4°237. sin1u - 20°2 = cos12u + 5°2 38. cos12u + 50°2 = sin12u - 20°239. sec13b + 10°2 = csc1b + 8°2 40. csc1b + 40°2 = sec1b - 20°2Determine whether each statement is true or false. See Example 4.

41. sin 50° 7 sin 40° 42. tan 28° … tan 40°

43. sin 46° 6 cos 46° 1Hint: cos 46° = sin 44°2 44. cos 28° 6 sin 28° 1Hint: sin 28° = cos 62°245. tan 41° 6 cot 41° 46. cot 30° 6 tan 40°

47. sec 60° 7 sec 30° 48. csc 20° 6 csc 30°

Give the exact value of each expression. See Example 5.

49. tan 30° 50. cot 30° 51. sin 30° 52. cos 30°

53. sec 30° 54. csc 30° 55. csc 45° 56. sec 45°

57. cos 45° 58. cot 45° 59. tan 45° 60. sin 45°

61. sin 60° 62. cos 60° 63. tan 60° 64. csc 60°

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552.1 Trigonometric Functions of Acute Angles

Concept Check Work each problem.

65. What value of A between 0° and 90° will produce the output shown on the graphing calculator screen?

66. A student was asked to give the exact value of sin 45°. Using a calculator, he gave the answer 0.7071067812. Explain why the teacher did not give him credit.

67. Find the equation of the line that passes through the origin and makes a 30° angle with the x-axis.

68. Find the equation of the line that passes through the origin and makes a 60° angle with the x-axis.

69. What angle does the line y = 23x make with the positive x-axis?70. What angle does the line y = 233 x make with the positive x-axis?71. Consider an equilateral triangle with each side having length 2k.

(a) What is the measure of each angle?

(b) Label one angle A. Drop a perpendicular from A to the side opposite A. Two 30° angles are formed at A, and two right tri-angles are formed. What is the length of the sides opposite the 30° angles?

(d) From the results of parts (a) – (c), complete the following statement:

In a 30°9 60° right triangle, the hypotenuse is always times as long as the shorter leg, and the longer leg has a length that is times as long as that of the shorter leg. Also, the shorter leg is opposite the angle, and the longer leg is opposite the angle.

2k2k

72. Consider a square with each side of length k.

(a) Draw a diagonal of the square. What is the measure of each angle formed by a side of the square and this diagonal?

(b) What is the length of the diagonal?

In a 45°9 45° right triangle, the hypotenuse has a length that is times as long as either leg.

Find the exact value of the variables in each figure.

73.

yw9

30° 60°x

74.

45°

60°

75.

90° 45°

45° 30°

76.

60°

45°

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56 CHAPTER 2 Acute Angles and Right Triangles

Find a formula for the area of each figure in terms of s.

77.

45°

78.

60° 60°

60°

79. With a graphing calculator, find the coordinates of the point of intersection of the graphs of y = x and y = 11 - x2. These coordinates are the cosine and sine of what angle between 0° and 90°?

80. Concept Check Suppose we know the length of one side and one acute angle of a 30°9 60° right triangle. Is it possible to determine the measures of all the sides and angles of the triangle?

Relating Concepts

For individual or collaborative investigation (Exercises 81–84)

The figure shows a 45° central angle in a circle with radius 4 units. To find the coordinates of point P on the circle, work Exercises 81–84 in order.

81. Sketch a line segment from P perpendicular to the x-axis.

82. Use the trigonometric ratios for a 45° angle to label the sides of the right triangle sketched in Exercise 81.

83. Which sides of the right triangle give the coordinates of point P? What are the coordinates of P?

84. The figure at the right shows a 60° central angle in a circle of radius 2 units. Follow the same proce-dure as in Exercises 81– 83 to find the coordinates of P in the figure.

45°4

60°2

2.2 Trigonometric Functions of Non-Acute Angles

Reference Angles Associated with every nonquadrantal angle in stand-ard position is an acute angle called its reference angle. A reference angle for an angle u, written u′, is the acute angle made by the terminal side of angle u and the x-axis.

■ Reference Angles■ Special Angles as

Reference Angles■ Determination of

Angle Measures with Special Reference Angles

Figure 10 shows several angles u (each less than one complete counterclock-wise revolution) in quadrants II, III, and IV, respectively, with the reference angle u′ also shown. In quadrant I, u and u′ are the same. If an angle u is negative or has measure greater than 360°, its reference angle is found by first finding its cotermi-nal angle that is between 0° and 360°, and then using the diagrams in Figure 10.

NOTE Reference angles are always positive and are between 0° and 90°.

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572.2 Trigonometric Functions of Non-Acute Angles

CAUTION A common error is to find the reference angle by using the terminal side of u and the y-axis. The reference angle is always found with reference to the x-axis.

EXAMPLE 1 Finding Reference Angles

Find the reference angle for each angle.

(a) 218° (b) 1387°

SOLUTION

(a) As shown in Figure 11(a), the positive acute angle made by the terminal side of this angle and the x-axis is

218° - 180° = 38°.

For u = 218°, the reference angle u′ = 38°.

38°

218°

218° – 180° = 38°

(a) (b)

Figure 11

53°307°

360° – 307° = 53°

(b) First find a coterminal angle between 0° and 360°. Divide 1387° by 360° to obtain a quotient of about 3.9. Begin by subtracting 360° three times (because of the whole number 3 in 3.9).

1387° - 3 # 360°= 1387° - 1080° Multiply.= 307° Subtract.

The reference angle for 307° (and thus for 1387°) is 360° - 307° = 53°. See Figure 11(b).

■✔ Now Try Exercises 5 and 9.

Figure 10

u in quadrant II

uu!

u in quadrant III

u in quadrant IV

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58 CHAPTER 2 Acute Angles and Right Triangles

Reference Angle U′ for U, where 0° * U * 360° *

yQ I

yQ II

yQ III

yQ IV

u! u!

u!u!

u! = 180° – uu! = 180° – u u! = u

u! = u – 180° u! = 360° – u

*The authors would like to thank Bethany Vaughn and Theresa Matick, of Vincennes Lincoln High School, for their suggestions concerning this table.

EXAMPLE 2 Finding Trigonometric Function Values of a Quadrant III Angle

Find exact values of the six trigonometric functions of 210°.

SOLUTION An angle of 210° is shown in Figure 12. The reference angle is

210° - 180° = 30°.

To find the trigonometric function values of 210°, choose point P on the ter-minal side of the angle so that the distance from the origin O to P is 2. (Any positive number would work, but 2 is most convenient.) By the results from

30°9 60° right triangles, the coordinates of point P become A -23 , -1 B , with x = -23 , y = -1, and r = 2. Then, by the definitions of the trigonometric functions, we obtain the following.

sin 210° = -12

= - 12

csc 210° = 2-1 = -2

cos 210° = -232

= - 232

sec 210° = 2-23 = - 2233

tan 210° = -1 -23 = 233 cot 210° = -23-1 = 23

■✔ Now Try Exercise 19.

Rationalize denominators as needed.

The preceding example suggests the following table for finding the refer-ence angle u′ for any angle u between 0° and 360°.

Special Angles as Reference Angles We can now find exact trigono-metric function values of angles with reference angles of 30°, 45°, or 60°.

Figure 12

210°

30°60°

y = –1r = 2x = –Ë3

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592.2 Trigonometric Functions of Non-Acute Angles

Notice in Example 2 that the trigonometric function values of 210° cor-respond in absolute value to those of its reference angle 30°. The signs are different for the sine, cosine, secant, and cosecant functions because 210° is a quadrant III angle. These results suggest a shortcut for finding the trigonometric function values of a non-acute angle, using the reference angle.

In Example 2, the reference angle for 210° is 30°. Using the trigonometric function values of 30°, and choosing the correct signs for a quadrant III angle, we obtain the same results.

We determine the values of the trigonometric functions for any nonquadran-tal angle u as follows. Keep in mind that all function values are positive when the terminal side is in Quadrant I, the sine and cosecant are positive in Quadrant II, the tangent and cotangent are positive in Quadrant III, and the cosine and secant are positive in Quadrant IV. In other cases, the function values are negative.

Finding Trigonometric Function Values for Any Nonquadrantal Angle U�

Step 1 If u 7 360°, or if u 6 0°, then find a coterminal angle by adding or subtracting 360° as many times as needed to obtain an angle greater than 0° but less than 360°.

Step 2 Find the reference angle u′.

Step 3 Find the trigonometric function values for reference angle u′.

Step 4 Determine the correct signs for the values found in Step 3. (Use the table of signs given earlier in the text or the paragraph above, if necessary.) This gives the values of the trigonometric functions for angle u.

NOTE To avoid sign errors when finding the trigonometric function val-ues of an angle, sketch it in standard position. Include a reference triangle complete with appropriate values for x, y, and r as done in Figure 12.

EXAMPLE 3 Finding Trigonometric Function Values Using Reference Angles

Find the exact value of each expression.

(a) cos1-240°2 (b) tan 675°SOLUTION

(a) Because an angle of -240° is coterminal with an angle of

-240° + 360° = 120°,

the reference angle is 180° - 120° = 60°, as shown in Figure 13(a). The cosine is negative in quadrant II.

cos1-240°2 = cos 120° Coterminal angle = -cos 60° Reference angle

= - 12

Evaluate.

(a)

Figure 13

u! = 60°

u = –240°

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60 CHAPTER 2 Acute Angles and Right Triangles

As shown in Figure 13(b), the reference angle is 360° - 315° = 45°. An angle of 315° is in quadrant IV, so the tangent will be negative.

tan 675°

= tan 315° Coterminal angle = - tan 45° Reference angle; quadrant-based sign choice = -1 Evaluate. ■✔ Now Try Exercises 37 and 39.

EXAMPLE 4 Using Function Values of Special Angles

Evaluate cos 120° + 2 sin2 60° - tan2 30°

SOLUTION Use the procedure explained earlier to determine cos 120° = - 12 .

Then use the values cos 120° = - 12 , sin 60° =232 , and tan 30° =

233 .

cos 120° + 2 sin2 60° - tan2 30°

= - 12

+ 2¢232

≤2 - ¢233

≤2 Substitute values. = - 1

2+ 2a 3

4b - 3

9 Apply the exponents.

= 23

Simplify.

■✔ Now Try Exercise 47.

EXAMPLE 5 Using Coterminal Angles to Find Function Values

Evaluate each function by first expressing it in terms of a function of an angle between 0° and 360°.

(a) cos 780° (b) cot1-405°2SOLUTION

(a) Subtract 360° as many times as necessary to obtain an angle between 0° and 360°, which gives the following.

cos 780°

= cos1780° - 2 # 360°2 Subtract 720°, which is 2 # 360°.= cos 60° Multiply first and then subtract.

= 12

Evaluate.

(b) Add 360° twice to obtain -405° + 21360°2 = 315° , which is located in quadrant IV and has reference angle 45°. The cotangent will be negative.

cot1-405°2 = cot 315° = -cot 45° = -1■✔ Now Try Exercises 27 and 31.

(b)

Figure 13

0u = 675° u! = 45°

(b) Subtract 360° to find an angle between 0° and 360° coterminal with 675°.

675° - 360° = 315°

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612.2 Trigonometric Functions of Non-Acute Angles

Determination of Angle Measures with Special Reference AnglesThe ideas discussed in this section can be used “in reverse” to find the measures of certain angles, given a trigonometric function value and an interval in which the angle must lie. We are most often interested in the interval 30°, 360°2.

EXAMPLE 6 Finding Angle Measures Given an Interval and a Function Value

Find all values of u, if u is in the interval 30°, 360°2 and cos u = - 222 .SOLUTION The value of cos u is negative, so u may lie in either quadrant II or III.

Because the absolute value of cos u is 222 , the reference angle u′ must be 45°. The two possible angles u are sketched in Figure 14.

180° - 45° = 135° Quadrant II angle u (from Figure 14 (a)) 180° + 45° = 225° Quadrant III angle u (from Figure 14 (b))

(a) (b)

Figure 14

u! = 45°

u in quadrant II

u = 135°

u in quadrant III

u = 225°

u! = 45°

■✔ Now Try Exercise 61.

CONCEPT PREVIEW Fill in the blanks to correctly complete each sentence.

1. The value of sin 240° is because 240° is in quadrant . (positive/negative)

The reference angle is , and the exact value of sin 240° is .

2. The value of cos 390° is because 390° is in quadrant . (positive/negative)

The reference angle is , and the exact value of cos 390° is .

3. The value of tan1-150°2 is because -150° is in quadrant (positive/negative)

. The reference angle is , and the exact value of tan1-150°2 is . 4. The value of sec 135° is because 135° is in quadrant . (positive/negative)

The reference angle is , and the exact value of sec 135° is .

2.2 Exercises

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62 CHAPTER 2 Acute Angles and Right Triangles

Concept Check Match each angle in Column I with its reference angle in Column II. Choices may be used once, more than once, or not at all. See Example 1.

5. 98° 6. 212°

7. -135° 8. -60°

9. 750° 10. 480°

A. 45° B. 60°

C. 82° D. 30°

E. 38° F. 32°

Find the exact value of each expression. See Example 3.

37. sin 1305° 38. sin 1500° 39. cos1-510°2 40. tan1-1020°241. csc1-855°2 42. sec1-495°2 43. tan 3015° 44. cot 2280°Evaluate each expression. See Example 4.

45. sin2 120° + cos2 120° 46. sin2 225° + cos2 225°

47. 2 tan2 120° + 3 sin2 150° - cos2 180° 48. cot2 135° - sin 30° + 4 tan 45°

49. sin2 225° - cos2 270° + tan2 60° 50. cot2 90° - sec2 180° + csc2 135°

51. cos2 60° + sec2 150° - csc2 210° 52. cot2 135° + tan4 60° - sin4 180°

Complete the table with exact trigonometric function values. Do not use a calculator. See Examples 2 and 3.

U sin U cos U tan U cot U sec U csc U

11. 30°12

232

2233

12. 45° 1 1

13. 60° 1

2 23 2 14. 120° 23

2 -23 223

15. 135° 222

- 222

-22 2216. 150°

- 23

2- 23

17. 210° - 12

233

23 -218. 240° - 23

2- 1

2 -2 - 223

Find exact values of the six trigonometric functions of each angle. Rationalize denomi-nators when applicable. See Examples 2, 3, and 5.

19. 300° 20. 315° 21. 405° 22. 420° 23. 480° 24. 495°

25. 570° 26. 750° 27. 1305° 28. 1500° 29. -300° 30. -390°

31. -510° 32. -1020° 33. -1290° 34. -855° 35. -1860° 36. -2205°

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632.2 Trigonometric Functions of Non-Acute Angles

Determine whether each statement is true or false. If false, tell why. See Example 4.

53. cos130° + 60°2 = cos 30° + cos 60° 54. sin 30° + sin 60° = sin130° + 60°255. cos 60° = 2 cos 30° 56. cos 60° = 2 cos2 30° - 1

57. sin2 45° + cos2 45° = 1 58. tan2 60° + 1 = sec2 60°

59. cos12 # 45°2 = 2 cos 45° 60. sin12 # 30°2 = 2 sin 30° # cos 30°Find all values of u, if u is in the interval 30°, 360°2 and has the given function value. See Example 6.

61. sin u = 12

62. cos u = 232

63. tan u = -2364. sec u = -22 65. cos u = 22

266. cot u = - 23

67. csc u = -2 68. sin u = - 232

69. tan u = 233

70. cos u = - 12

71. csc u = -22 72. cot u = -1Concept Check Find the coordinates of the point P on the circumference of each circle. (Hint: Sketch x- and y-axes, and interpret so that the angle is in standard position.)

73. P

150°6

74.

225°

75. Concept Check Does there exist an angle u with the function values cos u = 0.6 and sin u = - 0.8?

76. Concept Check Does there exist an angle u with the function values cos u = 23 and sin u = 34 ?

Suppose u is in the interval 190°, 180°2 . Find the sign of each of the following.77. cos

278. sin

279. sec1u + 180°2

80. cot1u + 180°2 81. sin1-u2 82. cos1-u2Concept Check Work each problem.

83. Why is sin u = sin1u + n # 360°2 true for any angle u and any integer n?84. Why is cos u = cos1u + n # 360°2 true for any angle u and any integer n?85. Without using a calculator, determine which of the following numbers is closest to

sin 115° : - 0.9, - 0.1, 0, 0.1, or 0.9.

86. Without using a calculator, determine which of the following numbers is closest to cos 115° : - 0.6, - 0.4, 0, 0.4, or 0.6.

87. For what angles u between 0° and 360° is cos u = sin u true?

88. For what angles u between 0° and 360° is cos u = - sin u true?

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64 CHAPTER 2 Acute Angles and Right Triangles

2.3 Approximations of Trigonometric Function Values

Calculator Approximations of Trigonometric Function Values We learned how to find exact function values for special angles and for angles having special reference angles earlier in this chapter. In this section we investigate how calculators provide approximations for function values of angles that do not sat-isfy these conditions. (Of course, they can also be used to find exact values such as cos 1-240°2 and tan 675°, as seen in Figure 15.)

■ Calculator Approximations of Trigonometric Function Values

■ Calculator Approximations of Angle Measures

■ An Application

CAUTION It is important to remember that when we are evaluating trig-onometric functions of angles given in degrees, the calculator must be in degree mode. An easy way to check this is to enter sin 90. If the displayed answer is 1, then the calculator is in degree mode.

Also remember that if the angle or the reference angle is not a special or quadrantal angle, then the value given by the calculator is an approximation. And even if the angle or reference angle is a special angle, the value given by the calculator will often be an approximation.

EXAMPLE 1 Finding Function Values with a Calculator

Approximate the value of each expression.

(a) sin 49° 12′ (b) sec 97.977° (c) 1

cot 51.4283° (d) sin1-246°2

SOLUTION See Figure 16. We give values to eight decimal places below.

(a) We may begin by converting 49° 12′ to decimal degrees.

49° 12′ = 49 1260°= 49.2°

However, some calculators allow direct entry of degrees, minutes, and seconds. (The method of entry varies among models.) Entering either sin149° 12′2 or sin 49.2° gives the same approximation.

sin 49° 12′ = sin 49.2° ≈ 0.75699506

(b) There are no dedicated calculator keys for the secant, cosecant, and cotan-gent functions. However, we can use reciprocal identities to evaluate them. Recall that sec u = 1cos u for all angles u, where cos u ≠ 0. Therefore, we use the reciprocal of the cosine function to evaluate the secant function.

sec 97.977° = 1cos 97.977°

≈ -7.20587921

1cot 51.4283°

= tan 51.4283° ≈ 1.25394815

(d) sin1-246°2 ≈ 0.91354546■✔ Now Try Exercises 11, 13, 17, and 21.

Degree mode

Figure 15

Degree mode

Figure 16

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652.3 Approximations of Trigonometric Function Values

Calculator Approximations of Angle Measures To find the measure of an angle having a certain trigonometric function value, calculators have three inverse functions (denoted sin−1 , cos−1 , and tan−1). If x is an appropriate num-ber, then sin−1 x, cos−1 x, or tan−1 x gives the measure of an angle whose sine, cosine, or tangent, respectively, is x. For applications in this chapter, these functions will return angles in quadrant I.

Figure 18

u > 0°

u < 0°

EXAMPLE 3 Finding Grade Resistance

When an automobile travels uphill or downhill on a highway, it experiences a force due to gravity. This force F in pounds is the grade resistance and is modeled by the equation

F = W sin u,

where u is the grade and W is the weight of the automobile. If the automo-bile is moving uphill, then u 7 0°; if downhill, then u 6 0°. See Figure 18. (Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.)

An Application

EXAMPLE 2 Using Inverse Trigonometric Functions to Find Angles

Find an angle u in the interval 30°, 90°2 that satisfies each condition.(a) sin u = 0.96770915 (b) sec u = 1.0545829SOLUTION

(a) Using degree mode and the inverse sine function, we find that an angle u having sine value 0.96770915 is 75.399995°. (There are infinitely many such angles, but the calculator gives only this one.)

u = sin-1 0.96770915 ≈ 75.399995°

See Figure 17.

(b) Use the identity cos u = 1sec u . If sec u = 1.0545829 , then

cos u = 11.0545829

Now, find u using the inverse cosine function. See Figure 17.

u = cos-1 a 11.0545829

b ≈ 18.514704°■✔ Now Try Exercises 31 and 35.

Degree mode

Figure 17

CAUTION Compare Examples 1(b) and 2(b).

To determine the secant of an angle, as in Example 1(b), we find the reciprocal of the cosine of the angle.

To determine an angle with a given secant value, as in Example 2(b), we find the inverse cosine of the reciprocal of the value.

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66 CHAPTER 2 Acute Angles and Right Triangles

CONCEPT PREVIEW Match each trigonometric function value or angle in Column I with its appropriate approximation in Column II.

2.3 Exercises

1. sin 83° 2. cos-1 0.45

3. tan 16° 4. cot 27°

5. sin-1 0.30 6. sec 18°

7. csc 80° 8. tan-1 30

9. csc-1 4 10. cot-1 30

A. 88.09084757° B. 63.25631605°

C. 1.909152433° D. 17.45760312°

E. 0.2867453858 F. 1.962610506

G. 14.47751219° H. 1.015426612

I. 1.051462224 J. 0.9925461516

Use a calculator to approximate the value of each expression. Give answers to six dec-imal places. In Exercises 21–28, simplify the expression before using the calculator. See Example 1.

11. sin 38° 42′ 12. cos 41° 24′ 13. sec 13° 15′

14. csc 145° 45′ 15. cot 183° 48′ 16. tan 421° 30′

17. sin1-312° 12′2 18. tan1-80° 06′2 19. csc1-317° 36′220. cot1-512° 20′2 21. 1

cot 23.4°22.

1sec 14.8°

23. cos 77°sin 77°

24. sin 33°cos 33°

25. cot190° - 4.72°226. cos190° - 3.69°2 27. 1

csc190° - 51°2 28. 1tan190° - 22°2

(a) Calculate F to the nearest 10 lb for a 2500-lb car traveling an uphill grade with u = 2.5°.

(b) Calculate F to the nearest 10 lb for a 5000-lb truck traveling a downhill grade with u = -6.1°.

SOLUTION

(a) F = W sin u Given model for grade resistance F = 2500 sin 2.5° Substitute given values. F ≈ 110 lb Evaluate.

(b) F = W sin u = 5000 sin1-6.1°2 ≈ -530 lbF is negative because the truck is moving downhill.

(c) F = W sin u = W sin 0° = W102 = 0 lbF = W sin u = W sin 90° = W112 = W lbThis agrees with intuition because if u = 0°, then there is level ground and gravity does not cause the vehicle to roll. If u were 90°, the road would be vertical and the full weight of the vehicle would be pulled downward by gravity, so F = W.

■✔ Now Try Exercises 69 and 71.

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672.3 Approximations of Trigonometric Function Values

Find a value of u in the interval 30°, 90°2 that satisfies each statement. Write each answer in decimal degrees to six decimal places. See Example 2.

29. tan u = 1.4739716 30. tan u = 6.4358841 31. sin u = 0.27843196

32. sin u = 0.84802194 33. cot u = 1.2575516 34. csc u = 1.3861147

35. sec u = 2.7496222 36. sec u = 1.1606249 37. cos u = 0.70058013

38. cos u = 0.85536428 39. csc u = 4.7216543 40. cot u = 0.21563481

Concept Check Answer each question.

41. A student, wishing to use a calculator to verify the value of sin 30°, enters the infor-mation correctly but gets a display of - 0.98803162. He knows that the display should be 0.5, and he also knows that his calculator is in good working order. What might the problem be?

42. At one time, a certain make of calculator did not allow the input of angles outside of a particular interval when finding trigonometric function values. For example, trying to find cos 2000° using the methods of this section gave an error message, despite the fact that cos 2000° can be evaluated. How would we use this calculator to find cos 2000°?

43. What value of A, to the nearest degree, between 0° and 90° will produce the output in the graphing cal-culator screen?

44. What value of A will produce the output (in degrees) in the graphing calculator screen? Give as many decimal places as shown on the calculator.

Use a calculator to evaluate each expression.

45. sin 35° cos 55° + cos 35° sin 55° 46. cos 100° cos 80° - sin 100° sin 80°

47. sin2 36° + cos2 36° 48. 2 sin 25° 13′ cos 25° 13′ - sin 50° 26′

49. cos 75° 29′ cos 14° 31′ - sin 75° 29′ sin 14° 31′

50. sin 28° 14′ cos 61° 46′ + cos 28° 14′ sin 61° 46′

Use a calculator to decide whether each statement is true or false. It may be that a true statement will lead to results that differ in the last decimal place due to rounding error.

51. sin 10° + sin 10° = sin 20° 52. cos 40° = 2 cos 20°

53. sin 50° = 2 sin 25° cos 25° 54. cos 70° = 2 cos2 35° - 1

55. cos 40° = 1 - 2 sin2 80° 56. 2 cos 38° 22′ = cos 76° 44′

57. sin 39° 48′ + cos 39° 48′ = 1 58. 12

sin 40° = sin c 12

140°2 d59. 1 + cot2 42.5° = csc2 42.5° 60. tan2 72° 25′ + 1 = sec2 72° 25′

61. cos130° + 20°2 = cos 30° cos 20° - sin 30° sin 20°62. cos130° + 20°2 = cos 30° + cos 20°Find two angles in the interval 30°, 360°2 that satisfy each of the following. Round answers to the nearest degree.

63. sin u = 0.92718385 64. sin u = 0.52991926

65. cos u = 0.71933980 66. cos u = 0.10452846

67. tan u = 1.2348971 68. tan u = 0.70020753

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68 CHAPTER 2 Acute Angles and Right Triangles

(Modeling) Grade Resistance Solve each problem. See Example 3.

69. Find the grade resistance, to the nearest ten pounds, for a 2100-lb car traveling on a 1.8° uphill grade.

70. Find the grade resistance, to the nearest ten pounds, for a 2400-lb car traveling on a -2.4° downhill grade.

71. A 2600-lb car traveling downhill has a grade resis-tance of -130 lb. Find the angle of the grade to the nearest tenth of a degree.

72. A 3000-lb car traveling uphill has a grade resistance of 150 lb. Find the angle of the grade to the nearest tenth of a degree.

73. A car traveling on a 2.7° uphill grade has a grade resistance of 120 lb. Determine the weight of the car to the nearest hundred pounds.

74. A car traveling on a -3° downhill grade has a grade resistance of -145 lb. Determine the weight of the car to the nearest hundred pounds.

75. Which has the greater grade resistance: a 2200-lb car on a 2° uphill grade or a 2000-lb car on a 2.2° uphill grade?

76. Complete the table for values of sin u , tan u , and pu180 to four decimal places.

U 0° 0.5° 1° 1.5° 2° 2.5° 3° 3.5° 4°sin U

tan U

180

(a) How do sin u, tan u, and pu180 compare for small grades u?

(b) Highway grades are usually small. Give two approximations of the grade resis-tance F = W sin u that do not use the sine function.

(c) A stretch of highway has a 4-ft vertical rise for every 100 ft of horizontal run. Use an approximation from part (b) to estimate the grade resistance, to the near-est pound, for a 2000-lb car on this stretch of highway.

(d) Without evaluating a trigonometric function, estimate the grade resistance, to the nearest pound, for an 1800-lb car on a stretch of highway that has a 3.75° grade.

(Modeling) Design of Highway Curves When highway curves are designed, the outside of the curve is often slightly elevated or inclined above the inside of the curve. See the figure. This inclination is the superelevation. For safety reasons, it is important that both the curve’s radius and superelevation be correct for a given speed limit. If an automobile is traveling at velocity V (in feet per second), the safe radius R, in feet, for a curve with superelevation u is modeled by the formula

R = V2

g1 ƒ + tan u2 ,where ƒ and g are constants. (Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.)

77. A roadway is being designed for automobiles traveling at 45 mph. If u = 3°, g = 32.2 , and ƒ = 0.14 , calculate R to the nearest foot. (Hint: 45 mph = 66 ft per sec)

78. Determine the radius of the curve, to the nearest foot, if the speed in Exercise 77 is increased to 70 mph.

79. How would increasing angle u affect the results? Verify your answer by repeating Exercises 77 and 78 with u = 4°.

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692.3 Approximations of Trigonometric Function Values

80. Refer to Exercise 77 and use the same values for ƒ and g. A highway curve has radius R = 1150 ft and a superelevation of u = 2.1°. What should the speed limit (in miles per hour) be for this curve?

(Modeling) Speed of Light When a light ray travels from one medium, such as air, to another medium, such as water or glass, the speed of the light changes, and the light ray is bent, or refracted, at the boundary between the two media. (This is why objects under water appear to be in a different position from where they really are.) It can be shown in physics that these changes are related by Snell’s law

c1c2

=sin u1sin u2

where c1 is the speed of light in the first medium, c2 is the speed of light in the second medium, and u1 and u2 are the angles shown in the figure. In Exercises 81 and 82, assume that c1 = 3 * 108 m per sec.

81. Find the speed of light in the second medium for each of the following.

(a) u1 = 46° , u2 = 31° (b) u1 = 39° , u2 = 28°

82. Find u2 for each of the following values of u1 and c2 . Round to the nearest degree.

(a) u1 = 40° , c2 = 1.5 * 108 m per sec (b) u1 = 62° , c2 = 2.6 * 108 m per sec

Medium 2

Medium 1 If this medium isless dense, lighttravels at a greaterspeed, c1.

If this medium ismore dense, lighttravels at a lesserspeed, c2.

(Modeling) Braking Distance If aerodynamic resistance is ignored, the braking dis-tance D (in feet) for an automobile to change its velocity from V1 to V2 (feet per second) can be modeled using the following equation.

D =1.051V1 2 - V2 22

64.41K1 + K2 + sin u2K1 is a constant determined by the efficiency of the brakes and tires, K2 is a constant determined by the rolling resistance of the automobile, and u is the grade of the highway. (Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.)

85. Compute the number of feet, to the nearest unit, required to slow a car from 55 mph to 30 mph while traveling uphill with a grade of u = 3.5°. Let K1 = 0.4 and K2 = 0.02. (Hint: Change miles per hour to feet per second.)

(Modeling) Fish’s View of the World The figure shows a fish’s view of the world above the surface of the water. (Source: Walker, J., “The Amateur Scientist,” Scientific American.) Suppose that a light ray comes from the horizon, enters the water, and strikes the fish’s eye.

83. Assume that this ray gives a value of 90° for angle u1 in the formula for Snell’s law. (In a practical situation, this angle would probably be a little less than 90°.) The speed of light in water is about 2.254 * 108 m per sec. Find angle u2 to the nearest tenth.

84. Refer to Exercise 83. Suppose an object is located at a true angle of 29.6° above the horizon. Find the apparent angle above the horizon to a fish.

Rayfrom zenith

Apparenthorizon

Window’scenter

Rayfrom

horizon

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70 CHAPTER 2 Acute Angles and Right Triangles

86. Repeat Exercise 85 with u = -2°.

87. How is braking distance affected by grade u?

88. An automobile is traveling at 90 mph on a highway with a downhill grade of u = -3.5°. The driver sees a stalled truck in the road 200 ft away and immediately applies the brakes. Assuming that a collision cannot be avoided, how fast (in miles per hour, to the nearest unit) is the car traveling when it hits the truck? (Use the same values for K1 and K2 as in Exercise 85.)

(Modeling) Measuring Speed by Radar Any offset between a stationary radar gun and a moving target creates a “cosine effect” that reduces the radar read-ing by the cosine of the angle between the gun and the vehicle. That is, the radar speed reading is the product of the actual speed and the cosine of the angle. (Source: Fischetti, M., “Working Knowledge,” Scientific American.)

89. Find the radar readings, to the nearest unit, for Auto A and Auto B shown in the figure.

Auto A

Auto B

Radar gun

10° angleActual speed: 70 mph

20° angleActual speed: 70 mph

POLICE

90. The speed reported by a radar gun is reduced by the cosine of angle u, shown in the figure, where r represents reduced speed and a represents actual speed. Use the figure to show why this “cosine effect” occurs.

Auto

Radar gun

(Modeling) Length of a Sag Curve When a highway goes downhill and then uphill, it has a sag curve. Sag curves are designed so that at night, headlights shine sufficiently far down the road to allow a safe stopping distance. See the figure. S and L are in feet.

u1u2

The minimum length L of a sag curve is determined by the height h of the car’s head-lights above the pavement, the downhill grade u1 6 0°, the uphill grade u2 7 0°, and the safe stopping distance S for a given speed limit. In addition, L is dependent on the vertical alignment of the headlights. Headlights are usually pointed upward at a slight angle a above the horizontal of the car. Using these quantities, for a 55 mph speed limit, L can be modeled by the formula

L =1u2 - u12S2

2001h + S tan a2 ,where S 6 L. (Source: Mannering, F. and W. Kilareski, Principles of Highway Engineer-ing and Traffic Analysis, Second Edition, John Wiley and Sons.)

91. Compute length L, to the nearest foot, if h = 1.9 ft, a = 0.9°, u1 = -3°, u2 = 4°, and S = 336 ft.

92. Repeat Exercise 91 with a = 1.5°.

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71CHAPTER 2 Quiz

Solve each problem.

1. Find exact values of the six trigonometric functions for angle A in the right triangle.

2440

2. Complete the table with exact trigonometric function values.

U sin U cos U tan U cot U sec U csc U

30° 45° 60°

3. Find the exact value of each variable in the figure.

36 z

x y

w45°30°

Chapter 2 Quiz (Sections 2.1–2.3)

4. Area of a Solar Cell A solar cell converts the energy of sunlight directly into electrical energy. The amount of energy a cell produces depends on its area. Suppose a solar cell is hexagonal, as shown in the first figure on the right. Express its area ( in terms of sin u and any side x. (Hint: Consider one of the six equilateral triangles from the hexagon. See the second figure on the right.) (Source: Kastner, B., Space Mathematics, NASA.)

xh x

Find exact values of the six trigonometric functions for each angle. Rationalize denomi-nators when applicable.

5. 135° 6. -150° 7. 1020°

Find all values of u, if u is in the interval 30°, 360°2 and has the given function value. 8. sin u = 23

2 9. sec u = -22

Use a calculator to approximate the value of each expression. Give answers to six deci-mal places.

10. sin 42° 18′ 11. sec1-212° 12′2Find a value of u in the interval 30°, 90°2 that satisfies each statement. Write each answer in decimal degrees to six decimal places.

12. tan u = 2.6743210 13. csc u = 2.3861147

Determine whether each statement is true or false.

14. sin160° + 30°2 = sin 60° + sin 30° 15. tan190° - 35°2 = cot 35°

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72 CHAPTER 2 Acute Angles and Right Triangles

2.4 Solutions and Applications of Right Triangles

Historical Background The beginnings of trigonometry can be traced back to antiquity. Figure 19 shows the Babylonian tablet Plimpton 322, which provides a table of secant values. The Greek mathematicians Hipparchus and Claudius Ptolemy developed a table of chords, which gives values of sines of angles between 0° and 90° in increments of 15 minutes. Until the advent of scien-tific calculators in the late 20th century, tables were used to find function values that we now obtain with the stroke of a key.

Applications of spherical trigonometry accompanied the study of astronomy for these ancient civilizations. Until the mid-20th century, spherical trigonometry was studied in undergraduate courses. See Figure 20.

An introduction to applications of the plane trigonometry studied in this text involves applying the ratios to sides of objects that take the shape of right triangles.

■ Historical Background■ Significant Digits■ Solving Triangles■ Angles of Elevation or

Depression

Figure 21

15 ft

18 ft

Significant Digits A number that represents the result of counting, or a number that results from theoretical work and is not the result of measurement, is an exact number. There are 50 states in the United States. In this statement, 50 is an exact number.

Most values obtained for trigonometric applications are measured values that are not exact. Suppose we quickly measure a room as 15 ft by 18 ft. See Figure 21. To calculate the length of a diagonal of the room, we can use the Pythagorean theorem.

d2 = 152 + 182 Pythagorean theorem d2 = 549 Apply the exponents and add. d = 2549 Square root property;

Choose the positive root. d ≈ 23.430749

Should this answer be given as the length of the diagonal of the room? Of course not. The number 23.430749 con-tains six decimal places, while the original data of 15 ft and 18 ft are accurate only to the nearest foot. In practice, the results of a calculation can be no more accurate than the least accurate number in the calculation. Thus, we should indicate that the diagonal of the 15-by-18-ft room is approximately 23 ft.

Figure 20 Plimpton 322

Figure 19

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732.4 Solutions and Applications of Right Triangles

If a wall measured to the nearest foot is 18 ft long, this actually means that the wall has length between 17.5 ft and 18.5 ft. If the wall is measured more accu-rately as 18.3 ft long, then its length is really between 18.25 ft and 18.35 ft. The results of physical measurement are only approximately accurate and depend on the precision of the measuring instrument as well as the aptness of the observer. The digits obtained by actual measurement are significant digits. The measure-ment 18 ft is said to have two significant digits; 18.3 ft has three significant digits.

In the following numbers, the significant digits are identified in color.

408 21.5 18.00 6.700 0.0025 0.09810 7300

Notice the following.

18.00 has four significant digits. The zeros in this number represent mea-sured digits accurate to the nearest hundredth.

The number 0.0025 has only two significant digits, 2 and 5, because the zeros here are used only to locate the decimal point.

The number 7300 causes some confusion because it is impossible to determine whether the zeros are measured values. The number 7300 may have two, three, or four significant digits. When presented with this situation, we assume that the zeros are not significant, unless the context of the problem indicates otherwise.

To determine the number of significant digits for answers in applications of angle measure, use the following table.

To perform calculations with measured numbers, start by identifying the number with the least number of significant digits. Round the final answer to the same number of significant digits as this number. Remember that the answer is no more accurate than the least accurate number in the calculation.

Figure 22

When we are solving triangles,a labeled sketch is an importantaid.

Significant Digits for Angles

Angle Measure to Nearest Examples

Write Answer to This Number of

Significant Digits

Degree 62°, 36° twoTen minutes, or nearest tenth of a degree 52° 30′, 60.4° threeMinute, or nearest hundredth of a degree 81° 48′, 71.25° fourTen seconds, or nearest thousandth of a degree 10° 52′ 20″, 21.264° five

Solving Triangles To solve a triangle means to find the measures of all the angles and sides of the triangle. As shown in Figure 22, we use a to represent the length of the side opposite angle A, b for the length of the side opposite angle B, and so on. In a right triangle, the letter c is reserved for the hypotenuse.

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74 CHAPTER 2 Acute Angles and Right Triangles

LOOKING AHEAD TO CALCULUSThe derivatives of the parametric equations x = ƒ1t2 and y = g1t2 often represent the rate of change of physical

quantities, such as velocities. When x

and y are related by an equation, the

derivatives are related rates because a change in one causes a related change

in the other. Determining these rates

in calculus often requires solving a

right triangle.

NOTE In Example 1, we could have found the measure of angle B first and then used the trigonometric function values of B to find the lengths of the unknown sides. A right triangle can usually be solved in several ways, each producing the correct answer.

To maintain accuracy, always use given information as much as pos-sible, and avoid rounding in intermediate steps.

Figure 23

34° 30!

c = 12.7 in.

A C

EXAMPLE 1 Solving a Right Triangle Given an Angle and a Side

Solve right triangle ABC, if A = 34° 30′ and c = 12.7 in.SOLUTION To solve the triangle, find the measures of the remaining sides and angles. See Figure 23. To find the value of a, use a trigonometric function involving the known values of angle A and side c. Because the sine of angle A is given by the quotient of the side opposite A and the hypotenuse, use sin A.

sin A = ac

sin A = side oppositehypotenuse

sin 34° 30′ = a12.7

A = 34° 30′, c = 12.7

a = 12.7 sin 34° 30′ Multiply by 12.7 and rewrite. a = 12.7 sin 34.5° Convert to decimal degrees. a ≈ 12.710.566406242 Use a calculator. a ≈ 7.19 in. Three significant digits

Assuming that 34° 30′ is given to the nearest ten minutes, we rounded the answer to three significant digits.

To find the value of b, we could substitute the value of a just calculated and the given value of c in the Pythagorean theorem. It is better, however, to use the information given in the problem rather than a result just calculated. If an error is made in finding a, then b also would be incorrect. And, rounding more than once may cause the result to be less accurate. To find b, use cos A.

cos A = bc

cos A = side adjacenthypotenuse

cos 34° 30′ = b12.7

A = 34° 30′, c = 12.7

b = 12.7 cos 34° 30′ Multiply by 12.7 and rewrite. b ≈ 10.5 in. Three significant digits

Once b is found, the Pythagorean theorem can be used to verify the results.All that remains to solve triangle ABC is to find the measure of angle B.

A + B = 90° A and B are complementary angles. 34° 30′ + B = 90° A = 34° 30′

B = 89° 60′ - 34° 30′ Rewrite 90°. Subtract 34° 30′. B = 55° 30′ Subtract degrees and minutes separately.

■✔ Now Try Exercise 25.

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752.4 Solutions and Applications of Right Triangles

Figure 24

c = 53.58 cm

a = 29.43 cm

EXAMPLE 2 Solving a Right Triangle Given Two Sides

Solve right triangle ABC, if a = 29.43 cm and c = 53.58 cm.SOLUTION We draw a sketch showing the given information, as in Figure 24. One way to begin is to find angle A using the sine function.

sin A = ac

sin A = side oppositehypotenuse

sin A = 29.4353.58

a = 29.43, c = 53.58

sin A ≈ 0.5492721165 Use a calculator.

A ≈ sin-110.54927211652 Use the inverse sine function. A ≈ 33.32° Four significant digits

A ≈ 33° 19′ 33.32° = 33° + 0.32160′2The measure of B is approximately

90° - 33° 19′ = 56° 41′. 90° = 89° 60′

We now find b from the Pythagorean theorem.

a2 + b2 = c2 Pythagorean theorem

29.432 + b2 = 53.582 a = 29.43, c = 53.58 b2 = 53.582 - 29.432 Subtract 29.432.

b = 22004.6915 Simplify on the right; square root property b ≈ 44.77 cm

■✔ Now Try Exercise 35.Choose the

positive square root.

CAUTION Be careful when interpreting the angle of depression. Both the angle of elevation and the angle of depression are measured between the line of sight and a horizontal line.

Angles of Elevation or Depression In applications of right triangles, the angle of elevation from point X to point Y (above X) is the acute angle formed by ray XY and a horizontal ray with endpoint at X. See Figure 25(a). The angle of depression from point X to point Y (below X) is the acute angle formed by ray XY and a horizontal ray with endpoint X. See Figure 25(b).

(b)

XAngle of

depression

Horizontal

Angle ofelevation

HorizontalX

(a)

Figure 25

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76 CHAPTER 2 Acute Angles and Right Triangles

EXAMPLE 3 Finding a Length Given the Angle of Elevation

At a point A, 123 ft from the base of a flagpole, the angle of elevation to the top of the flagpole is 26° 40′. Find the height of the flagpole.

SOLUTION

Step 1 See Figure 26. The length of the side adjacent to A is known, and the length of the side opposite A must be found. We will call it a.

Step 2 The tangent ratio involves the given values. Write an equation.

tan A =side oppositeside adjacent

Tangent ratio

tan 26° 40′ = a123

A = 26° 40′; side adjacent = 123

Step 3 a = 123 tan 26° 40′ Multiply by 123 and rewrite. a ≈ 12310.502218882 Use a calculator. a ≈ 61.8 ft Three significant digits

The height of the flagpole is 61.8 ft. ■✔ Now Try Exercise 53.

123 ft

26° 40!A

Figure 26

EXAMPLE 4 Finding an Angle of Depression

From the top of a 210-ft cliff, David observes a lighthouse that is 430 ft off-shore. Find the angle of depression from the top of the cliff to the base of the lighthouse.

SOLUTION As shown in Figure 27, the angle of depression is measured from a horizontal line down to the base of the lighthouse. The angle of depression and angle B, in the right triangle shown, are alternate interior angles whose measures are equal. We use the tangent ratio to solve for angle B.

tan B = 210430

Tangent ratio

B = tan-1a 210430b Use the inverse tangent function.

B ≈ 26° Two significant digits■✔ Now Try Exercise 55.

Figure 27

210 ft

430 ft

C B

Angle ofdepression

Solving an Applied Trigonometry Problem

Step 1 Draw a sketch, and label it with the given information. Label the quantity to be found with a variable.

Step 2 Use the sketch to write an equation relating the given quantities to the variable.

Step 3 Solve the equation, and check that the answer makes sense.

To solve applied trigonometry problems, follow the same procedure as solv-ing a triangle. Drawing a sketch and labeling it correctly in Step 1 is crucial.

George Polya (1887–1985)

Polya, a native of Budapest, Hungary, wrote more than 250 papers and a number of books. He proposed a general outline for solving applied problems in his classic book How to Solve It.

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772.4 Solutions and Applications of Right Triangles

Solve each right triangle. When two sides are given, give angles in degrees and minutes. See Examples 1 and 2.

13.

964 ma

36° 20!

14.

89.6 cm

ZX47.8°

CONCEPT PREVIEW Match each equation in Column I with the appropriate right tri-angle in Column II. In each case, the goal is to find the value of x.

2.4 Exercises

1. x = 5 cot 38°

2. x = 5 cos 38°

3. x = 5 tan 38°

4. x = 5 csc 38°

5. x = 5 sin 38°

6. x = 5 sec 38°

38°

38°x

C. 5

38°

D. 5

x38°

38°

F. x

38°

Concept Check Refer to the discussion of accuracy and significant digits in this section to answer the following.

7. Lake Ponchartrain Causeway The world’s longest bridge over a body of water (continu-ous) is the causeway that joins the north and south shores of Lake Ponchartrain, a salt-water lake that lies north of New Orleans, Louisiana. It consists of two parallel spans. The longer of the spans measures 23.83 mi. State the range represented by this number. (Source: www.worldheritage.org)

8. Mt. Everest When Mt. Everest was first surveyed, the surveyors obtained a height of 29,000 ft to the nearest foot. State the range represented by this number. (The survey-ors thought no one would believe a measurement of 29,000 ft, so they reported it as 29,002.) (Source: Dunham, W., The Mathematical Universe, John Wiley and Sons.)

9. Vehicular Tunnel The E. Johnson Memorial Tunnel in Colorado, which measures 8959 ft, is one of the longest land vehicular tunnels in the United States. What is the range of this number? (Source: World Almanac and Book of Facts.)

10. WNBA Scorer Women’s National Basketball Association player Maya Moore of the Minnesota Lynx received the 2014 award for the most points scored, 812. Is it appropriate to consider this number between 811.5 and 812.5? Why or why not? (Source: www.wnba.com)

11. If h is the actual height of a building and the height is measured as 58.6 ft, then *h - 58.6 * … .

12. If w is the actual weight of a car and the weight is measured as 1542 lb, then *w - 1542 * … .

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78 CHAPTER 2 Acute Angles and Right Triangles

15.

P 124 m

51.2°

16.

35.9 kmc

31° 40!

17. A

CB42.0892°

56.851 cmc

18.

3579.42 m

68.5142°

19. B c A

16.2 ft12.5 ft

20.

15.3 m

Ab C

4.80 m

Solve each right triangle. In each case, C = 90°. If angle information is given in degrees and minutes, give answers in the same way. If angle information is given in decimal degrees, do likewise in answers. When two sides are given, give angles in degrees and minutes. See Examples 1 and 2.

25. A = 28.0°, c = 17.4 ft 26. B = 46.0°, c = 29.7 m

27. B = 73.0°, b = 128 in. 28. A = 62.5°, a = 12.7 m

29. A = 61.0°, b = 39.2 cm 30. B = 51.7°, a = 28.1 ft

31. a = 13 m, c = 22 m 32. b = 32 ft, c = 51 ft

33. a = 76.4 yd, b = 39.3 yd 34. a = 958 m, b = 489 m

35. a = 18.9 cm, c = 46.3 cm 36. b = 219 m, c = 647 m

37. A = 53° 24′, c = 387.1 ft 38. A = 13° 47′, c = 1285 m

39. B = 39° 09′, c = 0.6231 m 40. B = 82° 51′, c = 4.825 cm

Concept Check Answer each question.

41. What is the meaning of the term angle of elevation?

42. Can an angle of elevation be more than 90°?

Concept Check Answer each question.

21. Can a right triangle be solved if we are given measures of its two acute angles and no side lengths? Why or why not?

22. If we are given an acute angle and a side in a right triangle, what unknown part of the triangle requires the least work to find?

23. Why can we always solve a right triangle if we know the measures of one side and one acute angle?

24. Why can we always solve a right triangle if we know the lengths of two sides?

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792.4 Solutions and Applications of Right Triangles

43. Why does the angle of depression DAB in the figure have the same measure as the angle of elevation ABC?

44. Why is angle CAB not an angle of depression in the figure for Exercise 43? C B

A D

AD is parallel to BC.

Solve each problem. See Examples 1– 4.

45. Height of a Ladder on a Wall A 13.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the top of the fire truck) if the ladder makes an angle of 43° 50′ with the horizontal.

46. Distance across a Lake To find the distance RS across a lake, a surveyor lays off length RT = 53.1 m, so that angle T = 32° 10′ and angle S = 57° 50′. Find length RS.

47. Height of a Building From a window 30.0 ft above the street, the angle of elevation to the top of the building across the street is 50.0° and the angle of depression to the base of this building is 20.0°. Find the height of the building across the street.

48. Diameter of the Sun To determine the diameter of the sun, an astronomer might sight with a transit (a device used by surveyors for measuring angles) first to one edge of the sun and then to the other, estimating that the included angle equals 32′. Assuming that the distance d from Earth to the sun is 92,919,800 mi, approximate the diameter of the sun.

Sun Earthd

NOT TOSCALE

49. Side Lengths of a Triangle The length of the base of an isosceles triangle is 42.36 in. Each base angle is 38.12°. Find the length of each of the two equal sides of the triangle. (Hint: Divide the triangle into two right triangles.)

50. Altitude of a Triangle Find the altitude of an isosceles triangle having base 184.2 cm if the angle opposite the base is 68° 44′.

13.5 m

43° 50!

Lake

30.0 ft20.0°

50.0°

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80 CHAPTER 2 Acute Angles and Right Triangles

Solve each problem. See Examples 3 and 4.

51. Height of a Tower The shadow of a vertical tower is 40.6 m long when the angle of elevation of the sun is 34.6°. Find the height of the tower.

52. Distance from the Ground to the Top of a Building The angle of depression from the top of a building to a point on the ground is 32° 30′. How far is the point on the ground from the top of the building if the building is 252 m high?

53. Length of a Shadow Suppose that the angle of elevation of the sun is 23.4°. Find the length of the shadow cast by a person who is 5.75 ft tall.

23.4°5.75 ft

54. Airplane Distance An airplane is flying 10,500 ft above level ground. The angle of depression from the plane to the base of a tree is 13° 50′. How far horizontally must the plane fly to be directly over the tree?

10,500 ft

13° 50!

55. Angle of Depression of a Light A company safety committee has recommended that a floodlight be mounted in a parking lot so as to illuminate the employee exit, as shown in the figure. Find the angle of depression of the light to the nearest minute.

39.82 ft

Employeeexit

51.74 ft

56. Height of a Building The angle of elevation from the top of a small building to the top of a nearby taller building is 46° 40′, and the angle of depression to the bottom is 14° 10′. If the shorter building is 28.0 m high, find the height of the taller building.

28.0 m

46° 40′

14° 10′

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812.4 Solutions and Applications of Right Triangles

57. Angle of Elevation of the Sun The length of the shadow of a building 34.09 m tall is 37.62 m. Find the angle of elevation of the sun to the nearest hundredth of a degree.

58. Angle of Elevation of the Sun The length of the shadow of a flagpole 55.20 ft tall is 27.65 ft. Find the angle of elevation of the sun to the nearest hundredth of a degree.

59. Angle of Elevation of the Pyramid of the Sun The Pyramid of the Sun is in the ancient Mexican city of Teotihuacan. The base is a square with sides about 700 ft long. The height of the pyramid is about 200 ft. Find the angle of elevation u of the edge indicated in the figure to two significant digits. (Hint: The base of the triangle in the figure is half the diagonal of the square base of the pyramid.) (Source: www.britannica.com)

60. Cloud Ceiling The U.S. Weather Bureau defines a cloud ceiling as the altitude of the lowest clouds that cover more than half the sky. To determine a cloud ceiling, a powerful searchlight projects a circle of light vertically on the bottom of the cloud. An observer sights the circle of light in the crosshairs of a tube called a clinometer. A pendant hanging vertically from the tube and resting on a protractor gives the angle of elevation. Find the cloud ceiling if the searchlight is located 1000 ft from the observer and the angle of elevation is 30.0° as measured with a clinometer at eye-height 6 ft. (Assume three significant digits.)

30.0°6 ft1000 ft

Cloud

ObserverSearchlight

61. Height of Mt. Everest The highest mountain peak in the world is Mt. Everest, located in the Himalayas. The height of this enormous mountain was determined in 1856 by surveyors using trigonometry long before it was first climbed in 1953. This difficult measurement had to be done from a great distance. At an altitude of 14,545 ft on a different mountain, the straight-line distance to the peak of Mt. Everest is 27.0134 mi and its angle of elevation is u = 5.82°. (Source: Dunham, W., The Mathematical Universe, John Wiley and Sons.)

14,545 ft

27.013

4 mi

(a) Approximate the height (in feet) of Mt. Everest.

(b) In the actual measurement, Mt. Everest was over 100 mi away and the curvature of Earth had to be taken into account. Would the curvature of Earth make the peak appear taller or shorter than it actually is?

62. Error in Measurement A degree may seem like a very small unit, but an error of one degree in measuring an angle may be very significant. For example, suppose a laser beam directed toward the visible center of the moon misses its assigned target by 30.0″. How far is it (in miles) from its assigned target? Take the distance from the surface of Earth to that of the moon to be 234,000 mi. (Source: A Sourcebook of Applications of School Mathematics by Donald Bushaw et al.)

700 ft700 ft

200 ft

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82 CHAPTER 2 Acute Angles and Right Triangles

2.5 Further Applications of Right Triangles

Historical Background Johann Müller, known as Regiomontanus (see Figure 28), was a fifteenth-century German astronomer whose best known book is On Triangles of Every Kind. He used the recently-invented printing process of Gutenberg to promote his research. In his excellent book Trigonometric Delights *, Eli Maor writes:

Regiomontanus was the first publisher of mathematical and astronomi-cal books for commercial use. In 1474 he printed his Ephemerides, tables listing the position of the sun, moon, and planets for each day from 1475 to 1506. This work brought him great acclaim; Christopher Columbus had a copy of it on his fourth voyage to the New World and used it to predict the famous lunar eclipse of February 29, 1504. The hostile natives had for some time refused Columbus’s men food and water, and he warned them that God would punish them and take away the moon’s light. His admo-nition was at first ridiculed, but when at the appointed hour the eclipse began, the terrified natives immediately repented and fell into submission.

■ Historical Background■ Bearing■ Further Applications

Expressing Bearing (Method 1)

When a single angle is given, it is understood that bearing is measured in a clockwise direction from due north.

Several sample bearings using Method 1 are shown in Figure 29.

Bearings of 32°, 164°, 229°, and 304°

Figure 29

32°

164°

229°

304°

Regiomontanus

Figure 28

Bearing We now investigate navigation problems. Bearing refers to the direction of motion of an object, such as a ship or airplane, or the direction of a second object at a distance relative to the ship or airplane.

We introduce two methods of measuring bearing.

CAUTION A correctly labeled sketch is crucial when solving applica-tions like those that follow. Some of the necessary information is often not directly stated in the problem and can be determined only from the sketch.

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832.5 Further Applications of Right Triangles

EXAMPLE 1 Solving a Problem Involving Bearing (Method 1)

Radar stations A and B are on an east-west line, 3.7 km apart. Station A detects a plane at C, on a bearing of 61°. Station B simultaneously detects the same plane, on a bearing of 331°. Find the distance from A to C.

SOLUTION Begin with a sketch showing the given information. See Figure 30. A line drawn due north is perpendicular to an east-west line, so right angles are formed at A and B. Angles CBA and CAB can be found as follows.

∠CBA = 331° - 270° = 61° and ∠CAB = 90° - 61° = 29°A right triangle is formed. The distance from A to C, denoted b in the figure, can be found using the cosine function for angle CAB.

cos 29° = b3.7

Cosine ratio

b = 3.7 cos 29° Multiply by 3.7 and rewrite. b ≈ 3.2 km Two significant digits

■✔ Now Try Exercise 23.

Expressing Bearing (Method 2)

Start with a north-south line and use an acute angle to show the direction, either east or west, from this line.

Figure 31 shows several sample bearings using this method. Either N or S always comes first, followed by an acute angle, and then E or W.

Figure 31

42°

N 42° E S 31° E

31°

SS 40° W

40°

52°

N 52° W

Figure 30

61°

A29°

61°

331°3.7 km

EXAMPLE 2 Solving a Problem Involving Bearing (Method 2)

A ship leaves port and sails on a bearing of N 47° E for 3.5 hr. It then turns and sails on a bearing of S 43° E for 4.0 hr. If the ship’s rate is 22 knots (nautical miles per hour), find the distance that the ship is from port.

SOLUTION Draw and label a sketch as in Figure 32. Choose a point C on a bearing of N 47° E from port at point A. Then choose a point B on a bearing of S 43° E from point C. Because north-south lines are parallel, angle ACD measures 47° by alternate inte-rior angles. The measure of angle ACB is

47° + 43° = 90°,

making triangle ABC a right triangle. Figure 32

APort

43°47°

S S

47°

b = 77nautical mi

a = 88nautical mi

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84 CHAPTER 2 Acute Angles and Right Triangles

Use the formula relating distance, rate, and time to find the distances in Figure 32 from A to C and from C to B.

b = 22 * 3.5 = 77 nautical mi Distance = rate * time

a = 22 * 4.0 = 88 nautical mi

Now find c, the distance from port at point A to the ship at point B.

a2 + b2 = c2 Pythagorean theorem 882 + 772 = c2 a = 88, b = 77

c = 2882 + 772 If a2 + b2 = c2 and c 7 0, then c = 2a2 + b2.

c ≈ 120 nautical mi Two significant digits■✔ Now Try Exercise 29.

EXAMPLE 3 Using Trigonometry to Measure a Distance

The subtense bar method is a method that surveyors use to determine a small distance d between two points P and Q. The subtense bar with length b is cen-tered at Q and situated perpendicular to the line of sight between P and Q. See Figure 33. Angle u is measured, and then the distance d can be determined.

Figure 33

b/2

P d Qu

Further Applications

(a) Find d when u = 1° 23′ 12″ and b = 2.0000 cm.(b) How much change would there be in the value of d if u measured 1″ larger?

SOLUTION

(a) From Figure 33, we obtain the following.

cot u

2= d b

Cotangent ratio

d = b2

cot u

2 Multiply and rewrite.

Let b = 2 . To evaluate u2 , we change u to decimal degrees.

1° 23′ 12″ ≈ 1.386666667°

Then d = 22

cot 1.386666667°

2≈ 82.634110 cm.

(b) If u is 1″ larger, then u = 1° 23′ 13″ ≈ 1.386944444°.

d = 22

cot 1.386944444°

2≈ 82.617558 cm

The difference is 82.634110 - 82.617558 = 0.016552 cm.■✔ Now Try Exercise 41.

Use cot u = 1tan u to evaluate.

Figure 32 (repeated)

APort

43°47°

S S

47°

b = 77nautical mi

a = 88nautical mi

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852.5 Further Applications of Right Triangles

EXAMPLE 4 Solving a Problem Involving Angles of Elevation

Francisco needs to know the height of a tree. From a given point on the ground, he finds that the angle of elevation to the top of the tree is 36.7°. He then moves back 50 ft. From the second point, the angle of elevation to the top of the tree is 22.2°. See Figure 34. Find the height of the tree to the nearest foot.

Figure 34

22.2°

50 ft

36.7°A C

D x

ALGEBRAIC SOLUTION

Figure 34 shows two unknowns: x, the distance from the center of the trunk of the tree to the point where the first observation was made, and h, the height of the tree. See Figure 35 in the Graphing Calculator Solution. Because nothing is given about the length of the hypotenuse of either triangle ABC or triangle BCD, we use a ratio that does not involve the hypotenuse — namely, the tangent.

In triangle ABC, tan 36.7° = hx or h = x tan 36.7°.

In triangle BCD, tan 22.2° = h50 + x or h = 150 + x2 tan 22.2°.

Each expression equals h, so the expressions must be equal.

x tan 36.7° = 150 + x2 tan 22.2° Equate expressions for h.

x tan 36.7° = 50 tan 22.2° + x tan 22.2° Distributive property

x tan 36.7° - x tan 22.2° = 50 tan 22.2° Write the x-terms on one side.

x1tan 36.7° - tan 22.2°2 = 50 tan 22.2° Factor out x.

x = 50 tan 22.2°tan 36.7° - tan 22.2°

Divide by the coefficient of x.

We saw above that h = x tan 36.7°. Substitute for x.

h = a 50 tan 22.2°tan 36.7° - tan 22.2° b tan 36.7°

Use a calculator.

tan 36.7° = 0.74537703 and tan 22.2° = 0.40809244

Thus,

tan 36.7° - tan 22.2° = 0.74537703 - 0.40809244 = 0.33728459

and h = a 5010.4080924420.33728459

b 0.74537703 ≈ 45.To the nearest foot, the height of the tree is 45 ft.

GRAPHING CALCULATOR SOLUTION*

In Figure 35, we have superimposed Figure 34 on coordinate axes with the origin at D. By definition, the tangent of the angle between the x-axis and the graph of a line with equation y = mx + b is the slope of the line, m. For line DB, m = tan 22.2°. Because b equals 0, the equation of line DB is

y1 = 1tan 22.2°2x.The equation of line AB is

y2 = 1tan 36.7°2x + b.Because b ≠ 0 here, we use the point A150, 02 and the point-slope form to find the equation.

y2 - y0 = m1x - x02 Point-slope form y2 - 0 = m1x - 502 x0 = 50, y0 = 0

y2 = tan 36.7°1x - 502 Lines y1 and y2 are graphed in Figure 36. The y-coordinate of the point of intersec-tion of the graphs gives the length of BC, or h. Thus, h ≈ 45 .

Figure 35

22.2°

50 ft

36.7°A C

D xx

Figure 36

−10

−20

150

■✔ Now Try Exercise 31.

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86 CHAPTER 2 Acute Angles and Right Triangles

CONCEPT PREVIEW Match the measure of bearing in Column I with the appropriate graph in Column II.

2.5 Exercises

1. 20°

2. S 70° W

3. 160°

4. S 20° W

5. N 70° W

6. 340°

7. 110°

8. 270°

9. 180°

10. N 70° E

The two methods of expressing bearing can be interpreted using a rectangular coordi-nate system. Suppose that an observer for a radar station is located at the origin of a coordinate system. Find the bearing of an airplane located at each point. Express the bearing using both methods.

11. 1-4, 02 12. 15, 02 13. 10, 42 14. 10, -2215. 1-5, 52 16. 1-3, -32 17. 12, -22 18. 12, 22Solve each problem. See Examples 1 and 2.

19. Distance Flown by a Plane A plane flies 1.3 hr at 110 mph on a bearing of 38°. It then turns and flies 1.5 hr at the same speed on a bearing of 128°. How far is the plane from its starting point?

128°

38°

A. N

20°

B. N

20°EW

C. N

20°

D. N

20°EW

E. N

70°

F. N

20°EW

G. N

70°EW

H. N

70°EW

I. N

180°EW

J. N

90°EW

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872.5 Further Applications of Right Triangles

20. Distance Traveled by a Ship A ship trav-els 55 km on a bearing of 27° and then travels on a bearing of 117° for 140 km. Find the distance from the starting point to the ending point.

21. Distance between Two Ships Two ships leave a port at the same time. The first ship sails on a bearing of 40° at 18 knots (nautical miles per hour) and the second on a bearing of 130° at 26 knots. How far apart are they after 1.5 hr?

22. Distance between Two Ships Two ships leave a port at the same time. The first ship sails on a bearing of 52° at 17 knots and the second on a bearing of 322° at 22 knots. How far apart are they after 2.5 hr?

23. Distance between Two Docks Two docks are located on an east-west line 2587 ft apart. From dock A, the bearing of a coral reef is 58° 22′. From dock B, the bearing of the coral reef is 328° 22′. Find the distance from dock A to the coral reef.

24. Distance between Two Lighthouses Two lighthouses are located on a north-south line. From lighthouse A, the bearing of a ship 3742 m away is 129° 43′. From light-house B, the bearing of the ship is 39° 43′. Find the distance between the lighthouses.

25. Distance between Two Ships A ship leaves its home port and sails on a bearing of S 61° 50′ E . Another ship leaves the same port at the same time and sails on a bearing of N 28° 10′ E . If the first ship sails at 24.0 mph and the second sails at 28.0 mph, find the distance between the two ships after 4 hr.

26. Distance between Transmitters Radio direction finders are set up at two points A and B, which are 2.50 mi apart on an east-west line. From A, it is found that the bearing of a signal from a radio transmitter is N 36° 20′ E , and from B the bear-ing of the same signal is N 53° 40′ W. Find the distance of the transmitter from B.

27. Flying Distance The bearing from A to C is S 52° E. The bearing from A to B is N 84° E. The bearing from B to C is S 38° W. A plane flying at 250 mph takes 2.4 hr to go from A to B. Find the distance from A to C.

28. Flying Distance The bearing from A to C is N 64° W. The bearing from A to B is S 82° W. The bearing from B to C is N 26° E. A plane flying at 350 mph takes 1.8 hr to go from A to B. Find the distance from B to C.

29. Distance between Two Cities The bearing from Winston-Salem, North Carolina, to Danville, Virginia, is N 42° E. The bearing from Danville to Goldsboro, North Carolina, is S 48° E. A car traveling at 65 mph takes 1.1 hr to go from Winston-Salem to Danville and 1.8 hr to go from Danville to Goldsboro. Find the distance from Winston-Salem to Goldsboro.

30. Distance between Two Cities The bearing from Atlanta to Macon is S 27° E, and the bearing from Macon to Augusta is N 63° E. An automobile traveling at 62 mph needs 1 14 hr to go from Atlanta to Macon and 1

34 hr to go from Macon to Augusta.

Find the distance from Atlanta to Augusta.

117°

27°

55 km 140 km

28° 10!

61° 50!

36° 20′

53° 40′

2.50 mi

Transmitter

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88 CHAPTER 2 Acute Angles and Right Triangles

Solve each problem. See Examples 3 and 4.

31. Height of a Pyramid The angle of elevation from a point on the ground to the top of a pyramid is 35° 30′. The angle of elevation from a point 135 ft farther back to the top of the pyramid is 21° 10′. Find the height of the pyramid.

21° 10′35° 30′

135 ft

32. Distance between a Whale and a Lighthouse A whale researcher is watching a whale approach directly toward a lighthouse as she observes from the top of this lighthouse. When she first begins watching the whale, the angle of depression to the whale is 15° 50′. Just as the whale turns away from the lighthouse, the angle of depression is 35° 40′. If the height of the lighthouse is 68.7 m, find the distance traveled by the whale as it approached the lighthouse.

68.7 m

15° 50′ 35° 40′

33. Height of an Antenna A scanner antenna is on top of the center of a house. The angle of elevation from a point 28.0 m from the center of the house to the top of the antenna is 27° 10′, and the angle of elevation to the bottom of the antenna is 18° 10′. Find the height of the antenna.

34. Height of Mt. Whitney The angle of elevation from Lone Pine to the top of Mt. Whitney is 10° 50′. A hiker, traveling 7.00 km from Lone Pine along a straight, level road toward Mt. Whitney, finds the angle of elevation to be 22° 40′. Find the height of the top of Mt. Whitney above the level of the road.

35. Find h as indicated in the figure. 36. Find h as indicated in the figure.

168 m

41.2°

52.5°

37. Distance of a Plant from a Fence In one area, the lowest angle of elevation of the sun in winter is 23° 20′. Find the minimum dis-tance x that a plant needing full sun can be placed from a fence 4.65 ft high.

38. Distance through a Tunnel A tunnel is to be built from A to B. Both A and B are visible from C. If AC is 1.4923 mi and BC is 1.0837 mi, and if C is 90°, find the measures of angles A and B.

392 ft29.5°

49.2°

23° 20!x

4.65 ft

Plant

A BTunnel

1.4923 mi 1.0837 mi

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892.5 Further Applications of Right Triangles

39. Height of a Plane above Earth Find the minimum height h above the surface of Earth so that a pilot at point A in the figure can see an object on the horizon at C, 125 mi away. Assume 4.00 * 103 mi as the radius of Earth.

40. Length of a Side of a Piece of Land A piece of land has the shape shown in the figure. Find the length x.

41. (Modeling) Distance between Two Points A variation of the subtense bar method that surveyors use to determine larger distances d between two points P and Q is shown in the figure. The subtense bar with length b is placed between points P and Q so that the bar is centered on and perpendicular to the line of sight between P and Q. Angles a and b are measured from points P and Q, respectively. (Source: Mueller, I. and K. Ramsayer, Introduction to Surveying, Frederick Ungar Publishing Co.)

b/2

P Qa b

(a) Find a formula for d involving a , b , and b.

(b) Use the formula from part (a) to determine d if a = 37′ 48″, b = 42′ 03″, and b = 2.000 cm.

42. (Modeling) Distance of a Shot Put A shot-putter trying to improve performance may wonder whether there is an optimal angle to aim for, or whether the velocity (speed) at which the ball is thrown is more important. The figure shows the path of a steel ball thrown by a shot-putter. The distance D depends on initial velocity v, height h, and angle u when the ball is released.

One model developed for this situation gives D as

D =v2 sin u cos u + v cos u 21v sin u22 + 64h

32 .

Typical ranges for the variables are v: 33–46 ft per sec; h: 6–8 ft; and u: 40°9 45°. (Source: Kreighbaum, E. and K. Barthels, Biomechanics, Allyn & Bacon.)

(a) To see how angle u affects distance D, let v = 44 ft per sec and h = 7 ft. Calcu-late D, to the nearest hundredth, for u = 40°, 42°, and 45°. How does distance D change as u increases?

(b) To see how velocity v affects distance D, let h = 7 and u = 42°. Calculate D, to the nearest hundredth, for v = 43, 44, and 45 ft per sec. How does distance D change as v increases?

125 miC A

NOT TOSCALE

198.4 m

52° 20! 30° 50!

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90 CHAPTER 2 Acute Angles and Right Triangles

43. (Modeling) Highway Curves A basic highway curve connecting two straight sections of road may be circular. In the figure, the points P and S mark the beginning and end of the curve. Let Q be the point of intersection where the two straight sections of highway leading into the curve would meet if extended. The radius of the curve is R, and the cen-tral angle u denotes how many degrees the curve turns. (Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.)

(a) If R = 965 ft and u = 37°, find the distance d between P and Q.(b) Find an expression in terms of R and u for the distance between points M and N.

44. (Modeling) Stopping Distance on a Curve Refer to Exercise 43. When an automobile travels along a cir-cular curve, objects like trees and buildings situated on the inside of the curve can obstruct the driver’s vision. These obstructions prevent the driver from seeing sufficiently far down the highway to ensure a safe

stopping distance. In the figure, the minimum distance d that should be cleared on the inside of the highway is modeled by the equation

d = R a1 - cos u2b .

(Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.)

(a) It can be shown that if u is measured in degrees, then u ≈ 57.3SR , where S is the safe stopping distance for the given speed limit. Compute d to the nearest foot for a 55 mph speed limit if S = 336 ft and R = 600 ft.

(b) Compute d to the nearest foot for a 65 mph speed limit given S = 485 ft and R = 600 ft.

The figure to the right indicates that the equation of a line passing through the point 1a, 02 and making an angle u with the x-axis is

y = 1tan u21x - a2.45. Find an equation of the line passing through the

point 125 , 02 that makes an angle of 35° with the x-axis.

46. Find an equation of the line passing through the point 15 , 02 that makes an angle of 15° with the x-axis.47. Show that a line bisecting the first and third quadrants satisfies the equation given in

the instructions.

48. Show that a line bisecting the second and fourth quadrants satisfies the equation given in the instructions.

49. The ray y = x, x Ú 0, contains the origin and all points in the coordinate system whose bearing is 45°. Determine an equation of a ray consisting of the origin and all points whose bearing is 240°.

50. Repeat Exercise 49 for a bearing of 150°.

P SQ

2u 2u

R R

NOT TO SCALE

0 (a, 0)x

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Key Terms

2.1 side oppositeside adjacent cofunctions

2.2 reference angle 2.4 exact number

significant digits

angle of elevation angle of depression

2.5 bearing

Quick ReviewConcepts Examples

A C

725

BHypotenuse Sideopposite

Side adjacent to A

sin A = 725

cos A = 2425

tan A = 724

csc A = 257

sec A = 2524

cot A = 247

Trigonometric Functions of Acute Angles

Right-Triangle-Based Definitions of Trigonometric FunctionsLet A represent any acute angle in standard position.

sin A = y r=

side oppositehypotenuse

csc A = r y=

hypotenuseside opposite

cos A = x r=

side adjacenthypotenuse

sec A = r x=

hypotenuseside adjacent

tan A = y x=

side oppositeside adjacent

cot A = x y=

side adjacentside opposite

Cofunction IdentitiesFor any acute angle A, cofunction values of complementary angles are equal.

sin A = cos 190° − A 2 cos A = sin 190° − A 2 sec A = csc 190° − A 2 csc A = sec 190° − A 2 tan A = cot 190° − A 2 cot A = tan 190° − A 2

Function Values of Special Angles

U sin U cos U tan U cot U sec U csc U

30° 12 232

233 23 2233 2

45°

222

1 1 22 2260°

232

12 23 233 2 2233

2.1

Chapter 2 Test Prep

CHAPTER 2 Test Prep

sin 55° = cos190° - 55°2 = cos 35°sec 48° = csc190° - 48°2 = csc 42°tan 72° = cot190° - 72°2 = cot 18°

60°

30°√3

√2

45°

30°9 60° right triangle 45°9 45° right triangle

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92 CHAPTER 2 Acute Angles and Right Triangles

Concepts Examples

Trigonometric Functions of Non-Acute Angles 2.2

Reference Angle U′ for U in 10°, 360° 2U in Quadrant I II III IV

U′ is u 180° - u u - 180° 360° - u

Finding Trigonometric Function Values for Any Nonquadrantal Angle U

Step 1 Add or subtract 360° as many times as needed to obtain an angle greater than 0° but less than 360°.

Step 2 Find the reference angle u′.

Step 3 Find the trigonometric function values for u′.

Step 4 Determine the correct signs for the values found in Step 3.

Quadrant I: For u = 25°, u′ = 25°Quadrant II: For u = 152°, u′ = 28°Quadrant III: For u = 200°, u′ = 20° Quadrant IV: For u = 320°, u′ = 40°

Find sin 1050°.

1050° - 21360°2 = 330° Coterminal angle in quadrant IVThe reference angle for 330° is u′ = 30°.

sin 1050°= -sin 30° Sine is negative in quadrant IV.

= - 12

sin 30° = 12

Approximations of Trigonometric Function Values2.3

Approximate each value.

cos 50° 15′ = cos 50.25° ≈ 0.63943900

csc 32.5° = 1sin 32.5°

≈ 1.86115900 csc u = 1sin u

Find an angle u in the interval 30° , 90°2 that satisfies each condition in color.

cos u ≈ 0.73677482 u ≈ cos-110.736774822 u ≈ 42.542600°

csc u ≈ 1.04766792

sin u ≈1

1.04766792 sin u = 1csc u

u ≈ sin-1 a 11.04766792

b u ≈ 72.65°

To approximate a trigonometric function value of an angle in degrees, make sure the calculator is in degree mode.

To find the corresponding angle measure given a trigono-metric function value, use an appropriate inverse function.

Solutions and Applications of Right Triangles2.4

Solving an Applied Trigonometry Problem

Step 1 Draw a sketch, and label it with the given informa-tion. Label the quantity to be found with a variable.

Find the angle of elevation of the sun if a 48.6-ft flag-pole casts a shadow 63.1 ft long.

Step 1 See the sketch. We must find u.

Shadow63.1 ft

Flagpole48.6 ft

Sun

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Concepts Examples

Step 2 Use the sketch to write an equation relating the given quantities to the variable.

Step 3 Solve the equation, and check that the answer makes sense.

Further Applications of Right Triangles

Expressing Bearing

Method 1 When a single angle is given, bearing is measured in a clockwise direction from due north.

Method 2 Start with a north-south line and use an acute angle to show direction, either east or west, from this line.

2.5

Step 2 tan u = 48.663.1

tan u ≈ 0.770206

Step 3 u = tan-1 0.770206 u ≈ 37.6°

The angle of elevation rounded to three significant digits is 37.6°, or 37° 40′.

Example: 220° Example: S 40° W

220°

40°

93CHAPTER 2 Review Exercises

Find exact values of the six trigonometric functions for each angle A.

1. A

1161

42 A

Chapter 2 Review Exercises

Find one solution for each equation. Assume that all angles involved are acute angles.

3. sin 4b = cos 5b 4. sec12u + 10°2 = csc14u + 20°2 5. tan15x + 11°2 = cot16x + 2°2 6. cos a3u

5+ 11°b = sin a 7u

10+ 40°b

Determine whether each statement is true or false. If false, tell why.

7. sin 46° 6 sin 58° 8. cos 47° 6 cos 58°

9. tan 60° Ú cot 40° 10. csc 22° … csc 68°

11. Explain why, in the figure, the cosine of angle A is equal to the sine of angle B.

C B

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94 CHAPTER 2 Acute Angles and Right Triangles

12. Which one of the following cannot be exactly determined using the methods of this chapter?

A. cos 135° B. cot1-45°2 C. sin 300° D. tan 140°Find exact values of the six trigonometric functions for each angle. Do not use a calcu-lator. Rationalize denominators when applicable.

13. 1020° 14. 120° 15. -1470° 16. -225°

Find all values of u, if u is in the interval 30°, 360°2 and u has the given function value.17. cos u = - 1

218. sin u = - 1

19. sec u = - 2233

20. cot u = -1

Use a calculator to approximate the value of each expression. Give answers to six deci-mal places.

25. sec 222° 30′ 26. sin 72° 30′ 27. csc 78° 21′

28. cot 305.6° 29. tan 11.7689° 30. sec 58.9041°

Use a calculator to find each value of u, where u is in the interval 30°, 90°2 . Give answers in decimal degrees to six decimal places.

31. sin u = 0.82584121 32. cot u = 1.1249386 33. cos u = 0.97540415

34. sec u = 1.2637891 35. tan u = 1.9633124 36. csc u = 9.5670466

Find two angles in the interval 30°, 360°2 that satisfy each of the following. Round answers to the nearest degree.

37. sin u = 0.73135370 38. tan u = 1.3763819

Determine whether each statement is true or false. If false, tell why. Use a calculator for Exercises 39 and 42.

39. sin 50° + sin 40° = sin 90° 40. 1 + tan2 60° = sec2 60°

41. sin 240° = 2 sin 120° # cos 120° 42. sin 42° + sin 42° = sin 84°43. A student wants to use a calculator to find the value of cot 25°. However, instead of

entering 1tan 25 , he enters tan-1 25. Assuming the calculator is in degree mode, will

this produce the correct answer? Explain.

44. Explain the process for using a calculator to find sec-1 10.

21. tan2 120° - 2 cot 240°

24. Find the sine, cosine, and tangent function values for each angle.

(a)

(–3, –3)

(b)

(1, –Ë3)

Evaluate each expression. Give exact values.

22. cos 60° + 2 sin2 30° 23. sec2 300° - 2 cos2 150°

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Solve each right triangle. In Exercise 46, give angles to the nearest minute. In Exer-cises 47 and 48, label the triangle ABC as in Exercises 45 and 46.

45.

a c = 748

58° 30!

46.

a = 129.7

b = 368.1

47. A = 39.72°, b = 38.97 m 48. B = 47° 53′, b = 298.6 m

Solve each problem. (Source for Exercises 49 and 50: Parker, M., Editor, She Does Math, Mathematical Association of America.)

49. Height of a Tree A civil engineer must determine the vertical height of the tree shown in the figure. The given angle was measured with a clinometer. Find the height of the leaning tree to the nearest whole number.

50 ft

70°

This is a picture of one type of clinometer, called an Abney hand level and clinometer. (Courtesy of Keuffel & Esser Co.)

50. (Modeling) Double Vision To correct mild double vision, a small amount of prism is added to a patient’s eyeglasses. The amount of light shift this causes is measured in prism diopters. A patient needs 12 prism diopters horizontally and 5 prism diop-ters vertically. A prism that corrects for both requirements should have length r and be set at angle u. Find the values of r and u in the figure.

51. Height of a Tower The angle of elevation from a point 93.2 ft from the base of a tower to the top of the tower is 38° 20′. Find the height of the tower.

93.2 ft

38° 20!

52. Height of a Tower The angle of depression from a television tower to a point on the ground 36.0 m from the bottom of the tower is 29.5°. Find the height of the tower.

29.5°

36.0 m

CHAPTER 2 Review Exercises

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96 CHAPTER 2 Acute Angles and Right Triangles

53. Length of a Diagonal One side of a rectangle measures 15.24 cm. The angle between the diagonal and that side is 35.65°. Find the length of the diagonal.

54. Length of Sides of an Isosceles Triangle An isosceles triangle has a base of length 49.28 m. The angle opposite the base is 58.746°. Find the length of each of the two equal sides.

55. Distance between Two Points The bearing of point B from point C is 254°. The bearing of point A from point C is 344°. The bearing of point A from point B is 32°. If the distance from A to C is 780 m, find the distance from A to B.

56. Distance a Ship Sails The bearing from point A to point B is S 55° E, and the bearing from point B to point C is N 35° E. If a ship sails from A to B, a distance of 81 km, and then from B to C, a distance of 74 km, how far is it from A to C?

57. Distance between Two Points Two cars leave an intersection at the same time. One heads due south at 55 mph. The other travels due west. After 2 hr, the bearing of the car headed west from the car headed south is 324°. How far apart are they at that time?

58. Find a formula for h in terms of k, A, and B. Assume A 6 B.

59. Create a right triangle problem whose solution is 3 tan 25°.

60. Create a right triangle problem whose solution can be found by evaluating u if sin u = 34 .

61. (Modeling) Height of a Satellite Artificial satellites that orbit Earth often use VHF signals to communicate with the ground. VHF signals travel in straight lines. The height h of the satellite above Earth and the time T that the satellite can communi-cate with a fixed location on the ground are related by the model

h = R ¢ 1cos 180TP - 1≤ ,where R = 3955 mi is the radius of Earth and P is the period for the satellite to orbit Earth. (Source: Schlosser, W., T. Schmidt-Kaler, and E. Milone, Challenges of Astronomy, Springer-Verlag.)

(a) Find h to the nearest mile when T = 25 min and P = 140 min. (Evaluate the cosine function in degree mode.)

(b) What is the value of h to the nearest mile if T is increased to 30 min?

62. (Modeling) Fundamental Surveying Problem The first fundamental problem of surveying is to determine the coordinates of a point Q given the coordinates of a point P, the distance between P and Q, and the bearing u from P to Q. See the figure. (Source: Mueller, I. and K. Ramsayer, Introduction to Surveying, Frederick Ungar Publishing Co.)

(a) Find a formula for the coordinates 1xQ , yQ2 of the point Q given u, the coordinates 1xP , yP2 of P, and the distance d between P and Q.(b) Use the formula found in part (a) to determine the coordinates 1xQ , yQ2 if 1xP , yP2 = 1123.62, 337.952, u = 17° 19′ 22″ , and d = 193.86 ft.

(xQ, yQ)

(xP, yP)P

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Chapter 2 Test

1. Find exact values of the six trigonometric functions for angle A in the right triangle.

513

2. Find the exact value of each variable in the figure.

z w

445° 30°

3. Find a solution for sin1u + 15°2 = cos12u + 30°2 . 4. Determine whether each statement is true or false. If false, tell why.

(a) sin 24° 6 sin 48° (b) cos 24° 6 cos 48°(c) cos160° + 30°2 = cos 60° # cos 30° - sin 60° # sin 30°

Solve each problem.

Find exact values of the six trigonometric functions for each angle. Rationalize denomi-nators when applicable.

5. 240° 6. -135° 7. 990°

Find all values of u, if u is in the interval 30°, 360°2 and has the given function value. 8. cos u = - 22

2 9. csc u = - 223

3 10. tan u = 1

15. Antenna Mast Guy Wire A guy wire 77.4 m long is attached to the top of an antenna mast that is 71.3 m high. Find the angle that the wire makes with the ground.

13. Find the value of u in the interval 30°, 90°4 in decimal degrees, ifsin u = 0.27843196.

Give the answer to six decimal places.

14. Solve the right triangle.

58° 30!

B a = 748 C

Solve each problem.

11. How would we find cot u using a calculator, if tan u = 1.6778490? Evaluate cot u.

12. Use a calculator to approximate the value of each expression. Give answers to six decimal places.

(a) sin 78° 21′ (b) tan 117.689° (c) sec 58.9041°

CHAPTER 2 Test

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98 CHAPTER 2 Acute Angles and Right Triangles

16. Height of a Flagpole To measure the height of a flagpole, Jan Marie found that the angle of elevation from a point 24.7 ft from the base to the top is 32° 10′. What is the height of the flagpole?

17. Altitude of a Mountain The highest point in Texas is Guadalupe Peak. The angle of depression from the top of this peak to a small miner’s cabin at an approximate elevation of 2000 ft is 26°. The cabin is located 14,000 ft horizontally from a point directly under the top of the mountain. Find the altitude of the top of the mountain to the nearest hundred feet.

18. Distance between Two Points Two ships leave a port at the same time. The first ship sails on a bearing of 32° at 16 knots (nautical miles per hour) and the second on a bearing of 122° at 24 knots. How far apart are they after 2.5 hr?

19. Distance of a Ship from a Pier A ship leaves a pier on a bearing of S 62° E and travels for 75 km. It then turns and continues on a bearing of N 28° E for 53 km. How far is the ship from the pier?

20. Find h as indicated in the figure.

168 m

41.2°

52.5°

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The speed of a planet revolving around its sun can be measured in linear and angular speed, both of which are discussed in this chapter covering radian measure of angles.

Radian Measure

Applications of Radian Measure

The Unit Circle and Circular Functions

Chapter 3 Quiz

Linear and Angular Speed

3.1

3.2

3.3

3.4

Radian Measure and the Unit Circle3

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100 CHAPTER 3 Radian Measure and the Unit Circle

3.1 Radian Measure

Radian Measure We have seen that angles can be measured in degrees. In more theoretical work in mathematics, radian measure of angles is preferred. Radian measure enables us to treat the trigonometric functions as functions with domains of real numbers, rather than angles.

Figure 1 shows an angle u in standard position, along with a circle of radius r. The vertex of u is at the center of the circle. Because angle u intercepts an arc on the circle equal in length to the radius of the circle, we say that angle u has a measure of 1 radian.

■ Radian Measure■ Conversions between

Degrees and Radians■ Trigonometric

Function Values of Angles in Radians

u = 1 radian

Figure 1

Radian

An angle with its vertex at the center of a circle that intercepts an arc on the circle equal in length to the radius of the circle has a measure of 1 radian.

It follows that an angle of measure 2 radians intercepts an arc equal in length to twice the radius of the circle, an angle of measure 12 radian intercepts an arc equal in length to half the radius of the circle, and so on. In general, if U is a central angle of a circle of radius r, and U intercepts an arc of length s, then the radian measure of U is sr . See Figure 2.

u = 2 radians

u = radian

C = 2Pr

u = 2p radians

Figure 2

The ratio sr is a pure number, where s and r are expressed in the same units. Thus, “radians” is not a unit of measure like feet or centimeters.

Conversions between Degrees and Radians The circumference of a circle—the distance around the circle—is given by C = 2pr, where r is the radius of the circle. The formula C = 2pr shows that the radius can be measured off 2p times around a circle. Therefore, an angle of 360°, which corresponds to a complete circle, intercepts an arc equal in length to 2p times the radius of the circle. Thus, an angle of 360° has a measure of 2p radians.

360° = 2P radians

An angle of 180° is half the size of an angle of 360°, so an angle of 180° has half the radian measure of an angle of 360°.

180° =12

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1013.1 Radian Measure

We can use the relationship 180° = p radians to develop a method for con-verting between degrees and radians as follows.

180° = P radians Degree/radian relationship

1° =P

180 radian Divide by 180. or 1 radian =

180°P

Divide by p.

Converting between Degrees and Radians

Multiply a degree measure by p

180 radian and simplify to convert to radians.

Multiply a radian measure by 180°p

and simplify to convert to degrees.

EXAMPLE 1 Converting Degrees to Radians

Convert each degree measure to radians.

(a) 45° (b) -270° (c) 249.8°

SOLUTION

(a) 45° = 45a p180

radianb = p4

radian Multiply by p180 radian.

(b) -270° = -270a p180

radianb = - 3p2

radians Multiply by p

180 radian. Write in lowest terms.

radianb ≈ 4.360 radians Nearest thousandth■✔ Now Try Exercises 11, 17, and 47.

This radian mode screen shows TI-84 Plus conversions for Example 1. Verify that the first two results are approximations for the exact values of p4 and -

3p2 .

EXAMPLE 2 Converting Radians to Degrees

Convert each radian measure to degrees.

(a) 9p4

(b) - 5p6

SOLUTION

(a) 9p4

radians = 9p4

a 180°pb = 405° Multiply by 180°p .

(b) - 5p6

radians = - 5p6

a 180°pb = -150° Multiply by 180°p .

■✔ Now Try Exercises 31, 35, and 59.

This degree mode screen shows how a TI-84 Plus calculator converts the radian measures in Example 2 to degree measures.

NOTE Replacing p with its approximate integer value 3 in the fractions above and simplifying gives a couple of facts to help recall the relationship between degrees and radians. Remember that these are only approximations.

1° ≈160

radian and 1 radian ≈ 60°

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102 CHAPTER 3 Radian Measure and the Unit Circle

One of the most important facts to remember when working with angles and their measures is summarized in the following statement.

NOTE Another way to convert a radian measure that is a rational multiple of p, such as 9p4 , to degrees is to substitute 180° for p. In Example 2(a), doing this would give the following.

9p4

radians =91180°2

4= 405°

Agreement on Angle Measurement Units

If no unit of angle measure is specified, then the angle is understood to be measured in radians.

For example, Figure 3(a) shows an angle of 30°, and Figure 3(b) shows an angle of 30 (which means 30 radians). An angle with measure 30 radians is coterminal with an angle of approximately 279°.

030

30 degrees

Note the difference between an angle of30 degrees and an angle of 30 radians.

30 radians

(b)(a)

Figure 3

The following table and Figure 4 on the next page give some equivalent angle measures in degrees and radians. Keep in mind that

180° = P radians.

Equivalent Angle Measures

Degrees Radians Degrees Radians

Exact Approximate Exact Approximate

0° 0 0 90°p

2 1.57

30°p

6 0.52 180° p 3.14

45°p

4 0.79 270°3p2

4.71

60°p

3 1.05 360° 2p 6.28

These exact values are rational multiples of p.

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1033.1 Radian Measure

Learn the equivalences in Figure 4 . They appear often in trigonometry.

90° = P2

60° = P3

45° = P4

30° = P6

0° = 0

330° = 11P6

315° = 7P4

300° = 5P3

270° = 3P2

240° = 4P3

225° = 5P4

210° = 7P6

180° = P

150° = 5P6

135° = 3P4

120° = 2P3

Figure 4

LOOKING AHEAD TO CALCULUSIn calculus, radian measure is much

easier to work with than degree mea-

sure. If x is measured in radians, then

the derivative of ƒ1x2 = sin x isƒ′1x2 = cos x.

However, if x is measured in degrees,

then the derivative of ƒ1x2 = sin x isƒ′1x2 = p

180 cos x.

Trigonometric Function Values of Angles in Radians Trigonometric function values for angles measured in radians can be found by first converting radian measure to degrees. (Try to skip this intermediate step as soon as possible, however, and find the function values directly from radian measure.)

EXAMPLE 3 Finding Function Values of Angles in Radian Measure

Find each function value.

(a) tan 2p3

(b) sin 3p2

SOLUTION

(a) First convert 2p3 radians to degrees.

tan 2p3

= tana 2p3

# 180°pb Multiply by 180°p to convert

radians to degrees.

= tan 120° Multiply.

= -23 tan 120° = - tan 60° = -23Consider the reference angle.(b) sin

3p2

= sin 270° = -1 3p2 radians = 270°

3# 180°pb Convert radians to degrees.

= cos1-240°2 Multiply. = - 1

2 cos1-240°2 = -cos 60° = - 12

■✔ Now Try Exercises 69, 79, and 83.

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104 CHAPTER 3 Radian Measure and the Unit Circle

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. An angle with its vertex at the center of a circle that intercepts an arc on the circle equal in length to the of the circle has measure 1 radian.

2. 360° = radians, and 180° = radians.

3. To convert to radians, multiply a degree measure by radian and simplify.

4. To convert to degrees, multiply a radian measure by and simplify.

CONCEPT PREVIEW Each angle u is an integer (e.g., 0, {1, {2,c) when measured in radians. Give the radian measure of the angle. (It helps to remember that p ≈ 3.)

10.

Convert each degree measure to radians. Leave answers as multiples of p. See Examples 1(a) and 1(b).

11. 60° 12. 30° 13. 90° 14. 120°

15. 150° 16. 270° 17. -300° 18. -315°

19. 450° 20. 480° 21. 1800° 22. 3600°

23. 0° 24. 180° 25. -900° 26. -1800°

27. Concept Check Explain the meaning of radian measure.

28. Concept Check Explain why an angle of radian measure t in standard position intercepts an arc of length t on a circle of radius 1.

Convert each radian measure to degrees. See Examples 2(a) and 2(b).

29. p

3 30.

8p3

31. 7p4

32. 2p3

33. 11p

6 34.

15p4

35. - p6

36. - 8p5

37. 7p10

38. 11p15

39. - 4p15

40. - 7p20

41. 17p20

42. 11p30

43. -5p 44. 15p

3.1 Exercises

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1053.1 Radian Measure

Convert each degree measure to radians. If applicable, round to the nearest thousandth. See Example 1(c).

45. 39° 46. 74° 47. 42.5° 48. 264.9°

49. 139° 10′ 50. 174° 50′ 51. 64.29° 52. 85.04°

53. 56° 25′ 54. 122° 37′ 55. -47.69° 56. -23.01°

Convert each radian measure to degrees. Write answers to the nearest minute. See Example 2(c).

57. 2 58. 5 59. 1.74 60. 3.06

61. 0.3417 62. 9.84763 63. -5.01095 64. -3.47189

65. Concept Check The value of sin 30 is not 12 . Why is this true?

66. Concept Check What is meant by an angle of one radian?

Find each exact function value. See Example 3.

67. sin p

3 68. cos

6 69. tan

4 70. cot

71. sec p

6 72. csc

4 73. sin

2 74. csc

75. tan 5p3

76. cot 2p3

77. sin 5p6

78. tan 5p6

79. cos 3p 80. sec p 81. sin a - 8p3b 82. cot a - 2p

83. sin a - 7p6b 84. cos a - p

6b 85. tan a - 14p

3b 86. csc a - 13p

87. Concept Check The figure shows the same angles measured in both degrees and radians. Complete the missing measures.

____°; radians

180°; ____ radians 0°; 0 radians

30°; ____ radian

60°; ____ radians

____°; radian

90°; radians

270°; radians

____°; radians4p__3

2p__3

p_2

p_4

3p__2

5p__3

3p__4

225°; ____ radians

210°; ____ radians

150°; ____ radians

____°; radians

315°; ____ radians

330°; ____ radians

88. Concept Check What is the exact radian measure of an angle measuring p degrees?

89. Concept Check Find two angles, one positive and one negative, that are coterminal with an angle of p radians.

90. Concept Check Give an expression that generates all angles coterminal with an angle of p2 radians. Let n represent any integer.

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106 CHAPTER 3 Radian Measure and the Unit Circle

Solve each problem.

91. Rotating Hour Hand on a Clock Through how many radians does the hour hand on a clock rotate in (a) 24 hr and (b) 4 hr?

92. Rotating Minute Hand on a Clock Through how many radians does the minute hand on a clock rotate in (a) 12 hr and (b) 3 hr?

93. Orbits of a Space Vehicle A space vehicle is orbiting Earth in a circular orbit. What radian measure corresponds to (a) 2.5 orbits and (b) 43 orbits?

94. Rotating Pulley A circular pulley is rotating about its center. Through how many radians does it turn in (a) 8 rotations and (b) 30 rotations?

95. Revolutions of a Carousel A stationary horse on a carousel makes 12 complete revolutions. Through what radian mea-sure angle does the horse revolve?

96. Railroad Engineering Some engineers use the term grade to represent 1100 of a right angle and express grade as a percent. For example, an angle of 0.9° would be referred to as a 1% grade. (Source: Hay, W., Railroad Engineering, John Wiley and Sons.)

(a) By what number should we multiply a grade (disregarding the % symbol) to convert it to radians?

(b) In a rapid-transit rail system, the maximum grade allowed between two stations is 3.5%. Express this angle in degrees and in radians.

Arc Length on a Circle The formula for finding the length of an arc of a circle follows directly from the definition of an angle u in radians, where u = sr .

In Figure 5, we see that angle QOP has measure 1 radian and intercepts an arc of length r on the circle. We also see that angle ROT has measure u radians and intercepts an arc of length s on the circle. From plane geometry, we know that the lengths of the arcs are proportional to the measures of their central angles.

sr= u

1 Set up a proportion.

Multiplying each side by r gives

s = r u. Solve for s.

3.2 Applications of Radian Measure■ Arc Length on a Circle■ Area of a Sector of a

Circle

rr O PU radians x

1 radian

Figure 5Arc Length

The length s of the arc intercepted on a circle of radius r by a central angle of measure u radians is given by the product of the radius and the radian measure of the angle.

s = r U, where U is in radians

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1073.2 Applications of Radian Measure

CAUTION When the formula s = r U is applied, the value of U MUST be expressed in radians, not degrees.

EXAMPLE 2 Finding the Distance between Two Cities

Latitude gives the measure of a central angle with vertex at Earth’s center whose initial side goes through the equator and whose terminal side goes through the given location. Reno, Nevada, is approximately due north of Los Angeles. The latitude of Reno is 40° N, and that of Los Angeles is 34° N. (The N in 34° N means north of the equator.) The radius of Earth is 6400 km. Find the north-south distance between the two cities.

SOLUTION As shown in Figure 7, the central angle between Reno and Los Angeles is

40° - 34° = 6°.

The distance between the two cities can be found using the formula s = r u, after 6° is converted to radians.

6° = 6 a p180b = p

30 radian

The distance between the two cities is given by s.

s = r u = 6400 a p30b ≈ 670 km Let r = 6400 and u = p30 .

■✔ Now Try Exercise 23.

EXAMPLE 1 Finding Arc Length Using s = r U

A circle has radius 18.20 cm. Find the length of the arc intercepted by a central angle having each of the following measures.

(a) 3p8

radians (b) 144°

SOLUTION

(a) As shown in Figure 6, r = 18.20 cm and u = 3p8 .

s = r u Arc length formula

s = 18.20 a 3p8b Let r = 18.20 and u = 3p8 .

s ≈ 21.44 cm Use a calculator.

(b) The formula s = r u requires that u be measured in radians. First, convert u to radians by multiplying 144° by p180 radian.

144° = 144 a p180b = 4p

5 radians Convert from degrees to radians.

The length s is found using s = r u.

s = r u = 18.20 a 4p5b ≈ 45.74 cm Let r = 18.20 and u = 4p5 .

r = 18.20 cm

s3P8

Figure 6

Be sure to use radians for u in s = r u. ■✔ Now Try Exercises 13 and 17.

Equator34°

40°

6° Los Angeles

Renos

6400 km

Figure 7

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108 CHAPTER 3 Radian Measure and the Unit Circle

39.72°

0.8725 ft

Figure 8

EXAMPLE 3 Finding a Length Using s = r U

A rope is being wound around a drum with radius 0.8725 ft. (See Figure 8.) How much rope will be wound around the drum if the drum is rotated through an angle of 39.72°?

SOLUTION The length of rope wound around the drum is the arc length for a circle of radius 0.8725 ft and a central angle of 39.72°. Use the formula s = r u, with the angle converted to radian measure. The length of the rope wound around the drum is approximated by s.

s = r u = 0.8725 c39.72 a p180b d ≈ 0.6049 ft

Convert to radian measure.

■✔ Now Try Exercise 35(a).

EXAMPLE 4 Finding an Angle Measure Using s = r U

Two gears are adjusted so that the smaller gear drives the larger one, as shown in Figure 9. If the smaller gear rotates through an angle of 225°, through how many degrees will the larger gear rotate?

SOLUTION First find the radian measure of the angle of rotation for the smaller gear, and then find the arc length on the smaller gear. This arc length will correspond to the arc length of the motion of the larger gear. Because

225° = 5p4 radians, for the smaller gear we have arc length

s = r u = 2.5 a 5p4b = 12.5p

4= 25p

8 cm.

The tips of the two mating gear teeth must move at the same linear speed, or the teeth will break. So we must have “equal arc lengths in equal times.” An arc with this length s on the larger gear corresponds to an angle measure u, in radians, where s = r u.

s = r u Arc length formula

25p

8= 4.8u Let s = 25p8 and r = 4.8 (for the larger gear).

125p192

= u 4.8 = 4810 =245 ; Multiply by

524 to solve for u.

Converting u back to degrees shows that the larger gear rotates through

125p192

a 180°pb ≈ 117°. Convert u = 125p192 to degrees.

■✔ Now Try Exercise 29.

2.5cm

4.8 cm

Figure 9

Area of a Sector of a Circle A sector of a circle is the portion of the interior of a circle intercepted by a central angle. Think of it as a “piece of pie.” See Figure 10. A complete circle can be thought of as an angle with measure 2p radians. If a central angle for a sector has measure u radians, then the sector makes up the fraction u2p of a complete circle. The area & of a complete circle with radius r is & = pr2. Therefore, we have the following.

Area & of a sector = u2p

1pr22 = 12

r2 u, where u is in radians.

The shadedregion is asector ofthe circle.

Figure 10

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1093.2 Applications of Radian Measure

CAUTION As in the formula for arc length, the value of U must be in radians when this formula is used to find the area of a sector.

Area of a Sector

(Video) math tests be like!! 🤣🤣 (4k memes) #fyp #viral

The area & of a sector of a circle of radius r and central angle u is given by the following formula.

& =12

r2 U, where U is in radians

EXAMPLE 5 Finding the Area of a Sector-Shaped Field

A center-pivot irrigation system provides water to a sector-shaped field with the measures shown in Figure 11. Find the area of the field.

SOLUTION First, convert 15° to radians.

15° = 15 a p180b = p

12 radian Convert to radians.

Now find the area of a sector of a circle.

& = 12

r2u Formula for area of a sector

& = 12

132122 a p12b Let r = 321 and u = p12 .

& ≈ 13,500 m2 Multiply. ■✔ Now Try Exercise 57.

15°

321

Figure 11

Center-pivot irrigation system

CONCEPT PREVIEW Find the exact length of each arc intercepted by the given central angle.

5P4

CONCEPT PREVIEW Find the radius of each circle.

3. 6P

3P4

7P4

14P

3.2 Exercises

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110 CHAPTER 3 Radian Measure and the Unit Circle

CONCEPT PREVIEW Find the measure of each central angle (in radians).

5. 3

6. 20

CONCEPT PREVIEW Find the area of each sector.

15P

CONCEPT PREVIEW Find the measure (in radians) of each central angle. The number inside the sector is the area.

3 sq units

10.

8 sq units

CONCEPT PREVIEW Find the measure (in degrees) of each central angle. The number inside the sector is the area.

11.

6P squnits

12.

96P squnits

Unless otherwise directed, give calculator approximations in answers in the rest of this exercise set.

Find the length to three significant digits of each arc intercepted by a central angle u in a circle of radius r. See Example 1.

13. r = 12.3 cm, u = 2p3 radians 14. r = 0.892 cm, u =11p10 radians

15. r = 1.38 ft, u = 5p6 radians 16. r = 3.24 mi, u =7p6 radians

17. r = 4.82 m, u = 60° 18. r = 71.9 cm, u = 135°

19. r = 15.1 in., u = 210° 20. r = 12.4 ft, u = 330°

21. Concept Check If the radius of a circle is doubled, how is the length of the arc intercepted by a fixed central angle changed?

22. Concept Check Radian measure simplifies many formulas, such as the formula for arc length, s = r u. Give the corresponding formula when u is measured in degrees instead of radians.

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1113.2 Applications of Radian Measure

Distance between Cities Find the distance in kilometers between each pair of cities, assuming they lie on the same north-south line. Assume that the radius of Earth is 6400 km. See Example 2.

23. Panama City, Panama, 9° N, and Pittsburgh, Pennsylvania, 40° N

24. Farmersville, California, 36° N, and Penticton, British Columbia, 49° N

25. New York City, New York, 41° N, and Lima, Peru, 12° S

26. Halifax, Nova Scotia, 45° N, and Buenos Aires, Argentina, 34° S

27. Latitude of Madison Madison, South Dakota, and Dallas, Texas, are 1200 km apart and lie on the same north-south line. The latitude of Dallas is 33° N. What is the latitude of Madison?

28. Latitude of Toronto Charleston, South Carolina, and Toronto, Canada, are 1100 km apart and lie on the same north-south line. The latitude of Charleston is 33° N. What is the latitude of Toronto?

Work each problem. See Examples 3 and 4.

29. Gear Movement Two gears are adjusted so that the smaller gear drives the larger one, as shown in the figure. If the smaller gear rotates through an angle of 300°, through how many degrees does the larger gear rotate?

30. Gear Movement Repeat Exercise 29 for gear radii of 4.8 in. and 7.1 in. and for an angle of 315° for the smaller gear.

31. Rotating Wheels The rotation of the smaller wheel in the figure causes the larger wheel to rotate. Through how many degrees does the larger wheel rotate if the smaller one rotates through 60.0°?

32. Rotating Wheels Repeat Exercise 31 for wheel radii of 6.84 in. and 12.46 in. and an angle of 150.0° for the smaller wheel.

33. Rotating Wheels Find the radius of the larger wheel in the figure if the smaller wheel rotates 80.0° when the larger wheel rotates 50.0°.

34. Rotating Wheels Repeat Exercise 33 if the smaller wheel of radius 14.6 in. rotates 120.0° when the larger wheel rotates 60.0°.

35. Pulley Raising a Weight Refer to the figure.

(a) How many inches will the weight in the figure rise if the pulley is rotated through an angle of 71° 50′?

(b) Through what angle, to the nearest minute, must the pulley be rotated to raise the weight 6 in.?

3.7 cm

7.1 cm

8.16cm

5.23cm

11.7cm r

9.27 in.

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112 CHAPTER 3 Radian Measure and the Unit Circle

36. Pulley Raising a Weight Find the radius of the pulley in the figure if a rotation of 51.6° raises the weight 11.4 cm.

37. Bicycle Chain Drive The figure shows the chain drive of a bicycle. How far will the bicycle move if the pedals are rotated through 180.0°? Assume the radius of the bicycle wheel is 13.6 in.

38. Car Speedometer The speedometer of Terry’s Honda CR-V is designed to be accu-rate with tires of radius 14 in.

(a) Find the number of rotations of a tire in 1 hr if the car is driven at 55 mph.

(b) Suppose that oversize tires of radius 16 in. are placed on the car. If the car is now driven for 1 hr with the speedometer reading 55 mph, how far has the car gone? If the speed limit is 55 mph, does Terry deserve a speeding ticket?

Suppose the tip of the minute hand of a clock is 3 in. from the center of the clock. For each duration, determine the dis-tance traveled by the tip of the minute hand. Leave answers as multiples of p.

39. 30 min 40. 40 min

41. 4.5 hr 42. 6 12

If a central angle is very small, there is little dif-ference in length between an arc and the inscribed chord. See the figure. Approximate each of the fol-lowing lengths by finding the necessary arc length. (Note: When a central angle intercepts an arc, the arc is said to subtend the angle.)

43. Length of a Train A railroad track in the desert is 3.5 km away. A train on the track subtends (horizontally) an angle of 3° 20′. Find the length of the train.

44. Repeat Exercise 43 for a railroad track 2.7 mi away and a train that subtends an angle of 2° 30′.

45. Distance to a Boat The mast of a boat is 32.0 ft high. If it subtends an angle of 2° 11′, how far away is it?

46. Repeat Exercise 45 for a boat mast 11.0 m high that subtends an angle of 1° 45′.

Find the area of a sector of a circle having radius r and central angle u. Express answers to the nearest tenth. See Example 5.

47. r = 29.2 m, u = 5p6 radians 48. r = 59.8 km, u =2p3 radians

49. r = 30.0 ft, u = p2 radians 50. r = 90.0 yd, u =5p6 radians

51. r = 12.7 cm, u = 81° 52. r = 18.3 m, u = 125°

53. r = 40.0 mi, u = 135° 54. r = 90.0 km, u = 270°

1.38 in.

4.72 in.

3 in.

12121212121212121212

1111111111222222

111111111111111111111

1111111111111110

489

Arc length ≈ length of inscribed chord

Inscribed chord

Arc

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1133.2 Applications of Radian Measure

Work each problem. See Example 5.

55. Angle Measure Find the measure (in radians) of a central angle of a sector of area 16 in.2 in a circle of radius 3.0 in.

56. Radius Length Find the radius of a circle in which a central angle of p6 radian determines a sector of area 64 m2.

57. Irrigation Area A center-pivot irrigation system provides water to a sector-shaped field as shown in the figure. Find the area of the field if u = 40.0° and r = 152 yd.

58. Irrigation Area Suppose that in Exercise 57 the angle is halved and the radius length is doubled. How does the new area compare to the original area? Does this result hold in general for any values of u and r?

59. Arc Length A circular sector has an area of 50 in.2. The radius of the circle is 5 in. What is the arc length of the sector?

60. Angle Measure In a circle, a sector has an area of 16 cm2 and an arc length of 6.0 cm. What is the measure of the central angle in degrees?

61. Measures of a Structure The figure illustrates Medicine Wheel, a Native American structure in northern Wyoming. There are 27 aboriginal spokes in the wheel, all equally spaced.

(a) Find the measure of each central angle in degrees and in radians in terms of p.

(b) If the radius of the wheel is 76.0 ft, find the circumference.

(d) Find the area of each sector formed by consecutive spokes.

62. Area Cleaned by a Windshield Wiper The Ford Model A, built from 1928 to 1931, had a single windshield wiper on the driver’s side. The total arm and blade was 10 in. long and rotated back and forth through an angle of 95°. The shaded region in the figure is the portion of the windshield cleaned by the 7-in. wiper blade. Find the area of the region cleaned to the nearest tenth.

63. Circular Railroad Curves In the United States, circular railroad curves are desig-nated by the degree of curvature, the central angle subtended by a chord of 100 ft. Suppose a portion of track has curvature 42.0°. (Source: Hay, W., Railroad Engineering, John Wiley and Sons.)

(a) What is the radius of the curve?

(b) What is the length of the arc determined by the 100-ft chord?

64. Land Required for a Solar-Power Plant A 300-megawatt solar-power plant requires approximately 950,000 m2 of land area to collect the required amount of energy from sunlight. If this land area is circular, what is its radius? If this land area is a 35° sector of a circle, what is its radius?

10in.

95°7 in.

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114 CHAPTER 3 Radian Measure and the Unit Circle

65. Area of a Lot A frequent problem in surveying city lots and rural lands adjacent to curves of highways and rail-ways is that of finding the area when one or more of the boundary lines is the arc of a circle. Find the area (to two significant digits) of the lot shown in the figure. (Source: Anderson, J. and E. Michael, Introduction to Surveying, McGraw-Hill.)

66. Nautical Miles Nautical miles are used by ships and airplanes. They are different from statute miles, where 1 mi = 5280 ft. A nautical mile is defined to be the arc length along the equator intercepted by a central angle AOB of 1′, as illustrated in the figure. If the equatorial radius of Earth is 3963 mi, use the arc length formula to approximate the number of statute miles in 1 nautical mile. Round the answer to two decimal places.

67. Circumference of Earth The first accurate estimate of the distance around Earth was done by the Greek astronomer Eratosthenes (276–195 b.c.), who noted that the noontime position of the sun at the summer solstice in the city of Syene differed by 7° 12′ from its noontime position in the city of Alexandria. (See the figure.) The dis-tance between these two cities is 496 mi. Use the arc length formula to estimate the radius of Earth. Then find the circumference of Earth. (Source: Zeilik, M., Introductory Astronomy and Astrophysics, Third Edition, Saunders College Publishers.)

496 mi

Syene

Alexandria

Shadow

7° 12'

Sun’s rays at noon

68. Longitude Longitude is the angular distance (expressed in degrees) East or West of the prime meridian, which goes from the North Pole to the South Pole through Greenwich, England. Arcs of 1° longitude are 110 km apart at the equator, and therefore 15° arcs subtend 15(110) km, or 1650 km, at the equator.

North Pole

South PolePrime meridian

Equator

15° E

1650 km

15° W30° W

Because Earth rotates 15° per hr, longitude is found by taking the difference between time zones multiplied by 15°. For example, if it is 12 noon where we are (in the United States) and 5 p.m. in Greenwich, we are located at longitude 5115°2, or 75° W.(a) What is the longitude at Greenwich, England?

(b) Use time zones to determine the longitude where you live.

60°

30 yd

40 yd

Nauticalmile

NOT TO SCALE

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1153.2 Applications of Radian Measure

69. Concept Check If the radius of a circle is doubled and the central angle of a sector is unchanged, how is the area of the sector changed?

70. Concept Check Give the formula for the area of a sector when the angle is mea-sured in degrees.

Volume of a Solid Multiply the area of the base by the height to find a formula for the volume V of each solid.

71.

72.

Outside radius is r1,inside radius is r2.

Relating Concepts

For individual or collaborative investigation (Exercises 73—76)

(Modeling) Measuring Paper Curl Manufacturers of paper determine its quality by its curl. The curl of a sheet of paper is measured by holding it at the center of one edge and comparing the arc formed by the free end to arcs on a chart lying flat on a table. Each arc in the chart corresponds to a number d that gives the depth of the arc. See the figure. (Source: Tabakovic, H., J. Paullet, and R. Bertram, “Measuring the Curl of Paper,” The College Mathematics Journal, Vol. 30, No. 4.)

Chart

Paper

Hand

h r

To produce the chart, it is necessary to find a function that relates d to the length of arc L. Work Exercises 73–76 in order, to determine that function. Refer to the figure on the right.

73. Express L in terms of r and u, and then solve for r.

74. Use a right triangle to relate r, h, and u. Solve for h.

75. Express d in terms of r and h. Then substitute the answer from Exercise 74 for h. Factor out r.

76. Use the answer from Exercise 73 to substitute for r in the result from Exercise 75. This result is a formula that gives d for specific values of u.

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116 CHAPTER 3 Radian Measure and the Unit Circle

3.3 The Unit Circle and Circular Functions

We have defined the six trigonometric functions in such a way that the domain of each function was a set of angles in standard position. These angles can be measured in degrees or in radians. In advanced courses, such as calculus, it is necessary to modify the trigonometric functions so that their domains consist of real numbers rather than angles. We do this by using the relationship between an angle u and an arc of length s on a circle.

■ Circular Functions■ Values of the Circular

Functions■ Determining a Number

with a Given Circular Function Value

■ Applications of Circular Functions

■ Function Values as Lengths of Line Segments

Circular Functions In Figure 12, we start at the point 11, 02 and measure an arc of length s along the circle. If s 7 0, then the arc is measured in a counter-clockwise direction, and if s 6 0, then the direction is clockwise. (If s = 0, then no arc is measured.) Let the endpoint of this arc be at the point 1x, y2. The circle in Figure 12 is the unit circle—it has center at the origin and radius 1 unit (hence the name unit circle). Recall from algebra that the equation of this circle is

x2 + y2 = 1. The unit circle

The radian measure of u is related to the arc length s. For u measured in radians and for r and s measured in the same linear units, we know that

s = r u.

When the radius has measure 1 unit, the formula s = r u becomes s = u. Thus, the trigonometric functions of angle u in radians found by choosing a point 1x, y2 on the unit circle can be rewritten as functions of the arc length s, a real number. When interpreted this way, they are called circular functions.

x = cos sy = sin s

(x, y)

(1, 0)

(0, 1)

(–1, 0)

(0, –1)

Arc of length s

The unit circle x2 + y2 = 1

Figure 12

Circular Functions

The following functions are defined for any real number s represented by a directed arc on the unit circle.

sin s = y cos s = x tan s =yx 1x 3 0 2

csc s =1y 1 y 3 0 2 sec s = 1

x 1x 3 0 2 cot s = x

y 1 y 3 0 2

The unit circle is symmetric with respect to the x-axis, the y-axis, and the origin. If a point 1a, b2 lies on the unit circle, so do 1a, -b2, 1-a, b2, and 1-a, -b2. Furthermore, each of these points has a reference arc of equal mag-nitude. For a point on the unit circle, its reference arc is the shortest arc from the point itself to the nearest point on the x-axis. (This concept is analogous to the reference angle concept.) Using the concept of symmetry makes determining sines and cosines of the real numbers identified in Figure 13* a relatively simple procedure if we know the coordinates of the points labeled in quadrant I.

*The authors thank Professor Marvel Townsend of the University of Florida for her suggestion to include Figure 13.

LOOKING AHEAD TO CALCULUSIf you plan to study calculus, you

must become very familiar with radian

measure. In calculus, the trigonometric

or circular functions are always under-

stood to have real number domains.

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1173.3 The Unit Circle and Circular Functions

For example, the quadrant I real number p3 is associated with the point Q12 , 232 R on the unit circle. Therefore, we can use symmetry to identify the coor-dinates of points having p3 as reference arc.

(0, 1)

(1, 0)

(0, –1)

(–1, 0) 00°180°

60°90°

150°

210°

300°

360°

315°

330°

2PP

135°

120°

225°

240°270°

45°

30°

12( , )√32

12( , – )√3212(– , – )√32

( , )√22√22

12( , – )√32

12( , )√32

12(– , )√32

(– , )√22√22

( , – )√22√22

12(– , – )√32

(– , – )√22√22

12(– , )√32

3P2

5P3

7P4

4P3

5P4

7P6

5P6

3P4

2P3

11P6

The unit circle x2 + y2 = 1

Figure 13

Symmetry and Function Values for Real Numbers with Reference Arc P3 s

Quadrant of s

Symmetry Type and Corresponding Point

cos s

sin s

3I not applicable; ¢1

2 , 232

≤ 12

232

p - p3

= 2P3

II y-axis; ¢ - 12

, 232

≤ - 12

232

p + p3

= 4P3

III origin; ¢ - 12

, - 232

≤ - 12

- 232

2p - p3

= 5P3

IV x-axis; ¢12

, - 232

≤ 12

- 232

NOTE Because cos s = x and sin s = y, we can replace x and y in the equation of the unit circle x2 + y2 = 1 and obtain the following.

cos2 s + sin2 s = 1 Pythagorean identity

The ordered pair 1x, y2 represents a point on the unit circle, and therefore-1 … x … 1 and -1 … y … 1,−1 " cos s " 1 and −1 " sin s " 1.

For any value of s, both sin s and cos s exist, so the domain of these functions is the set of all real numbers.

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118 CHAPTER 3 Radian Measure and the Unit Circle

For tan s, defined as yx , x must not equal 0. The only way x can equal 0 is

when the arc length s is p2 , - p2 ,

3p2 , -

3p2 , and so on. To avoid a 0 denominator,

the domain of the tangent function must be restricted to those values of s that satisfy

s 3 12n + 1 2 P2

, where n is any integer.

The definition of secant also has x in the denominator, so the domain of secant is the same as the domain of tangent. Both cotangent and cosecant are defined with a denominator of y. To guarantee that y ≠ 0, the domain of these functions must be the set of all values of s that satisfy

s 3 nP, where n is any integer.

Domains of the Circular Functions

The domains of the circular functions are as follows.

Sine and Cosine Functions: 1−H, H 2Tangent and Secant Functions:5s∣ s 3 12n + 1 2 P

2 , where n is any integer6

Cotangent and Cosecant Functions:5s∣ s 3 nP, where n is any integer6Values of the Circular Functions The circular functions of real numbers

correspond to the trigonometric functions of angles measured in radians. Let us assume that angle u is in standard position, superimposed on the unit circle. See Figure 14. Suppose that u is the radian measure of this angle. Using the arc length formula

s = r u with r = 1, we have s = u.

Thus, the length of the intercepted arc is the real number that corresponds to the radian measure of u. We use the trigonometric function definitions to obtain the following.

sin u =yr

=y1

= y = sin s, cos u = xr

= x1

= x = cos s, and so on.

As shown here, the trigonometric functions and the circular functions lead to the same function values, provided that we think of the angles as being in radian measure. This leads to the following important result.

(0, 1)

(1, 0)(–1, 0)

(0, –1)

x2 + y2 = 1

r = 1

y s = U

(cos s, sin s) = (x, y)

Figure 14

Evaluating a Circular Function

Circular function values of real numbers are obtained in the same manner as trigonometric function values of angles measured in radians. This applies both to methods of finding exact values (such as reference angle analysis) and to calculator approximations. Calculators must be in radian mode when they are used to find circular function values.

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1193.3 The Unit Circle and Circular Functions

EXAMPLE 1 Finding Exact Circular Function Values

Find the exact values of sin 3p2 , cos

3p2 , and

tan 3p2 .

SOLUTION Evaluating a circular function at

the real number 3p2 is equivalent to evaluating it

at 3p2 radians. An angle of

3p2 radians intersects

the unit circle at the point 10, -12, as shown in Figure 15. Because

sin s = y, cos s = x, and tan s =yx

it follows that

sin 3p2

= -1, cos 3p2

= 0, and tan 3p2

is undefined.

■✔ Now Try Exercises 11 and 13.

(0, –1)

(0, 1)

(–1, 0)

(1, 0)

U = 3P2

Figure 15

EXAMPLE 2 Finding Exact Circular Function Values

Find each exact function value using the specified method.

(a) Use Figure 13 to find the exact values of cos 7p4 and sin 7p4 .

(b) Use Figure 13 and the definition of the tangent to find the exact value of

tan A - 5p3 B.(c) Use reference angles and radian-to-degree conversion to find the exact value

of cos 2p3 .

SOLUTION

(a) In Figure 13, we see that the real number 7p4 corresponds to the unit circle

point Q222 , - 222 R .cos

7p4

= 222 and sin

7p4

= - 222

(b) Moving around the unit circle 5p3 units in the negative direction yields the same ending point as moving around p3 units in the positive direction. Thus,

- 5p3 corresponds to Q12 , 232 R.tan a - 5p

3b = tan p

3=23212

= 232

, 12

= 232

# 21

= 23tan s = yxSimplify this complex fraction.

(c) An angle of 2p3 radians corresponds to an angle of 120°. In standard position, 120° lies in quadrant II with a reference angle of 60°.

Cosine is negative in quadrant II.

cos 2p3

= cos 120° = -cos 60° = - 12

Reference angle

■✔ Now Try Exercises 17, 23, 27, and 31.

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120 CHAPTER 3 Radian Measure and the Unit Circle

EXAMPLE 3 Approximating Circular Function Values

Find a calculator approximation for each circular function value.

(a) cos 1.85 (b) cos 0.5149 (c) cot 1.3209 (d) sec1-2.92342SOLUTION

(a) cos 1.85 ≈ -0.2756 Use a calculator in radian mode.

(b) cos 0.5149 ≈ 0.8703 Use a calculator in radian mode.

(c) As before, to find cotangent, secant, and cosecant function values, we must use the appropriate reciprocal functions. To find cot 1.3209, first find tan 1.3209 and then find the reciprocal.

cot 1.3209 = 1tan 1.3209

≈ 0.2552 Tangent and cotangent are reciprocals.

(d) sec1-2.92342 = 1cos1-2.92342 ≈ -1.0243 Cosine and secant are reciprocals.

■✔ Now Try Exercises 33, 39, and 43.

Radian mode

This is how the TI-84 Plus calculator displays the results of Example 3, fixed to four decimal places.

CAUTION Remember, when used to find a circular function value of a real number, a calculator must be in radian mode.

Determining a Number with a Given Circular Function Value We can reverse the process of Example 3 and use a calculator to determine an angle measure, given a trigonometric function value of the angle. Remember that the keys marked sin−1, cos−1, and tan−1 do not represent reciprocal functions. They enable us to find inverse function values.

For reasons explained in a later chapter, the following statements are true.

For all x in 3-1, 14, a calculator in radian mode returns a single value in C - p2 , p2 D for sin-1 x .For all x in 3-1, 14, a calculator in radian mode returns a single value in 30, p4 for cos-1 x .For all real numbers x, a calculator in radian mode returns a single value in A - p2 , p2 B for tan-1 x .

EXAMPLE 4 Finding Numbers Given Circular Function Values

Find each value as specified.

(a) Approximate the value of s in the interval 30, p2 4 if cos s = 0.9685.(b) Find the exact value of s in the interval 3p, 3p2 4 if tan s = 1.SOLUTION

(a) Because we are given a cosine value and want to determine the real number in 30, p2 4 that has this cosine value, we use the inverse cosine function of a calculator. With the calculator in radian mode, we find s as follows.

s = cos-110.96852 ≈ 0.2517

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1213.3 The Unit Circle and Circular Functions

See Figure 16. The screen indicates that the real number in C 0, p2 D having cosine equal to 0.9685 is 0.2517.

(b) Recall that tan p4 = 1, and in quadrant III tan s is positive.

tan ap + p4b = tan 5p

4= 1

Thus, s = 5p4 . See Figure 17. ■✔ Now Try Exercises 63 and 71.

Radian mode

Figure 16x

(1, 0)

P =

3p2

5P4

( , )√22√22

(– , – )√22√22

Figure 17

Applications of Circular Functions

EXAMPLE 5 Modeling the Angle of Elevation of the Sun

The angle of elevation u of the sun in the sky at any latitude L is calculated with the formula

sin u = cos D cos L cos v + sin D sin L,

where u = 0 corresponds to sunrise and u = p2 occurs if the sun is directly over-head. The Greek letter v (lowercase omega) is the number of radians that Earth has rotated through since noon, when v = 0. D is the declination of the sun, which varies because Earth is tilted on its axis. (Source: Winter, C., R. Sizmann, and L. L. Vant-Hull, Editors, Solar Power Plants, Springer-Verlag.)

Sacramento, California, has latitude L = 38.5°, or 0.6720 radian. Find the angle of elevation u of the sun at 3 p.m. on February 29, 2012, where at that time D ≈ -0.1425 and v ≈ 0.7854.

SOLUTION Use the given formula for sin u.

sin u = cos D cos L cos v + sin D sin L sin u = cos1-0.14252 cos10.67202 cos10.78542 + sin1-0.14252 sin10.67202

Let D = -0.1425, L = 0.6720, and v = 0.7854. sin u ≈ 0.4593426188

u ≈ 0.4773 radian, or 27.3° Use inverse sine.■✔ Now Try Exercise 89.

This screen supports the result in Example 4(b) with calculator approximations.

Function Values as Lengths of Line Segments The diagram shown in Figure 18 illustrates a correspondence that ties together the right triangle ratio definitions of the trigonometric functions and the unit circle interpretation. The arc SR is the first-quadrant portion of the unit circle, and the standard-position angle POQ is designated u. By definition, the coordinates of P are 1cos u, sin u2. The six trigonometric functions of u can be interpreted as lengths of line seg-ments found in Figure 18.

For cos u and sin u, use right triangle POQ and right triangle ratios.

cos U =side adjacent to u

hypotenuse=

OQOP

=OQ1

= OQ

sin U =side opposite u

hypotenuse=

PQOP

=PQ1

= PQ

x(1, 0)RQ

UT(0, 1)

Figure 18

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122 CHAPTER 3 Radian Measure and the Unit Circle

For tan u and sec u, use right triangle VOR in Figure 18 (repeated below in the margin) and right triangle ratios.

tan U =side opposite u

side adjacent to u= VR

OR= VR

1= VR

sec U =hypotenuse

side adjacent to u= OV

OR= OV

1= OV

For csc u and cot u, first note that US and OR are parallel. Thus angle SUO is equal to u because it is an alternate interior angle to angle POQ, which is equal to u. Use right triangle USO and right triangle ratios.

csc SUO = csc U =hypotenuse

side opposite u= OU

OS= OU

1= OU

cot SUO = cot U =side adjacent to uside opposite u

= USOS

= US1

= US

Figure 19 uses color to illustrate the results found above.

(1, 0)

UT(0, 1)

cos U = OQ

(1, 0)

UT(0, 1)

sin U = PQ tan U = VR

(1, 0)

UT(0, 1)

(a) (b) (c)

(1, 0)

UT(0, 1)

sec U = OV

(1, 0)

UT(0, 1)

csc U = OU

(1, 0)

UT(0, 1)

cot U = US (d) (e) (f)

Figure 19

EXAMPLE 6 Finding Lengths of Line Segments

Figure 18 is repeated in the margin. Suppose that angle TVU measures 60°. Find the exact lengths of segments OQ, PQ, VR, OV, OU, and US.

SOLUTION Angle TVU has the same measure as angle OVR because they are vertical angles. Therefore, angle OVR measures 60°. Because it is one of the acute angles in right triangle VOR, u must be its complement, measuring 30°.

OQ = cos 30° = 232

OV = sec 30° = 2233

x(1, 0)RQ

UT(0, 1)

Figure 18 (repeated) Use the equations found in Figure 19, with u = 30°.

PQ = sin 30° = 12

OU = csc 30° = 2

VR = tan 30° = 233

US = cot 30° = 23■✔ Now Try Exercise 93.

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1233.3 The Unit Circle and Circular Functions

CONCEPT PREVIEW Fill in the blanks to complete the coordinates for each point indicated in the first quadrant of the unit circle in Exercise 1. Then use it to find each exact circular function value in Exercises 2–5, and work Exercise 6.

(__, __)

00°

60°90°

45°

30°

3.3 Exercises

2. cos 0 3. sin p

4. sin p

3 5. tan

6. Find s in the interval 30, p2 4 if cos s = 12 .

CONCEPT PREVIEW Each figure shows an angle u in standard position with its ter-minal side intersecting the unit circle. Evaluate the six circular function values of u.

( , )2√2 2√21

(– , )8171517 1

( , – )513 1213

10.

12(– , – )√32

Find the exact values of (a) sin s, (b) cos s, and (c) tan s for each real number s. See Example 1.

11. s = p2

12. s = p 13. s = 2p

14. s = 3p 15. s = -p 16. s = - 3p2

Find each exact function value. See Example 2.

17. sin 7p6

18. cos 5p3

19. tan 3p4

20. sec 2p3

21. csc 11p

6 22. cot

5p6

23. cos a - 4p3b 24. tan a - 17p

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124 CHAPTER 3 Radian Measure and the Unit Circle

25. cos 7p4

26. sec 5p4

27. sin a - 4p3b 28. sin a - 5p

29. sec 23p

6 30. csc

13p3

31. tan 5p6

32. cos 3p4

Find a calculator approximation to four decimal places for each circular function value. See Example 3.

33. sin 0.6109 34. sin 0.8203 35. cos1-1.1519236. cos1-5.28252 37. tan 4.0203 38. tan 6.475239. csc1-9.49462 40. csc 1.3875 41. sec 2.844042. sec1-8.34292 43. cot 6.0301 44. cot 3.8426

Concept Check Without using a calculator, decide whether each function value is positive or negative. (Hint: Consider the radian measures of the quadrantal angles, and remember that p ≈ 3.14.)

55. cos 2 56. sin1-12 57. sin 558. cos 6 59. tan 6.29 60. tan1-6.292Find the approximate value of s, to four decimal places, in the interval C 0, p2 D that makes each statement true. See Example 4(a).

61. tan s = 0.2126 62. cos s = 0.7826 63. sin s = 0.9918

64. cot s = 0.2994 65. sec s = 1.0806 66. csc s = 1.0219

0.4 radian0.2 radian

0.60.2

0.4

0.6

0.8

0.2 1 radian

0.8 radian0.6 radian

Concept Check The figure displays a unit circle and an angle of 1 radian. The tick marks on the circle are spaced at every two-tenths radian. Use the figure to estimate each value.

45. cos 0.8 46. cos 0.6 47. sin 2

48. sin 5.4 49. sin 3.8 50. cos 3.2

51. a positive angle whose cosine is -0.65

52. a positive angle whose sine is -0.95

53. a positive angle whose sine is 0.7

54. a positive angle whose cosine is 0.3

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1253.3 The Unit Circle and Circular Functions

Find the exact value of s in the given interval that has the given circular function value. See Example 4(b).

67. cp2

, p d ; sin s = 12

68. cp2

, p d ; cos s = - 12

69. cp, 3p2d ; tan s = 23 70. cp, 3p

2d ; sin s = - 1

71. c 3p2

, 2p d ; tan s = -1 72. c 3p2

, 2p d ; cos s = 232

Find the exact values of s in the given interval that satisfy the given condition.

73. 30, 2p2; sin s = - 232

74. 30, 2p2; cos s = - 12

75. 30, 2p2; cos2 s = 12

76. 30, 2p2; tan2 s = 377. 3-2p, p2; 3 tan2 s = 1 78. 3-p, p2; sin2 s = 1

Suppose an arc of length s lies on the unit circle x2 + y2 = 1, starting at the point 11, 02 and terminating at the point 1x, y2. (See Figure 12.) Use a calculator to find the approximate coordinates for 1x, y2 to four decimal places. (Hint: x = cos s and y = sin s.)79. s = 2.5 80. s = 3.4 81. s = -7.4 82. s = -3.9

Concept Check For each value of s, use a calculator to find sin s and cos s, and then use the results to decide in which quadrant an angle of s radians lies.

83. s = 51 84. s = 49 85. s = 65 86. s = 79

Concept Check Each graphing calculator screen shows a point on the unit circle. Find the length, to four decimal places, of the shortest arc of the circle from 11, 02 to the point. 87. x2 + y2 = 1

−1.55

−2.35

1.55

2.35

88. x2 + y2 = 1

−1.55

−2.35

1.55

2.35

(Modeling) Solve each problem. See Example 5.

89. Elevation of the Sun Refer to Example 5.

(a) Repeat the example for New Orleans, which has latitude L = 30°.(b) Compare the answers. Do they agree with intuition?

90. Length of a Day The number of daylight hours H at any location can be calculated using the formula

cos10.1309H2 = - tan D tan L, where D and L are defined as in Example 5. Use this trigonometric equation to

calculate the shortest and longest days in Minneapolis, Minnesota, if its latitude L = 44.88°, the shortest day occurs when D = -23.44°, and the longest day occurs when D = 23.44°. Remember to convert degrees to radians. Round the answer to the nearest tenth. (Source: Winter, C., R. Sizmann, and L. L. Vant-Hull, Editors, Solar Power Plants, Springer-Verlag.)

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126 CHAPTER 3 Radian Measure and the Unit Circle

91. Maximum Temperatures Because the values of the circular functions repeat every 2p, they may be used to describe phenomena that repeat periodically. For example, the maximum afternoon temperature in a given city might be modeled by

t = 60 - 30 cos ap6

xb , where t represents the maximum afternoon temperature in degrees Fahrenheit in

month x, with x = 0 representing January, x = 1 representing February, and so on. Find the maximum afternoon temperature, to the nearest degree, for each of the following months.

(a) January (b) April (c) May

(d) June (e) August (f ) October

92. Temperature in Fairbanks Suppose the temperature in Fairbanks is modeled by

T1x2 = 37 sin c 2p365

1x - 1012 d + 25, where T1x2 is the temperature in degrees Fahrenheit on day x, with x = 1 corre-

sponding to January 1 and x = 365 corresponding to December 31. Use a calculator to estimate the temperature, to the nearest degree, on the following days. (Source: Lando, B. and C. Lando, “Is the Graph of Temperature Variation a Sine Curve?,” The Mathematics Teacher, vol. 70.)

(a) March 1 (day 60) (b) April 1 (day 91) (c) Day 150

(d) June 15 (e) September 1 (f ) October 31

Refer to Figures 18 and 19, and work each problem. See Example 6.

93. Suppose that angle u measures 60°. Find the exact length of each segment.(a) OQ (b) PQ (c) VR

(d) OV (e) OU (f ) US

94. Repeat Exercise 93 for u = 38°. Give lengths as approximations to four significant digits.

Convert each degree measure to radians.

1. 225° 2. -330°

Convert each radian measure to degrees.

3. 5p3

4. - 7p6

A central angle of a circle with radius 300 in. intercepts an arc of 450 in. (These measures are accurate to the nearest inch.) Find each measure.

5. the radian measure of the angle 6. the area of the sector

Find each exact circular function value.

7. cos 7p4

8. sin a - 5p6b 9. tan 3p

10. Find the exact value of s in the interval 3p2 , p4 if sin s = 232 .

Chapter 3 Quiz (Sections 3.1—3.3)

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1273.4 Linear and Angular Speed

3.4 Linear and Angular Speed

Linear Speed There are situations when we need to know how fast a point on a circular disk is moving or how fast the central angle of such a disk is changing. Some examples occur with machinery involving gears or pulleys or the speed of a car around a curved portion of highway.

Suppose that point P moves at a constant speed along a circle of radius r and center O. See Figure 20. The measure of how fast the position of P is changing is the linear speed. If v represents linear speed, then

speed = distancetime

, or v =st

where s is the length of the arc traced by point P at time t. (This formula is just a restatement of r = dt with s as distance, v as rate (speed), and t as time.)

■ Linear Speed■ Angular Speed

O B

P moves ata constantspeed alongthe circle.

Figure 20

Angular Speed Refer to Figure 20. As point P in the figure moves along the circle, ray OP rotates around the origin. Because ray OP is the terminal side of angle POB, the measure of the angle changes as P moves along the circle. The measure of how fast angle POB is changing is its angular speed. Angular speed, symbolized v, is given as

V =U

t , where U is in radians.

Here u is the measure of angle POB at time t. As with earlier formulas in this chapter, U must be measured in radians, with V expressed in radians per unit of time.

The length s of the arc intercepted on a circle of radius r by a central angle of measure u radians is s = r u. Using this formula, the formula for linear speed, v = st , can be written in several useful forms.

v = st

Formula for linear speed

v = r ut

s = r u

v = r # ut

abc = a # bc v = rv v = ut

As an example of linear and angular speeds, consider the following. The human joint that can be flexed the fastest is the wrist, which can rotate through 90°, or p2 radians, in 0.045 sec while holding a tennis racket. The angular speed of a human wrist swinging a tennis racket is

v = ut

Formula for angular speed

v =p2

0.045 Let u = p2 and t = 0.045.

v ≈ 35 radians per sec. Use a calculator.

Formulas for Angular and Linear Speed

Angular Speed V

Linear Speed v

V =U

(v in radians per unit time t, u in radians)

v =st

v =r Ut

v = rV

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128 CHAPTER 3 Radian Measure and the Unit Circle

If the radius (distance) from the tip of the racket to the wrist joint is 2 ft, then the speed at the tip of the racket is

v = rv Formula for linear speed v ≈ 21352 Let r = 2 and v = 35. v = 70 ft per sec, or about 48 mph. Use a calculator.

In a tennis serve the arm rotates at the shoulder, so the final speed of the racket is considerably greater. (Source: Cooper, J. and R. Glassow, Kinesiology, Second Edition, C.V. Mosby.)

EXAMPLE 1 Using Linear and Angular Speed Formulas

Suppose that point P is on a circle with radius 10 cm, and ray OP is rotating with angular speed p18 radian per sec.

(a) Find the angle generated by P in 6 sec.

(b) Find the distance traveled by P along the circle in 6 sec.

SOLUTION

(a) Solve for u in the angular speed formula v = ut , and substitute the known quantities v = p18 radian per sec and t = 6 sec.

u = vt Angular speed formula solved for u

u = p18

162 Let v = p18 and t = 6. u = p

3 radians Multiply.

(b) To find the distance traveled by P, use the arc length formula s = r u with r = 10 cm and, from part (a), u = p3 radians.

s = r u = 10 ap3b = 10p

3 cm Let r = 10 and u = p3 .

v = rv = 10 a p18b = 5p

9 cm per sec Linear speed formula

■✔ Now Try Exercise 7.

EXAMPLE 2 Finding Angular Speed of a Pulley and Linear Speed of a Belt

A belt runs a pulley of radius 6 cm at 80 revolutions per min. See Figure 21.

(a) Find the angular speed of the pulley in radians per second.

(b) Find the linear speed of the belt in centimeters per second.

6 cm

Figure 21

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1293.4 Linear and Angular Speed

SOLUTION

(a) The angular speed 80 revolutions per min can be converted to radians per second using the following facts.

1 revolution = 2p radians and 1 min = 60 sec

We multiply by the corresponding unit fractions. Here, just as with the unit circle, the word unit means 1, so multiplying by a unit fraction is equivalent to multiplying by 1. We divide out common units in the same way that we divide out common factors.

v = 80 revolutions1 min

# 2p radians1 revolution

# 1 min60 sec

v = 160p radians60 sec

Multiply. Divide out common units.

v = 8p3

radians per sec Angular speed

(b) The linear speed v of the belt will be the same as that of a point on the cir-cumference of the pulley.

v = rv = 6 a 8p3b = 16p ≈ 50 cm per sec Linear speed

■✔ Now Try Exercise 47.

EXAMPLE 3 Finding Linear Speed and Distance Traveled by a Satellite

A satellite traveling in a circular orbit 1600 km above the surface of Earth takes 2 hr to make an orbit. The radius of Earth is approximately 6400 km. See Figure 22.

(a) Approximate the linear speed of the satellite in kilometers per hour.

(b) Approximate the distance the satellite travels in 4.5 hr.

SOLUTION

(a) The distance of the satellite from the center of Earth is approximately

r = 1600 + 6400 = 8000 km.

The angular speed 1 orbit per 2 hr can be converted to radians per hour using the fact that 1 orbit = 2p radians.

v = 1 orbit2 hr

# 2p radians1 orbit

= p radians per hr Angular speed

Earth

Satellite

NOT TO SCALE

1600km

6400km

Figure 22

We now use the formula for linear speed with r = 8000 km and v = p radi-ans per hr.

v = rv = 8000p ≈ 25,000 km per hr Linear speed

(b) To approximate the distance traveled by the satellite, we use s = vt.

s = vt Formula for arc length s = 8000p14.52 Let v = 8000p and t = 4.5. s ≈ 110,000 km Multiply. Approximate to two significant digits.

Unit fraction: 2p radians1 orbit = 1

This is similar to the distance formula d = rt.

■✔ Now Try Exercise 45.

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130 CHAPTER 3 Radian Measure and the Unit Circle

CONCEPT PREVIEW Fill in the blank to correctly complete each sentence. As necessary, refer to the figure that shows point P moving at a constant speed along the unit circle.

3.4 Exercises

O B (1, 0)

P moves ata constantspeed alongthe unit circle.

(0, 1)

(0, −1)

(−1, 0)

Suppose that point P is on a circle with radius r, and ray OP is rotating with angular speed v. Use the given values of r, v, and t to do the following. See Example 1.

(a) Find the angle generated by P in time t.

(b) Find the distance traveled by P along the circle in time t.

7. r = 20 cm, v = p12 radian per sec, t = 6 sec

8. r = 30 cm, v = p10 radian per sec, t = 4 sec

9. r = 8 in., v = p3 radians per min, t = 9 min

10. r = 12 ft, v = 8p radians per min, t = 5 min

Use the formula v = ut to find the value of the missing variable.

11. v = 2p3 radians per sec, t = 3 sec 12. v =p4 radian per min, t = 5 min

13. v = 0.91 radian per min, t = 8.1 min 14. v = 4.3 radians per min, t = 1.6 min

15. u = 3p4 radians, t = 8 sec 16. u =2p5 radians, t = 10 sec

17. u = 3.871 radians, t = 21.47 sec 18. u = 5.225 radians, t = 2.515 sec

19. u = 2p9 radian, v =5p27 radian per min

20. u = 3p8 radians, v =p24 radian per min

1. The measure of how fast the position of point P is changing is the .

2. The measure of how fast angle POB is changing is the .

3. If the angular speed of point P is 1 radian per sec, then P will move around the entire unit circle in sec.

4. If the angular speed of point P is p radians per sec, then the linear speed is unit(s) per sec.

5. An angular speed of 1 revolution per min on the unit circle is equivalent to an angu-lar speed, v, of radians per min.

6. If P is rotating with angular speed p2 radians per sec, then the distance traveled by P in 10 sec is units.

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1313.4 Linear and Angular Speed

The formula v = ut can be rewritten as u = vt. Substituting vt for u converts s = r u to s = rvt. Use the formula s = rvt to find the value of the missing variable.

27. r = 6 cm, v = p3 radians per sec, t = 9 sec

28. r = 9 yd, v = 2p5 radians per sec, t = 12 sec

29. s = 6p cm, r = 2 cm, v = p4 radian per sec

30. s = 12p5 m , r =32 m , v =

2p5 radians per sec

31. s = 3p4 km, r = 2 km, t = 4 sec

32. s = 8p9 m , r =43 m , t = 12 sec

Use the formula v = rv to find the value of the missing variable.

21. r = 12 m, v = 2p3 radians per sec

22. r = 8 cm, v = 9p5 radians per sec

23. v = 9 m per sec, r = 5 m

24. v = 18 ft per sec, r = 3 ft

25. v = 12 m per sec, v = 3p2 radians per sec

26. v = 24.93 cm per sec, v = 0.3729 radian per sec

Solve each problem. See Examples 1–3.

43. Speed of a Bicycle The tires of a bicycle have radius 13.0 in. and are turning at the rate of 215 revolutions per min. See the figure. How fast is the bicycle traveling in miles per hour? (Hint: 5280 ft = 1 mi)

13.0 in.

Find the angular speed v for each of the following.

33. the hour hand of a clock 34. the second hand of a clock

35. the minute hand of a clock 36. a gear revolving 300 times per min

Find the linear speed v for each of the following.

37. the tip of the minute hand of a clock, if the hand is 7 cm long

38. the tip of the second hand of a clock, if the hand is 28 mm long

39. a point on the edge of a flywheel of radius 2 m, rotating 42 times per min

40. a point on the tread of a tire of radius 18 cm, rotating 35 times per min

41. the tip of a propeller 3 m long, rotating 500 times per min (Hint: r = 1.5 m)

42. a point on the edge of a gyroscope of radius 83 cm, rotating 680 times per min

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132 CHAPTER 3 Radian Measure and the Unit Circle

44. Hours in a Martian Day Mars rotates on its axis at the rate of about 0.2552 radian per hr. Approximately how many hours are in a Martian day (or sol)? (Source: World Almanac and Book of Facts.)

Opposite sides of Mars

45. Angular and Linear Speeds of Earth The orbit of Earth about the sun is almost circular. Assume that the orbit is a circle with radius 93,000,000 mi. Its angular and linear speeds are used in designing solar-power facilities.

(a) Assume that a year is 365 days, and find the angle formed by Earth’s movement in one day.

(b) Give the angular speed in radians per hour.

46. Angular and Linear Speeds of Earth Earth revolves on its axis once every 24 hr. Assuming that Earth’s radius is 6400 km, find the following.

(a) angular speed of Earth in radians per hour

(b) linear speed at the North Pole or South Pole

(d) approximate linear speed at Salem, Oregon (halfway from the equator to the North Pole)

47. Speeds of a Pulley and a Belt The pulley shown has a radius of 12.96 cm. Suppose it takes 18 sec for 56 cm of belt to go around the pulley.

(a) Find the linear speed of the belt in centimeters per second.

(b) Find the angular speed of the pulley in radians per second.

48. Angular Speeds of Pulleys The two pulleys in the figure have radii of 15 cm and 8 cm, respectively. The larger pulley rotates 25 times in 36 sec. Find the angular speed of each pulley in radians per second.

49. Radius of a Spool of Thread A thread is being pulled off a spool at the rate of 59.4 cm per sec. Find the radius of the spool if it makes 152 revolutions per min.

50. Time to Move along a Railroad Track A railroad track is laid along the arc of a circle of radius 1800 ft. The circular part of the track subtends a central angle of 40°. How long (in seconds) will it take a point on the front of a train traveling 30.0 mph to go around this portion of the track?

51. Angular Speed of a Motor Propeller The propeller of a 90-horsepower outboard motor at full throttle rotates at exactly 5000 revolutions per min. Find the angular speed of the propeller in radians per second.

52. Linear Speed of a Golf Club The shoulder joint can rotate at 25.0 radians per sec. If a golfer’s arm is straight and the distance from the shoulder to the club head is 5.00 ft, find the linear speed of the club head from shoulder rotation. (Source: Cooper, J. and R. Glassow, Kinesiology, Second Edition, C.V. Mosby.)

12.96cm

15 cm 8 cm

93,000,000 mi

Sun

Earth

NOT TOSCALE

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Chapter 3 Test Prep

Key Terms

3.1 radian circumference

3.2 latitude sector of a circle subtend

degree of curvature nautical mile statute mile longitude

3.3 unit circle circular functions reference arc

3.4 linear speed v angular speed v unit fraction

Quick ReviewConcepts Examples

u = 1 radian

Convert 135° to radians.

135° = 135 a p180

radianb = 3p4

radians

Convert - 5p3 radians to degrees.

- 5p3

radians = - 5p3

a180°pb = -300°

Radian Measure

An angle with its vertex at the center of a circle that inter-cepts an arc on the circle equal in length to the radius of the circle has a measure of 1 radian.

180° = P radians Degree/Radian Relationship

Converting between Degrees and Radians

Multiply a degree measure by p180 radian and simplify to convert to radians.

Multiply a radian measure by 180°p and simplify to con-vert to degrees.

3.1

Applications of Radian Measure

Arc LengthThe length s of the arc intercepted on a circle of radius r by a central angle of measure u radians is given by the product of the radius and the radian measure of the angle.

s = r U, where U is in radians

Area of a SectorThe area & of a sector of a circle of radius r and central angle u is given by the following formula.

& =12

r2 U, where U is in radians

3.2

Find the central angle u in the figure.

u = sr= 3

4 radian 0 r = 4

s = 3u

Find the area & of the sector in the figure above.

& = 12

1422 a34b = 6 sq units

133CHAPTER 3 Test Prep

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134 CHAPTER 3 Radian Measure and the Unit Circle

Concepts Examples

Use the unit circle to find each value.

sin 5p6

= 12

cos 3p2

= 0

tan p

4=222222

= 1

csc 7p4

= 1

- 222 = -22 sec

7p6

= 1

- 232 = - 2233 cot p

12232

= 233

sin 0 = 0

cos p

2= 0

Find the exact value of s in C 0, p2 D if cos s = 232 .In C 0, p2 D , the arc length s = p6 is associated with the

point Q232 , 12R . The first coordinate iscos s = cos p

6= 23

2 .

Thus we have s = p6 .

The Unit Circle and Circular Functions3.3

A belt runs a machine pulley of radius 8 in. at 60 revolu-tions per min.

(a) Find the angular speed v in radians per minute.

v = 60 revolutions1 min

# 2p radians1 revolution

v = 120p radians per min

(b) Find the linear speed v in inches per minute.

v = rv

v = 81120p2 v = 960p in. per min

Circular FunctionsStart at the point 11, 02 on the unit circle x2 + y2 = 1 and measure off an arc of length + s + along the circle, moving counterclockwise if s is positive and clockwise if s is nega-tive. Let the endpoint of the arc be at the point 1x, y2. The six circular functions of s are defined as follows. (Assume that no denominators are 0.)

sin s = y cos s = x tan s =yx

csc s =1y

sec s =1x

cot s =xy

The Unit Circle

00°180°

60°90°

150°

210°

300°

360°

315°

330°

2PP

135°

120°

225°

240°270°

45°

30°

3P2

5P3

7P4

4P3

5P4

7P6

5P6

3P4

2P3

11P6

(0, 1)

(1, 0)

(0, –1)

(–1, 0)x

( , )√32

12( , – )√3212(– , – )√32

( , )√22√22

12( , – )√32

( , )√32

(– , )√32(– , )√22√22

( , – )√22√22

12(– , – )√32

(– , – )√22√22

(– , )√32

The unit circle x2 + y2 = 1

Linear and Angular Speed

Formulas for Angular and Linear Speed

3.4

Angular Speed V Linear Speed v

V =U

(v in radians per unit time t, u in radians)

v =st

v =r Ut

v = rV

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135CHAPTER 3 Review Exercises

Chapter 3 Review Exercises

2 in.

12121212121212121212

1111111111222222

111111111111111111111

1111111111111110

489

Concept Check Work each problem.

1. What is the meaning of “an angle with measure 2 radians”?

2. Consider each angle in standard position having the given radian measure. In what quadrant does the terminal side lie?

(a) 3 (b) 4 (c) -2 (d) 7

3. Find three angles coterminal with an angle of 1 radian.

4. Give an expression that generates all angles coterminal with an angle of p6 radian. Let n represent any integer.

Convert each degree measure to radians. Leave answers as multiples of p.

5. 45° 6. 120° 7. 175° 8. 330° 9. 800° 10. 1020°

Convert each radian measure to degrees.

11. 5p4

12. 9p10

13. 8p3

14. 6p5

15. - 11p18

16. - 21p5

Solve each problem. Use a calculator as necessary.

21. Arc Length The radius of a circle is 15.2 cm. Find the length of an arc of the circle

intercepted by a central angle of 3p4 radians.

22. Arc Length Find the length of an arc intercepted by a central angle of 0.769 radian on a circle with radius 11.4 cm.

23. Angle Measure Find the measure (in degrees) of a central angle that intercepts an arc of length 7.683 cm in a circle of radius 8.973 cm.

24. Angle Measure Find the measure (in radians) of a central angle whose sector has

area 50p3 cm2 in a circle of radius 10 cm.

25. Area of a Sector Find the area of a sector of a circle having a central angle of 21° 40′ in a circle of radius 38.0 m.

26. Area of a Sector A central angle of 7p4 radians forms a sector of a circle. Find the area of the sector if the radius of the circle is 28.69 in.

27. Diameter of the Moon The dis-tance to the moon is approximately 238,900 mi. Use the arc length formula to estimate the diameter d of the moon if angle u in the fig-ure is measured to be 0.5170°.

28. Concept Check Using s = r u and & = 12 r2 u, express & in terms of s and u.

29. Concept Check The hour hand of a wall clock measures 6 in. from its tip to the center of the clock.

(a) Through what angle (in radians) does the hour hand pass between 1 o’clock and 3 o’clock?

(b) What distance does the tip of the hour hand travel during the time period from 1 o’clock to 3 o’clock?

Suppose the tip of the minute hand of a clock is 2 in. from the center of the clock. For each duration, determine the distance traveled by the tip of the minute hand. Leave an-swers as multiples of p.

17. 15 min 18. 20 min 19. 3 hr 20. 8 hr

NOT TO SCALE

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136 CHAPTER 3 Radian Measure and the Unit Circle

30. Concept Check What would happen to the central angle for a given arc length of a circle if the circle’s radius were doubled? (Assume everything else is unchanged.)

Distance between Cities Assume that the radius of Earth is 6400 km.

31. Find the distance in kilometers between cities on a north-south line that are on lati-tudes 28° N and 12° S, respectively.

32. Two cities on the equator have longitudes of 72° E and 35° W, respectively. Find the distance between the cities.

Concept Check Find the measure of each central angle u (in radians) and the area of each sector.

33.

1.5

34.

Find each exact function value.

35. tan p

3 36. cos

2p3

37. sin a - 5p6b

38. tan a - 7p3b 39. csc a - 11p

6b 40. cot1-13p2

Concept Check Without using a calculator, determine which of the two values is greater.

41. tan 1 or tan 2 42. sin 1 or tan 1 43. cos 2 or sin 2

44. Concept Check Match each domain in Column II with the appropriate circular function pair in Column I.

I II

(a) sine and cosine A. 1-∞, ∞2(b) tangent and secant B. 5s + s ≠ np, where n is any integer6(c) cotangent and cosecant C. 5s + s ≠ 12n + 12 p2 , where n is any integer6

Find a calculator approximation to four decimal places for each circular function value.

45. sin 1.0472 46. tan 1.2275 47. cos1-0.2443248. cot 3.0543 49. sec 7.3159 50. csc 4.8386

Find the approximate value of s, to four decimal places, in the interval C 0, p2 D that makes each statement true.

51. cos s = 0.9250 52. tan s = 4.0112 53. sin s = 0.4924

54. csc s = 1.2361 55. cot s = 0.5022 56. sec s = 4.5600

Find the exact value of s in the given interval that has the given circular function value.

57. c 0, p2d ; cos s = 22

2 58. cp

2 , p d ; tan s = -23

59. cp, 3p2d ; sec s = - 223

3 60. c 3p

2 , 2p d ; sin s = - 1

M03_LHSD7642_11_AIE_C03_pp099-138.indd 136 02/11/15 3:36 pm

137CHAPTER 3 Review Exercises

0.2

1.4 1.4

Figure A

Figure B

65. (Modeling) Archaeology An archaeology professor believes that an unearthed fragment is a piece of the edge of a circular ceremonial plate and uses a formula that will give the radius of the original plate using measurements from the fragment, shown in Figure A. Measurements are in inches.

In Figure B, a is 12 the length of chord NP, and b is the distance from the mid-point of chord NP to the circle. According to the formula, the radius r of the circle, OR, is given by

r = a2 + b22b

What is the radius of the original plate from which the fragment came?

66. (Modeling) Phase Angle of the Moon Because the moon orbits Earth, we observe different phases of the moon during the period of a month. In the figure, t is the phase angle.

Suppose that point P is on a circle with radius r, and ray OP is rotating with angular speed v. Use the given values of r, v, and t to do the following.

(a) Find the angle generated by P in time t.

(b) Find the distance traveled by P along the circle in time t.

61. r = 15 cm, v = 2p3 radians per sec, t = 30 sec

62. r = 45 ft, v = p36 radian per min, t = 12 min

Solve each problem.

63. Linear Speed of a Flywheel Find the linear speed of a point on the edge of a fly-wheel of radius 7 cm if the flywheel is rotating 90 times per sec.

64. Angular Speed of a Ferris Wheel A Ferris wheel has radius 25 ft. A person takes a seat, and then the wheel turns 5p6 radians.

(a) How far is the person above the ground to the nearest foot?

(b) If it takes 30 sec for the wheel to turn 5p6 radians, what is the angular speed of the wheel?

EarthSun

Moon

The phase F of the moon is modeled by

F1t2 = 12

11 - cos t2 and gives the fraction of the moon’s face that is illuminated by the sun. (Source:

Duffet-Smith, P., Practical Astronomy with Your Calculator, Cambridge University Press.) Evaluate each expression and interpret the result.

(a) F102 (b) F ap2b (c) F1p2 (d) F a3p

M03_LHSD7642_11_AIE_C03_pp099-138.indd 137 02/11/15 3:36 pm

138 CHAPTER 3 Radian Measure and the Unit Circle

Convert each degree measure to radians.

1. 120° 2. -45° 3. 5° (to the nearest thousandth)

Convert each radian measure to degrees.

4. 3p4

5. - 7p6

6. 4 (to the nearest minute)

7. A central angle of a circle with radius 150 cm intercepts an arc of 200 cm. Find each measure.

(a) the radian measure of the angle (b) the area of a sector with that central angle

8. Rotation of Gas Gauge Arrow The arrow on a car’s gasoline gauge is 12 in. long. See the figure. Through what angle does the arrow rotate when it moves 1 in. on the gauge?

Find each exact function value.

9. sin 3p4

10. cos a - 7p6b 11. tan 3p

12. sec 8p3

13. tan p 14. cos 3p2

15. Determine the six exact circular function values of s in the figure.

16. Give the domains of the six circular functions.

17. (a) Use a calculator to approximate s in the interval C 0, p2 D if sin s = 0.8258 .(b) Find the exact value of s in the interval C 0, p2 D if cos s = 12 .

18. Angular and Linear Speed of a Point Suppose that point P is on a circle with radius 60 cm, and ray OP is rotating with angular speed p12 radian per sec.

(a) Find the angle generated by P in 8 sec.

(b) Find the distance traveled by P along the circle in 8 sec.

19. Orbital Speed of Jupiter It takes Jupiter 11.86 yr to complete one orbit around the sun. See the fig-ure. If Jupiter’s average distance from the sun is 483,800,000 mi, find its orbital speed (speed along its orbital path) in miles per second. (Source: World Almanac and Book of Facts.)

20. Ferris Wheel A Ferris wheel has radius 50.0 ft. A person takes a seat, and then the wheel turns 2p3 radians.

(a) How far is the person above the ground?

(b) If it takes 30 sec for the wheel to turn 2p3 radians, what is the angular speed of the wheel?

Chapter 3 Test

Empty Full

483,800,000mi

Sun Jupiter

NOT TO SCALE

s = 76P

x2 + y2 = 1

0 (1, 0)(–1, 0)

(0, 1)

(0, –1)

M03_LHSD7642_11_AIE_C03_pp099-138.indd 138 02/11/15 3:36 pm

139

Phenomena that repeat in a regular pattern, such as average monthly temperature, fractional part of the moon’s illumination, and high and low tides, can be modeled by periodic functions.

Graphs of the Sine and Cosine Functions

Translations of the Graphs of the Sine and Cosine Functions

Chapter 4 Quiz

Graphs of the Tangent and Cotangent Functions

Graphs of the Secant and Cosecant Functions

Summary Exercises on Graphing Circular Functions

Harmonic Motion

4.1

4.2

4.3

4.4

4.5

Graphs of the Circular Functions4

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140 CHAPTER 4 Graphs of the Circular Functions

Periodic Functions Phenomena that repeat with a predictable pattern, such as tides, phases of the moon, and hours of daylight, can be modeled by sine and cosine functions. These functions are periodic. The periodic graph in Figure 1 represents a normal heartbeat.

4.1 Graphs of the Sine and Cosine Functions■ Periodic Functions■ Graph of the Sine

Function■ Graph of the Cosine

Function■ Techniques for

Graphing, Amplitude, and Period

■ Connecting Graphs with Equations

■ A Trigonometric Model

Periodic Function

A periodic function is a function ƒ such that

ƒ 1x 2 = ƒ 1x + np 2 ,for every real number x in the domain of ƒ, every integer n, and some posi-tive real number p. The least possible positive value of p is the period of the function.

LOOKING AHEAD TO CALCULUSPeriodic functions are used throughout

calculus, so it is important to know

their characteristics. One use of

these functions is to describe the

location of a point in the plane using

polar coordinates, an alternative to rectangular coordinates.

The circumference of the unit circle is 2p, so the least value of p for which the sine and cosine functions repeat is 2p. Therefore, the sine and cosine func-tions are periodic functions with period 2P. For every positive integer n,

sin x = sin 1x + n # 2P 2 and cos x = cos 1x + n # 2P 2 .

Figure 1

Figure 2

(–1, 0) (1, 0)

( x, y) =

(0, –1)

(0, 1)

0 0p

(cos s, sin s)

The unit circlex2 + y2 = 1

3p2

Periodic functions are defined as follows.

Graph of the Sine Function We have seen that for a real number s, the point on the unit circle corresponding to s has coordinates 1cos s, sin s2. See Figure 2. Trace along the circle to verify the results shown in the table.

As s Increases from sin s cos s

0 to p2 Increases from 0 to 1 Decreases from 1 to 0p2 to p Decreases from 1 to 0 Decreases from 0 to -1

p to 3p2 Decreases from 0 to -1 Increases from -1 to 03p2 to 2p Increases from -1 to 0 Increases from 0 to 1

To avoid confusion when graphing the sine function, we use x rather than s. This corresponds to the letters in the xy-coordinate system. Selecting key values of x and finding the corresponding values of sin x leads to the table in Figure 3.

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1414.1 Graphs of the Sine and Cosine Functions

To obtain the traditional graph in Figure 3, we plot the points from the table, use symmetry, and join them with a smooth curve. Because y = sin x is periodic with period 2p and has domain 1-∞, ∞2, the graph continues in the same pat-tern in both directions. This graph is a sine wave, or sinusoid.

The sine function is related to the unit circle. Its domain consists of real numbers corresponding to angle measures (or arc lengths) on the unit circle. Its range corresponds to y-coordinates (or sine values) on the unit circle.

Consider the unit circle in Figure 2 and assume that the line from the origin to some point on the circle is part of the pedal of a bicycle, with a foot placed on the circle itself. As the pedal is rotated from 0 radians on the horizontal axis through various angles, the angle (or arc length) giving the pedal’s location and its corresponding height from the horizontal axis given by sin x are used to cre-ate points on the sine graph. See Figure 4 on the next page.

NOTE A function ƒ is an odd function if for all x in the domain of ƒ,

ƒ 1−x 2 = −ƒ 1x 2 .The graph of an odd function is symmetric with respect to the origin. This means that if 1x, y2 belongs to the function, then 1-x, -y2 also belongs to the function. For example, Ap2 , 1 B and A - p2 , -1 B are points on the graph of y = sin x, illustrating the property sin1-x2 = -sin x.

Figure 3

0–1

–2

f(x) = sin x, –2P ≤ x ≤ 2P

–2p 2p–p p–

–

3p2

Sine Function f 1x 2 = sin xDomain: 1-∞, ∞2 Range: 3-1, 14

x y

0 0p6

222

232

p2 1

p 03p2 -1

2p 0

The graph is continuous over its entire domain, 1-∞, ∞2. Its x-intercepts have x-values of the form np, where n is an integer.

Its period is 2p.

The graph is symmetric with respect to the origin, so the function is an odd function. For all x in the domain, sin1-x2 = -sin x.

f(x) = sin x

−4

11p4

−

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142 CHAPTER 4 Graphs of the Circular Functions

LOOKING AHEAD TO CALCULUSThe discussion of the derivative of a

function in calculus shows that for the

sine function, the slope of the tangent

line at any point x is given by cos x.

For example, look at the graph of

y = sin x and notice that a tangent line at x = {p2 , {

3p2 , {

5p2 , . . . will be

horizontal and thus have slope 0. Now

look at the graph of y = cos x and see that for these values, cos x = 0.

Cosine Function f 1x 2 = cos xDomain: 1-∞, ∞2 Range: 3-1, 14

x y

0 1p

6 232

4 222

312

2 0

p -13p2 0

2p 1

The graph is continuous over its entire domain, 1-∞, ∞2. Its x-intercepts have x-values of the form 12n + 12p2 , where n is an integer. Its period is 2p.

The graph is symmetric with respect to the y-axis, so the function is an even function. For all x in the domain, cos1-x2 = cos x.

0–1

–2

f(x) = cos x, –2p ≤ x ≤ 2p

–2p 2p–p p–– 3p2

3p2

Figure 5

f(x) = cos x

−4

11p4

−

Figure 4

(1, 0)

(0, –1)

(0, 1)

–1

y = sin x

3p2

2p3

3p2

2p3

7p6

7p6 –

, 1( )

3p2

, –1( )

, –( )12 √32, ( )12√32

, ––( )12√32

NOTE A function ƒ is an even function if for all x in the domain of ƒ,

ƒ 1−x 2 = ƒ 1x 2 .The graph of an even function is symmetric with respect to the y-axis. This means that if 1x, y2 belongs to the function, then 1-x, y2 also belongs to the function. For example, Ap2 , 0 B and A - p2 , 0 B are points on the graph of y = cos x, illustrating the property cos1-x2 = cos x.

Graph of the Cosine Function The graph of y = cos x in Figure 5 is the graph of the sine function shifted, or translated, P2 units to the left.

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1434.1 Graphs of the Sine and Cosine Functions

The calculator graphs of ƒ1x2 = sin x in Figure 3 and ƒ1x2 = cos x in Figure 5 are shown in the ZTrig viewing windowc - 11p

4 ,

11p4d by 3-4, 44 A11p4 ≈ 8.639379797 B

of the TI-84 Plus calculator, with Xscl = p2 and Yscl = 1. (Other models have different trigonometry viewing windows.)■

The graph of y = 2 sin x is shown in blue, and that of y = sin x is shown in red. Compare to Figure 6.

−4

11p4

−

x 0 p2 p3p2 2p

sin x 0 1 0 -1 02 sin x 0 2 0 -2 0

EXAMPLE 1 Graphing y = a sin x

Graph y = 2 sin x, and compare to the graph of y = sin x.

SOLUTION For a given value of x, the value of y is twice what it would be for y = sin x. See the table of values. The change in the graph is the range, which becomes 3-2, 24. See Figure 6, which also includes a graph of y = sin x.

Figure 6

0–1

–2

y = sin x

Period: 2p y = 2 sin x

–2p

–p p–

–

3p2

The amplitude of a periodic function is half the difference between the maximum and minimum values. It describes the height of the graph both above and below a horizontal line passing through the “middle” of the graph. Thus, for the basic sine function y = sin x (and also for the basic cosine function y = cos x), the amplitude is computed as follows.

31 - 1-124 = 12

122 = 1 Amplitude of y = sin xFor y = 2 sin x, the amplitude is

32 - 1-224 = 12

142 = 2. Amplitude of y = 2 sin xWe can think of the graph of y = a sin x as a vertical stretching of the

graph of y = sin x when a + 1 and a vertical shrinking when 0 * a * 1.■✔ Now Try Exercise 15.

Amplitude

The graph of y = a sin x or y = a cos x, with a ≠ 0, will have the same shape as the graph of y = sin x or y = cos x, respectively, except with range 3- 'a ' , 'a ' 4. The amplitude is 'a ' .

Techniques for Graphing, Amplitude, and Period The examples that follow show graphs that are “stretched” or “compressed” (shrunk) either vertically, horizontally, or both when compared with the graphs of y = sin x or y = cos x.

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144 CHAPTER 4 Graphs of the Circular Functions

While the coefficient a in y = a sin x or y = a cos x affects the amplitude of the graph, the coefficient of x in the argument affects the period. Consider y = sin 2x. We can complete a table of values for the interval 30, 2p4.

NOTE One method to divide an interval into four equal parts is as follows.

Step 1 Find the midpoint of the interval by adding the x-values of the end-points and dividing by 2.

Step 2 Find the quarter points (the midpoints of the two intervals found in Step 1) using the same procedure.

EXAMPLE 2 Graphing y = sin bx

Graph y = sin 2x, and compare to the graph of y = sin x.

SOLUTION In this function the coefficient of x is 2, so b = 2 and the period is 2p2 = p. Therefore, the graph will complete one period over the interval 30, p4.

We can divide the interval 30, p4 into four equal parts by first finding its midpoint: 12 10 + p2 = p2 . The quarter points are found next by determining the midpoints of the two intervals C 0, p2 D and Cp2 , p D .

a0 + p2b = p

4 and

ap2

+ pb = 3p4

Quarter points

12 Ap2 + p B = 12 A3p2 B = 3p4

x 0 p4

3p4

p 5p4

3p2

7p4

sin 2 x 0 1 0 -1 0 1 0 -1 0

Note that one complete cycle occurs in p units, not 2p units. Therefore, the period here is p, which equals 2p2 . Now consider y = sin 4x. Look at the next table.

x 0 p8

3p8

5p8

3p4

7p8

sin 4 x 0 1 0 -1 0 1 0 -1 0

These values suggest that one complete cycle is achieved in p2 or 2p4 units, which

is reasonable because

sin a4 # p2b = sin 2p = 0.

In general, the graph of a function of the form y = sin bx or y = cos bx, for b + 0, will have a period different from 2P when b 3 1.

To see why this is so, remember that the values of sin bx or cos bx will take on all possible values as bx ranges from 0 to 2p. Therefore, to find the period of either of these functions, we must solve the following three-part inequality.

0 … bx … 2p bx ranges from 0 to 2p.

0 … x … 2pb

Divide each part by the positive number b.

Thus, the period is 2Pb . By dividing the interval C 0, 2pb D into four equal parts, we obtain the values for which sin bx or cos bx is -1, 0, or 1. These values will give minimum points, x-intercepts, and maximum points on the graph. (If a function has b 6 0, then identities can be used to rewrite the function so that b 7 0.)

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1454.1 Graphs of the Sine and Cosine Functions

The interval 30, p4 is divided into four equal parts using these x-values. 0,

4 ,

2 ,

3p4

, p

Left First-quarter Midpoint Third-quarter Right endpoint point point endpoint

We plot the points from the table of values given at the top of the previous page, and join them with a smooth sinusoidal curve. More of the graph can be sketched by repeating this cycle, as shown in Figure 7. The amplitude is not changed.

Figure 7

0.51

0–0.5–1

y = sin x

y = sin 2x

2p–p p– –3p2

3p2

We can think of the graph of y = sin bx as a horizontal stretching of the graph of y = sin x when 0 * b * 1 and a horizontal shrinking when b + 1.

■✔ Now Try Exercise 27.

Period

For b 7 0, the graph of y = sin bx will resemble that of y = sin x, but with period 2pb . Also, the graph of y = cos bx will resemble that of y = cos x, but with period 2pb .

EXAMPLE 3 Graphing y = cos bx

Graph y = cos 23 x over one period.

SOLUTION The period is

2p23

= 2p , 23

= 2p # 32

= 3p. To divide by a fraction, multiply by its reciprocal.

We divide the interval 30, 3p4 into four equal parts to obtain the x-values 0, 3p4 , 3p2 ,

9p4 , and 3p that yield minimum points, maximum points, and x-intercepts.

We use these values to obtain a table of key points for one period.

This screen shows a graph of the function in Example 3. By choosing

Xscl = 3p4 , the x-intercepts, maxima, and minima coincide with tick marks on the x-axis.

−2

3p0

The amplitude is 1 because the maximum value is 1, the minimum value is -1, and 12 31 - 1-124 = 12 122 = 1. We plot these points and join them with a smooth curve. The graph is shown in Figure 8.

■✔ Now Try Exercise 25.

x 0 3p43p2

9p4 3p

23 x 0

p 3p2 2p

cos 23 x 1 0 -1 0 1

Figure 8

0–1–2

y = cos x23

3p2

3p4

9p4

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146 CHAPTER 4 Graphs of the Circular Functions

NOTE Look at the middle row of the table in Example 3. Dividing C 0, 2pb D into four equal parts gives the values 0, p2 , p,

3p2 , and 2p for this row, result-

ing here in values of -1, 0, or 1. These values lead to key points on the graph, which can be plotted and joined with a smooth sinusoidal curve.

Guidelines for Sketching Graphs of Sine and Cosine Functions

To graph y = a sin bx or y = a cos bx, with b 7 0, follow these steps.

Step 1 Find the period, 2pb . Start at 0 on the x-axis, and lay off a distance of 2pb .

Step 2 Divide the interval into four equal parts. (See the Note preceding Example 2.)

Step 3 Evaluate the function for each of the five x-values resulting from Step 2. The points will be maximum points, minimum points, and x-intercepts.

Step 4 Plot the points found in Step 3, and join them with a sinusoidal curve having amplitude 'a ' .

Step 5 Draw the graph over additional periods as needed.

EXAMPLE 4 Graphing y = a sin bx

Graph y = -2 sin 3x over one period using the preceding guidelines.

SOLUTION

Step 1 For this function, b = 3, so the period is 2p3 . The function will be graphed over the interval C 0, 2p3 D .

Step 2 Divide the interval C 0, 2p3 D into four equal parts to obtain the x-values 0, p6 ,

p3 , p2 , and

2p3 .

Step 3 Make a table of values determined by the x-values from Step 2.

Figure 9

–1

–2 y = –2 sin 3x

2p3

x 0 p6p3

2p3

3x 0 p2

p 3p2

sin 3x 0 1 0 -1 0−2 sin 3x 0 -2 0 2 0

Step 4 Plot the points 10, 02, Ap6 , -2 B , Ap3 , 0 B , Ap2 , 2 B , and A2p3 , 0 B , and join them with a sinusoidal curve having amplitude 2. See Figure 9.

Step 5 The graph can be extended by repeating the cycle.

Notice that when a is negative, the graph of y = a sin bx is a reflection across the x-axis of the graph of y = ∣a ∣ sin bx.

■✔ Now Try Exercise 29.

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1474.1 Graphs of the Sine and Cosine Functions

EXAMPLE 5 Graphing y = a cos bx (Where b Is a Multiple of P)

Graph y = -3 cos px over one period.

SOLUTION

Step 1 Here b = p and the period is 2pp = 2, so we will graph the function over the interval 30, 24.

Step 2 Dividing 30, 24 into four equal parts yields the x-values 0, 12 , 1, 32 , and 2.Step 3 Make a table using these x-values.

x 0 12 132 2

Px 0 p2 p3p2 2p

cos Px 1 0 -1 0 1−3 cos Px -3 0 3 0 -3

Step 4 Plot the points 10, -32, A12 , 0 B, 11, 32, A32 , 0 B, and 12, -32, and join them with a sinusoidal curve having amplitude ' -3 ' = 3. See Figure 10.

Step 5 The graph can be extended by repeating the cycle.

Notice that when b is an integer multiple of P, the first coordinates of the x-intercepts of the graph are rational numbers.

■✔ Now Try Exercise 37.

Figure 10

–3

–2

–1

y = –3 cos Px

EXAMPLE 6 Determining an Equation for a Graph

Determine an equation of the form y = a cos bx or y = a sin bx, where b 7 0, for the given graph.

SOLUTION This graph is that of a cosine function that is reflected across its horizontal axis, the x-axis. The amplitude is half the distance between the max-imum and minimum values.

32 - 1-224 = 12

142 = 2 The amplitude 'a ' is 2.Because the graph completes a cycle on the interval 30, 4p4, the period is 4p. We use this fact to solve for b.

4p = 2pb

Period = 2pb

4pb = 2p Multiply each side by b.

b = 12

Divide each side by 4p.

An equation for the graph is

y = -2 cos 12

x-axis reflection Horizontal stretch

■✔ Now Try Exercise 41.

–3–2–1

4p3p2pp

Connecting Graphs with Equations

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148 CHAPTER 4 Graphs of the Circular Functions

A Trigonometric Model Sine and cosine functions may be used to model many real-life phenomena that repeat their values in a cyclical, or peri odic, manner. Average temperature in a certain geographic location is one such example.

EXAMPLE 7 Interpreting a Sine Function Model

The average temperature (in °F) at Mould Bay, Canada, can be approximated by the function

ƒ1x2 = 34 sin cp6

1x - 4.32 d ,where x is the month and x = 1 corresponds to January, x = 2 to February, and so on.

(a) To observe the graph over a two-year interval, graph ƒ in the window 30, 254 by 3-45, 454.

(b) According to this model, what is the average temperature during the month of May?

SOLUTION

(a) The graph of ƒ1x2 = 34 sin Cp6 1x - 4.32 D is shown in Figure 11. Its ampli-tude is 34, and the period is

2pp6

= 2p , p6

= 2p # 6p

= 12. Simplify the complex fraction.

Function ƒ has a period of 12 months, or 1 year, which agrees with the changing of the seasons.

−45

250

Figure 11

(b) May is the fifth month, so the average temperature during May is

ƒ152 = 34 sin cp6

15 - 4.32 d ≈ 12°F. Let x = 5 in the given function.See the display at the bottom of the screen in Figure 11.

(c) From the graph, it appears that the average annual temperature is about 0°F because the graph is centered vertically about the line y = 0.

■✔ Now Try Exercise 57.

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. The amplitude of the graphs of the sine and cosine functions is , and the period of each is .

2. For the x-values 0 to p2 , the graph of the sine function and that of the (rises/falls) cosine function . (rises/falls)

4.1 Exercises

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1494.1 Graphs of the Sine and Cosine Functions

Concept Check Match each function with its graph in choices A–F.

7. y = -sin x 8. y = -cos x 9. y = sin 2x

10. y = cos 2x 11. y = 2 sin x 12. y = 2 cos x

x0 pp

–1

0 pp2

–2

0 2pp

x0 2pp

–1

x0 2pp

–1

–2

2pp

Graph each function over the interval 3-2p, 2p4. Give the amplitude. See Example 1.13. y = 2 cos x 14. y = 3 sin x 15. y =

sin x

16. y = 34

cos x 17. y = -cos x 18. y = -sin x

19. y = -2 sin x 20. y = -3 cos x 21. y = sin1-x222. Concept Check In Exercise 21, why is the graph the same as that of y = -sin x?

3. The graph of the sine function crosses the x-axis for all numbers of the form , where n is an integer.

4. The domain of both the sine and cosine functions (in interval form) is , and the range is .

5. The least positive number x for which cos x = 0 is .

6. On the interval 3p, 2p4, the function values of the cosine function increase from to .

Graph each function over a two-period interval. Give the period and amplitude. See Examples 2–5.

23. y = sin 12

x 24. y = sin 23

x 25. y = cos 34

26. y = cos 13

x 27. y = sin 3x 28. y = cos 2x

29. y = 2 sin 14

x 30. y = 3 sin 2x 31. y = -2 cos 3x

32. y = -5 cos 2x 33. y = cos px 34. y = -sin px

35. y = -2 sin 2px 36. y = 3 cos 2px 37. y =12

cos p

2 x

38. y = - 23

sin p

4 x 39. y = p sin px 40. y = -p cos px

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150 CHAPTER 4 Graphs of the Circular Functions

Connecting Graphs with Equations Determine an equation of the form y = a cos bx or y = a sin bx, where b 7 0, for the given graph. See Example 6.

41.

–1

–2

–3

2pp

3p2

42.

–1

–2

–3

2pp0

3p2

43.

–1

–2

–3

2pp0 3p2

44.

–1

–2

–3

2pp0 3p2

45.

–1

–2

–3

2pp0

3p2

46.

–1

–2

–3

2pp0

3p2

(Modeling) Solve each problem.

47. Average Annual Temperature Scientists believe that the average annual tempera-ture in a given location is periodic. The average temperature at a given place dur-ing a given season fluctuates as time goes on, from colder to warmer, and back to colder. The graph shows an idealized description of the temperature (in °F) for approximately the last 150 thousand years of a particular location.

(a) Find the highest and lowest temperatures recorded.

(b) Use these two numbers to find the amplitude.

(d) What is the trend of the temperature now?

48. Blood Pressure Variation The graph gives the variation in blood pressure for a typical person. Systolic and diastolic pressures are the upper and lower limits of the periodic changes in pressure that pro-duce the pulse. The length of time between peaks is the period of the pulse.

(a) Find the systolic and diastolic pressures.

(b) Find the amplitude of the graph.

Years ago

Average Annual Temperature (Idealized)

150,

000

100,

000

50,0

80°

65° °F

50°

Time (in seconds)

Blood Pressure Variation

Pres

sure

(in

mer

cury

)

0.8 1.6

Systolicpressure

Diastolicpressure

120

Period =0.8 sec

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1514.1 Graphs of the Sine and Cosine Functions

Hana: High, +40 min, +0.1 ft; Low, +18 min, -0.2 ft

Makena: High, +1:21, -0.5 ft; Low, +1:09, -0.2 ft

Maalaea: High, +1:52, -0.1 ft; Low, +1:19, -0.2 ft

Lahaina: High, +1:18, -0.2 ft; Low, +1:01, -0.1 ft

(Modeling) Tides for Kahului Harbor The chart shows the tides for Kahului Harbor (on the island of Maui, Hawaii). To identify high and low tides and times for other Maui areas, the following adjustments must be made.

JANUARY

19 20 21 226

A.M.Noon 6

P.M.6

A.M.Noon 6

P.M.6

A.M.Noon 6

P.M.6

A.M.Noon 6

P.M.

Feet

Source: Maui News. Original chart prepared byEdward K. Noda and Associates.

(Modeling) Solve each problem.

54. Activity of a Nocturnal Animal Many activities of living organisms are periodic. For example, the graph at the right below shows the time that a certain nocturnal animal begins its evening activity.

(a) Find the amplitude of this graph.

(b) Find the period.

Apr Jun Aug Oct Dec Feb Apr

6:307:007:308:00

4:004:305:005:30

Tim

e P.

Month

Activity of a Nocturnal Animal

55. Atmospheric Carbon Dioxide At Mauna Loa, Hawaii, atmospheric carbon dioxide levels in parts per million (ppm) were measured regularly, beginning in 1958. The function

L1x2 = 0.022x2 + 0.55x + 316 + 3.5 sin 2pxcan be used to model these levels, where x is in years and x = 0 corresponds to 1960. (Source: Nilsson, A., Greenhouse Earth, John Wiley and Sons.)

(a) Graph L in the window 315, 454 by 3325, 3854.(b) When do the seasonal maximum and minimum carbon dioxide levels occur?

Use the graph to approximate each answer.

49. The graph is an example of a periodic function. What is the period (in hours)?

50. What is the amplitude?

51. At what time on January 20 was low tide at Kahului? What was the height then?

52. Repeat Exercise 51 for Maalaea.

53. At what time on January 22 was high tide at Lahaina? What was the height then?

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152 CHAPTER 4 Graphs of the Circular Functions

56. Atmospheric Carbon Dioxide Refer to Exercise 55. The carbon dioxide content in the atmosphere at Barrow, Alaska, in parts per million (ppm) can be modeled by the function

C1x2 = 0.04x2 + 0.6x + 330 + 7.5 sin 2px,where x = 0 corresponds to 1970. (Source: Zeilik, M. and S. Gregory, Introductory Astronomy and Astrophysics, Brooks/Cole.)

(a) Graph C in the window 35, 504 by 3320, 4504.(b) What part of the function causes the amplitude of the oscillations in the graph

of C to be larger than the amplitude of the oscillations in the graph of L in Exercise 55, which models Hawaii?

57. Average Daily Temperature The temperature in Anchorage, Alaska, can be approx-imated by the function

T1x2 = 37 + 21 sin c 2p365

1x - 912 d ,where T1x2 is the temperature in degrees Fahrenheit on day x, with x = 1 corre-sponding to January 1 and x = 365 corresponding to December 31. Use a calculator to estimate the temperature on the following days. (Source: World Almanac and Book of Facts.)

(a) March 15 (day 74) (b) April 5 (day 95) (c) Day 200

(d) June 25 (e) October 1 (f) December 31

58. Fluctuation in the Solar Constant The solar constant S is the amount of energy per unit area that reaches Earth’s atmosphere from the sun. It is equal to 1367 watts per m2 but varies slightly throughout the seasons. This fluctuation ∆S in S can be calculated using the formula

∆S = 0.034S sin c 2p182.5 - N2365.25

d .In this formula, N is the day number covering a four-year period, where N = 1 cor-responds to January 1 of a leap year and N = 1461 corresponds to December 31 of the fourth year. (Source: Winter, C., R. Sizmann, and L. L.Vant-Hull, Editors, Solar Power Plants, Springer-Verlag.)

(a) Calculate ∆S for N = 80, which is the spring equinox in the first year.(b) Calculate ∆S for N = 1268, which is the summer solstice in the fourth year.(c) What is the maximum value of ∆S?(d) Find a value for N where ∆S is equal to 0.

Musical Sound Waves Pure sounds produce single sine waves on an oscilloscope. Find the amplitude and period of each sine wave graph. On the vertical scale, each square represents 0.5. On the horizontal scale, each square represents 30° or p6 .

59. 60.

61. Concept Check Compare the graphs of y = sin 2x and y = 2 sin x over the interval 30, 2p4. Can we say that, in general, sin bx = b sin x for b 7 0? Explain.62. Concept Check Compare the graphs of y = cos 3x and y = 3 cos x over the interval 30, 2p4. Can we say that, in general, cos bx = b cos x for b 7 0? Explain.

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1534.2 Translations of the Graphs of the Sine and Cosine Functions

Relating Concepts

For individual or collaborative investigation (Exercises 63–66)

Connecting the Unit Circle and Sine Graph Using a TI-84 Plus calculator, adjust the settings to correspond to the following screens.

MODE FORMAT Y = EDITOR

4.2 Translations of the Graphs of the Sine and Cosine Functions

Horizontal Translations The graph of the function

y = ƒ 1x − d 2is translated horizontally compared to the graph of y = ƒ1x2. The translation is d units to the right if d 7 0 and 'd ' units to the left if d 6 0. See Figure 12.

With circular functions, a horizontal translation is a phase shift. In the function y = ƒ1x - d2, the expression x - d is the argument.

■ Horizontal Translations■ Vertical Translations■ Combinations of

Translations■ A Trigonometric Model

yy = f(x)

–3 0x

y = f (x – 4)y = f (x + 3)

Horizontal translations of y = f (x)

Figure 12

Graph the two equations (which are in parametric form), and watch as the unit circle and the sine function are graphed simultaneously. Press the TRACE key once to obtain the screen shown on the left below. Then press the up-arrow key to obtain the screen shown on the right below. The screen on the left gives a unit circle interpretation of cos 0 = 1 and sin 0 = 0. The screen on the right gives a rectangular coordinate graph interpretation of sin 0 = 0.

−2.5

−1.38

2.5

6.67

−2.5

−1.38

2.5

6.67

63. On the unit circle graph, let T = 2. Find X and Y, and interpret their values.

64. On the sine graph, let T = 2. What values of X and Y are displayed? Interpret these values with an equation in X and Y.

65. Now go back and redefine Y2T as cos1T2. Graph both equations. On the cosine graph, let T = 2. What values of X and Y are displayed? Interpret these values with an equation in X and Y.

66. Explain the relationship between the coordinates of the unit circle and the coor-dinates of the sine and cosine graphs.

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154 CHAPTER 4 Graphs of the Circular Functions

EXAMPLE 1 Graphing y = sin 1x − d 2Graph y = sin Ax - p3 B over one period.SOLUTION Method 1 For the argument x - p3 to result in all possible values throughout one period, it must take on all values between 0 and 2p, inclusive. To find an interval of one period, we solve the following three-part inequality.

0 … x - p3

… 2p Three-part inequality

3… x … 7p

3 Add p3 to each part.

Use the method described in the previous section to divide the interval Cp3 , 7p3 D into four equal parts, obtaining the following x-values.

These are key x-values.

3 ,

5p6

, 4p3

, 11p

6 ,

7p3

A table of values using these x-values follows.

x p35p6

4p3

11p6

7p3

x − P3 0p2 p

3p2 2p

sin 1x − P3 2 0 1 0 -1 0We join the corresponding points with a smooth curve to obtain the solid blue graph shown in Figure 13. The period is 2p, and the amplitude is 1.

CAUTION In Example 1, the argument of the function is Ax - p3 B. The parentheses are important here. If the function had been

y = sin x - p3

the graph would be that of y = sin x translated vertically p3 units down.

The graph can be extended through additional periods by repeating the given

portion of the graph, as necessary.

Figure 13

y = sin x

1y = sin x – ( )

–1

5p6

7p3

4p3

11p6

– 2p

Method 2 We can also graph y = sin Ax - p3 B by using a horizontal trans-lation of the graph of y = sin x. The argument x - p3 indicates that the graph will be translated p3 units to the right (the phase shift) compared to the graph of y = sin x. See Figure 13.

To graph a function using this method, first graph the basic circular func-tion, and then graph the desired function using the appropriate translation.

■✔ Now Try Exercise 39.

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1554.2 Translations of the Graphs of the Sine and Cosine Functions

We join the corresponding points with a smooth curve to obtain the solid blue graph shown in Figure 14. The period is 2p, and the amplitude is 3.

Method 2 Write y = 3 cos Ax + p4 B in the form y = a cos1x - d2.y = 3 cos ax + p

4b , or y = 3 cos c x - a - p

4b d Rewrite to subtract - p4 .

This result shows that d = - p4 . Because - p4 is negative, the phase shift is 0 - p4 0 = p4 unit to the left. The graph is the same as that of y = 3 cos x (the red

graph in the calculator screen shown in the margin), except that it is translated p4 unit to the left (the blue graph). ■✔ Now Try Exercise 41.

EXAMPLE 2 Graphing y = a cos 1x − d 2Graph y = 3 cos Ax + p4 B over one period.SOLUTION Method 1 We first solve the following three-part inequality.

0 … x + p4

… 2p Three-part inequality

- p4

… x … 7p4

Subtract p4 from each part.

Dividing this interval into four equal parts gives the following x-values. We use them to make a table of key points.

- p4

, p

4 ,

3p4

, 5p4

, 7p4

Key x-values

Figure 15

–2

y = –2 cos (3x + P)

–p p– –2p3

2p3

– p6

Figure 14

y = 3 cos x

–3

3y = 3 cos (x + )4P

3p4

5p4

7p4

–

y2 = 3 cosx

−4

11p4

−

x - p4p4

3p4

5p4

7p4

x + P4 0p2 p

3p2 2p

cos 1x + P4 2 1 0 -1 0 13 cos 1x + P4 2 3 0 -3 0 3

These x-values lead to maximum points, minimum points, and x-intercepts.

EXAMPLE 3 Graphing y = a cos 3b 1x − d 2 4Graph y = -2 cos13x + p2 over two periods.SOLUTION Method 1 We first solve the three-part inequality

0 … 3x + p … 2p

to find the interval C - p3 , p3 D . Dividing this interval into four equal parts gives the points A - p3 , -2 B , A - p6 , 0 B , 10, 22, Ap6 , 0 B , and Ap3 , -2 B . We plot these points and join them with a smooth curve. By graphing an additional half period to the left and to the right, we obtain the graph shown in Figure 15.

Method 2 First write the equation in the form y = a cos 3b1x - d24.y = -2 cos13x + p2, or y = -2 cos c3 ax + p

3b d Rewrite by factoring out 3.

Then a = -2, b = 3, and d = - p3 . The amplitude is $ -2 $ = 2, and the period is 2p3 (because the value of b is 3). The phase shift is 0 - p3 0 = p3 units to the left compared to the graph of y = -2 cos 3x. Again, see Figure 15.

■✔ Now Try Exercise 47.

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156 CHAPTER 4 Graphs of the Circular Functions

Vertical Translations The graph of a function of the form

y = c + ƒ 1x 2is translated vertically compared to the graph of y = ƒ1x2. See Figure 16. The translation is c units up if c 7 0 and is ' c ' units down if c 6 0.

Figure 16

y = –5 + f (x)

y = f (x)

y = 3 + f (x)6

–4

Vertical translations of y = f (x)

y2 = −2 cos 3x

−3

11p4

−

EXAMPLE 4 Graphing y = c + a cos bx

Graph y = 3 - 2 cos 3x over two periods.

SOLUTION We use Method 1. The values of y will be 3 greater than the cor-responding values of y in y = -2 cos 3x. This means that the graph of y = 3 - 2 cos 3x is the same as the graph of y = -2 cos 3x, vertically translated 3 units up. The period of y = -2 cos 3x is 2p3 , so the key points have these x-values.

0, p

6 , p

3 , p

2 ,

2p3

Key x-values

Use these x-values to make a table of points.

x 0 p6p3

2p3

cos 3x 1 0 -1 0 12 cos 3x 2 0 -2 0 23 − 2 cos 3x 1 3 5 3 1

The key points are shown on the graph in Figure 17, along with more of the graph, which is sketched using the fact that the function is periodic.

Figure 17

y = 3 – 2 cos 3x1

y = 3

–2p3

2p3

–p3

p3p2

–p6

–

■✔ Now Try Exercise 51.

CAUTION If we use Method 2 to graph the function y = 3 - 2 cos 3x in Example 4, we must first graph

y = -2 cos 3x

and then apply the vertical translation 3 units up. To begin, use the fact that a = -2 and b = 3 to determine that the amplitude is 2, the period is 2p3 , and the graph is the reflection of the graph of y = 2 cos 3x across the x-axis. Then, because c = 3, translate the graph of y = -2 cos 3x up 3 units. See Figure 17.

If the vertical translation is applied first, then the reflection must be across the line y = 3, not across the x-axis.

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1574.2 Translations of the Graphs of the Sine and Cosine Functions

Combinations of Translations

Further Guidelines for Sketching Graphs of Sine and Cosine Functions

To graph y= c+ a sin 3b 1x− d 2 4 or y= c+ a cos 3b 1x− d 2 4 , with b 7 0, follow these steps.

Method 1

Step 1 Find an interval whose length is one period 2pb by solving the three-part inequality 0 … b1x - d2 … 2p.

Step 2 Divide the interval into four equal parts to obtain five key x-values.

Step 3 Evaluate the function for each of the five x-values resulting from Step 2. The points will be maximum points, minimum points, and points that intersect the line y = c (“middle” points of the wave).

Step 4 Plot the points found in Step 3, and join them with a sinusoidal curve having amplitude 'a ' .

Step 5 Draw the graph over additional periods, as needed.

Method 2

Step 1 Graph y = a sin bx or y = a cos bx. The amplitude of the function is 'a ' , and the period is 2pb .

Step 2 Use translations to graph the desired function. The vertical transla-tion is c units up if c 7 0 and ' c ' units down if c 6 0. The horizontal translation (phase shift) is d units to the right if d 7 0 and 'd ' units to the left if d 6 0.

EXAMPLE 5 Graphing y = c + a sin 3b 1x − d 2 4Graph y = -1 + 2 sin14x + p2 over two periods.SOLUTION We use Method 1. We must first write the expression on the right side of the equation in the form c + a sin 3b1x - d24.y = -1 + 2 sin14x + p2, or y = -1 + 2 sin c4 ax + p

4b d Rewrite by factoring out 4.

Step 1 Find an interval whose length is one period.

0 … 4 ax + p4b … 2p Three-part inequality

0 … x + p4 … p

2 Divide each part by 4.

- p4

… x … p4

Subtract p4 from each part.

Step 2 Divide the interval C - p4 , p4 D into four equal parts to obtain these x-values.- p

4 , - p

8 , 0,

8 , p

4 Key x-values

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158 CHAPTER 4 Graphs of the Circular Functions

Step 3 Make a table of values.

Figure 18

y = –1

–2

–3

y = –1 + 2 sin(4x + P)

–1

–p4

–p2

3p8

–3p8

–

EXAMPLE 6 Modeling Temperature with a Sine Function

The maximum average monthly temperature in New Orleans, Louisiana, is 83°F, and the minimum is 53°F. The table shows the average monthly temperatures. The scatter diagram for a two-year interval in Figure 19 strongly suggests that the temperatures can be modeled with a sine curve.

Figure 19

0 25

Source: World Almanac and Book of Facts.

Month °F Month °F

Jan 53 July 83

Feb 56 Aug 83

Mar 62 Sept 79

Apr 68 Oct 70

May 76 Nov 61

June 81 Dec 55

(a) Using only the maximum and minimum temperatures, determine a function of the form

ƒ1x2 = a sin 3b1x - d24 + c, where a, b, c, and d are constants,that models the average monthly temperature in New Orleans. Let x represent the month, with January corresponding to x = 1.

(b) On the same coordinate axes, graph ƒ for a two-year period together with the actual data values found in the table.

x - p4 - p8 0

x + P4 0p8

3p8

4 1x + P4 2 0 p2 p 3p2 2psin 34 1x + P4 2 4 0 1 0 -1 02 sin 34 1x + P4 2 4 0 2 0 -2 0−1 + 2 sin 14 x + P 2 -1 1 -1 -3 -1

Steps 4 and 5 Plot the points found in the table and join them with a sinusoidal curve. Figure 18 shows the graph, extended to the right and left to include two full periods.

■✔ Now Try Exercise 57.

A Trigonometric Model For natural phenomena that occur in periodic patterns (such as seasonal temperatures, phases of the moon, heights of tides) a sinusoidal function will provide a good approximation of a set of data points.

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1594.2 Translations of the Graphs of the Sine and Cosine Functions

SOLUTION

(a) We use the maximum and minimum average monthly temperatures to find the amplitude a.

a = 83 - 532

= 15 Amplitude a

The average of the maximum and minimum temperatures is a good choice for c. The average is

83 + 532

= 68. Vertical translation c

Because temperatures repeat every 12 months, b can be found as follows.

12 = 2pb

Period = 2pb

b = p6

Solve for b.

The coldest month is January, when x = 1, and the hottest month is July, when x = 7. A good choice for d is 4 because April, when x = 4, is located at the midpoint between January and July. Also, notice that the average monthly temperature in April is 68°F, which is the value of the vertical translation, c. The average monthly temperature in New Orleans is modeled closely by the following equation.

ƒ1x2 = a sin 3b1x - d24 + c ƒ1x2 = 15 sin cp

6 1x - 42 d + 68 Substitute for a, b, c, and d.

(b) Figure 20 shows two iterations of the data points from the table, along with the graph of y = 15 sin Cp6 1x - 42 D + 68. The graph of y = 15 sin p6 x + 68 is shown for comparison.

(c) We used the given data for a two-year period and the sine regression capabil-ity of a graphing calculator to produce the model

ƒ1x2 = 15.35 sin10.52x - 2.132 + 68.89described in Figure 21(a). Its graph along with the data points is shown in Figure 21(b).

■✔ Now Try Exercise 61.

Figure 20

0 25

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. The graph of y = sin Ax + p4 B is obtained by shifting the graph of y = sin x unit(s) to the . (right/left)

2. The graph of y = cos Ax - p6 B is obtained by shifting the graph of y = cos x unit(s) to the .

(right/left)

4.2 Exercises

(a)

0 25

(b)

Figure 21

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160 CHAPTER 4 Graphs of the Circular Functions

3. The graph of y = 4 sin x is obtained by stretching the graph of y = sin x vertically by a factor of .

4. The graph of y = -3 sin x is obtained by stretching the graph of y = sin x by a fac-tor of and reflecting across the -axis.

5. The graph of y = 6 + 3 sin x is obtained by shifting the graph of y = 3 sin x unit(s) .

(up/down)

6. The graph of y = -5 + 2 cos x is obtained by shifting the graph of y = 2 cos x unit(s) .

(up/down)

7. The graph of y = 3 + 5 cos Ax + p5 B is obtained by shifting the graph of y = cos x unit(s) horizontally to the , stretching it vertically by a factor

(right/left) of , and then shifting it unit(s) vertically . (up/down)

8. Repeat Exercise 7 for y = -2 + 3 cos Ax - p6 B .Concept Check Match each function with its graph in choices A– I. (One choice will not be used.)

9. y = sin ax - p4b 10. y = sin ax + p

4b 11. y = cos ax - p

12. y = cos ax + p4b 13. y = 1 + sin x 14. y = -1 + sin x

15. y = 1 + cos x 16. y = -1 + cos x

A. y

–1

– 7p4

3p4

0–1

2pp

–2

0–1

2pp

–10

9p4

5p4

0–1

–2

2pp

G. y

–1

– 7p4

3p4

H. y

–1

9p4

5p4

0–1

2pp

17. The graphs of y = sin x + 1 and y = sin1x + 12 are NOT the same. Explain why this is so.

18. Concept Check Refer to Exercise 17. Which one of the two graphs is the same as that of y = 1 + sin x?

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1614.2 Translations of the Graphs of the Sine and Cosine Functions

Concept Check Match each function in Column I with the appropriate description in Column II.

19. y = 3 sin12x - 4220. y = 2 sin13x - 4221. y = -4 sin13x - 2222. y = -2 sin14x - 32

A. amplitude = 2, period = p2 , phase shift =34

B. amplitude = 3, period = p, phase shift = 2

C. amplitude = 4, period = 2p3 , phase shift =23

D. amplitude = 2, period = 2p3 , phase shift =43

Concept Check Fill in each blank with the word right or the word left.

23. If the graph of y = cos x is translated p2 units horizontally to the , it will coincide with the graph of y = sin x.

24. If the graph of y = sin x is translated p2 units horizontally to the , it will coincide with the graph of y = cos x.

Connecting Graphs with Equations Each function graphed is of the form y = c + cos x, y = c + sin x, y = cos1x - d2, or y = sin1x - d2, where d is the least possible positive value. Determine an equation of the graph.

25.

0–1

–2

–3

2pp

26.

0–1

–2

2pp

27.

–2

–1

28.

–2

2p3

5p3

– –1

Find the amplitude, the period, any vertical translation, and any phase shift of the graph of each function. See Examples 1–5.

29. y = 2 sin1x + p2 30. y = 3 sin ax + p2b

31. y = - 14

cos a12

x + p2b 32. y = - 1

2 sin a1

2 x + pb

33. y = 3 cos cp2

ax - 12b d 34. y = -cos cp ax - 1

3b d

35. y = 2 - sin a3x - p5b 36. y = -1 + 1

2 cos12x - 3p2

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162 CHAPTER 4 Graphs of the Circular Functions

Graph each function over a two-period interval. See Examples 1 and 2.

37. y = cos ax - p2b 38. y = sin ax - p

4b 39. y = sin ax + p

40. y = cos ax + p3b 41. y = 2 cos ax - p

3b 42. y = 3 sin ax - 3p

Graph each function over a one-period interval. See Example 3.

43. y = 32

sin c 2 ax + p4b d 44. y = - 1

2 cos c 4 ax + p

2b d

45. y = -4 sin12x - p2 46. y = 3 cos14x + p247. y = 1

2 cos a 1

2 x - p

4b 48. y = - 1

4 sin a3

4 x + p

Graph each function over a two-period interval. See Example 4.

49. y = -3 + 2 sin x 50. y = 2 - 3 cos x 51. y = -1 - 2 cos 5x

52. y = 1 - 23

sin 34

x 53. y = 1 - 2 cos 12

x 54. y = -3 + 3 sin 12

55. y = -2 + 12

sin 3x 56. y = 1 + 23

cos 12

Graph each function over a one-period interval. See Example 5.

57. y = -3 + 2 sin ax + p2b 58. y = 4 - 3 cos1x - p2

59. y = 12

+ sin c 2 ax + p4b d 60. y = - 5

2+ cos c 3 ax - p

6b d

(Modeling) Solve each problem. See Example 6.

61. Average Monthly Temperature The average monthly temperature (in °F) in Seattle, Washington, is shown in the table.

(a) Plot the average monthly temperature over a two-year period, letting x = 1 correspond to January of the first year. Do the data seem to indicate a translated sine graph?

(b) The highest average monthly temperature is 66°F in August, and the lowest average monthly temperature is 41°F in January. Their average is 53.5°F. Graph the data together with the line y = 53.5. What does this line represent with regard to temperature in Seattle?

(d) Determine a function of the form ƒ1x2 = a sin3b1x - d24 + c, where a, b, c, and d are constants, that models the data.

(e) Graph ƒ together with the data on the same coordinate axes. How well does ƒ model the given data?

(f ) Use the sine regression capability of a graphing calculator to find the equation of a sine curve that fits these data.

Source: World Almanac and Book of Facts.

Month °F Month °F

Jan 41 July 65

Feb 43 Aug 66

Mar 46 Sept 61

Apr 50 Oct 53

May 56 Nov 45

June 61 Dec 41

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1634.2 Translations of the Graphs of the Sine and Cosine Functions

62. Average Monthly Temperature The average monthly temperature (in °F) in Phoenix, Arizona, is shown in the table.

(a) Predict the average annual temperature.

(b) Plot the average monthly temperature over a two-year period, letting x = 1 correspond to January of the first year.

(c) Determine a function of the form ƒ1x2 = a cos3b1x - d24 + c, where a, b, c, and d are constants, that models the data.

(d) Graph ƒ together with the data on the same co-ordinate axes. How well does ƒ model the data?

(e) Use the sine regression capability of a graphing calculator to find the equation of a sine curve that fits these data (two years).

Source: World Almanac and Book of Facts.

Month °F Month °F

Jan 54 July 93

Feb 58 Aug 91

Mar 63 Sept 86

Apr 70 Oct 75

May 79 Nov 62

June 89 Dec 54

(Modeling) Monthly Temperatures A set of temperature data (in °F) is given for a par-ticular location. (Source: www.weatherbase.com)

(a) Plot the data over a two-year interval.

(b) Use sine regression to determine a model for the two-year interval. Let x = 1 rep-resent January of the first year.

63. Average Monthly Temperature, Buenos Aires, Argentina

Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec

77.2 74.7 70.5 63.9 57.7 52.2 51.6 54.9 57.6 63.9 69.1 73.8

64. Average High Temperature, Buenos Aires, Argentina

Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec

86.7 83.7 79.5 72.9 66.2 60.1 58.8 63.1 66.0 72.5 77.5 82.6

(Modeling) Fractional Part of the Moon Illuminated The tables give the fractional part of the moon that is illuminated during the month indicated. (Source: http://aa.usno.navy.mil)

(a) Plot the data for the month.

(b) Use sine regression to determine a model for the data.

65. January 2015

Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Fraction 0.84 0.91 0.96 0.99 1.00 0.99 0.96 0.92 0.86 0.79 0.70 0.62 0.52 0.42 0.33 0.23

Day 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

Fraction 0.15 0.08 0.03 0.00 0.01 0.04 0.10 0.19 0.28 0.39 0.50 0.61 0.71 0.80 0.87

66. November 2015

Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Fraction 0.73 0.63 0.53 0.43 0.34 0.25 0.18 0.11 0.06 0.02 0.00 0.00 0.02 0.06

Day 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Fraction 0.12 0.19 0.28 0.39 0.49 0.61 0.71 0.81 0.90 0.96 0.99 1.00 0.98 0.93 0.87 0.79

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http://www.weatherbase.com

http://aa.usno.navy.mil

164 CHAPTER 4 Graphs of the Circular Functions

1. Give the amplitude, period, vertical translation, and phase shift of the function

y = 3 - 4 sin A2x + p2 B . Chapter 4 Quiz (Sections 4.1–4.2)

Graph each function over a two-period interval. Give the period and amplitude.

2. y = -4 sin x 3. y = - 12

cos 2x 4. y = 3 sin px

5. y = -2 cos ax + p4b 6. y = 2 + sin12x - p2 7. y = -1 + 1

2 sin x

Connecting Graphs with Equations Each function graphed is of the form y = a cos bx or y = a sin bx, where b 7 0. Determine an equation of the graph.

–2

2pp

–1

pp2

10.

–10

2pp

(Modeling) Average Monthly Temperature The average temperature (in °F) at a certain location can be approximated by the function

ƒ1x2 = 12 sin cp6

1x - 3.92 d + 72,where x = 1 represents January, x = 2 represents February, and so on.

11. What is the average temperature in April?

12. What is the lowest average monthly temper-ature? What is the highest?

4.3 Graphs of the Tangent and Cotangent Functions

Graph of the Tangent Function Consider the table of selected points accompanying the graph of the tangent function in Figure 22 on the next page.

These points include special values between - p2 and p

2 . The tangent function

is undefined for odd multiples of p

2 and, thus, has vertical asymptotes for such values. A vertical asymptote is a vertical line that the graph approaches but does not intersect. As the x-values get closer and closer to the line, the function values increase or decrease without bound. Furthermore, because

tan1-x2 = - tan x, Odd functionthe graph of the tangent function is symmetric with respect to the origin.

■ Graph of the Tangent Function

■ Graph of the Cotangent Function

■ Techniques for Graphing

■ Connecting Graphs with Equations

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1654.3 Graphs of the Tangent and Cotangent Functions

The tangent function has period p. Because tan x = sin xcos x , tangent values are 0 when sine values are 0, and are undefined when cosine values are 0. As x-values increase from - p2 to

2 , tangent values range from -∞ to ∞ and increase through-out the interval. Those same values are repeated as x increases from

2 to 3p2 , from

3p2 to

5p2 , and so on. The graph of y = tan x from -

3p2 to

3p2 is shown in Figure 23.

x y = tan x

- p3 -23 ≈ -1.7- p4 -1- p6 - 233 ≈ -0.6

0 0p6

233 ≈ 0.6

p4 1p3 23 ≈ 1.7

Figure 22

–2

2y = tan x

–P2

–p6

–p3

x–p –p

Tangent Function f 1x 2 = tan xDomain: 5x ' x ≠ 12n + 12 p2 , where n is any integer6 Range: 1-∞, ∞2

x y

- p2 undefined

- p4 -10 0p4 1p2 undefined

The graph is discontinuous at values of x of the form x = 12n + 12 p2 and has vertical asymptotes at these values.

Its x-intercepts have x-values of the form np.

Its period is p.

There are no minimum or maximum values, so its graph has no amplitude.

The graph is symmetric with respect to the origin, so the function is an odd function. For all x in the domain, tan1-x2 = - tan x.

Figure 24

–2

f(x) = tan x, – < x <

–p2

–p4

p4p2

−4

11p4

f(x) = tan x

11p4

−

Figure 23

–2

y = tan x Period: p

p–p

3p2

– –p2

p4p2

3p2

–p4

The graph continues in this pattern.

Graph of the Cotangent Function A similar analysis for selected points between 0 and p for the graph of the cotangent function yields the graph in Figure 25 on the next page. Here the vertical asymptotes are at x-values that are integer multiples of p. Because

cot1-x2 = -cot x, Odd functionthis graph is also symmetric with respect to the origin.

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166 CHAPTER 4 Graphs of the Circular Functions

x y = cot xp6 23 ≈ 1.7p4 1p3

233 ≈ 0.6

p2 0

2p3 - 233 ≈ -0.6

3p4 -1

5p6 -23 ≈ -1.7

Figure 25

5p6

p6 3p

y = cot x

2p3

–2

–1

3p4

The cotangent function also has period p. Cotangent values are 0 when cosine values are 0, and are undefined when sine values are 0. As x-values increase from 0 to p, cotangent values range from ∞ to -∞ and decrease throughout the inter-val. Those same values are repeated as x increases from p to 2p, from 2p to 3p, and so on. The graph of y = cot x from -p to p is shown in Figure 26.

Cotangent Function f 1x 2 = cot xDomain: 5x ' x ≠ np, where n is any integer6 Range: 1-∞, ∞2

x y

0 undefinedp4 1p2 0

3p4

-1p undefined

The graph is discontinuous at values of x of the form x = np and has ver-tical asymptotes at these values.

Its x-intercepts have x-values of the form 12n + 12 p2 . Its period is p.

There are no minimum or maximum values, so its graph has no amplitude.

The graph is symmetric with respect to the origin, so the function is an odd function. For all x in the domain, cot1-x2 = -cot x.

f(x) = cot x, 0 < x < P

–1

xpp

4p2

−4

11p4

f(x) = cot x

11p4

−

Figure 27

The tangent function can be graphed directly with a graphing calculator, using the tangent key. To graph the cotangent function, however, we must use one of the identities

cot x = 1tan x

or cot x = cos xsin x

because graphing calculators generally do not have cotangent keys.■

Figure 26

y = cot x Period: p

–p p–p2

The graph continues in this pattern.

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1674.3 Graphs of the Tangent and Cotangent Functions

Techniques for Graphing

Guidelines for Sketching Graphs of Tangent and Cotangent Functions

To graph y = a tan bx or y = a cot bx, with b 7 0, follow these steps.

Step 1 Determine the period, pb . To locate two adjacent vertical asymptotes, solve the following equations for x:

For y = a tan bx: bx = - p2 and bx =p

2 .

For y = a cot bx: bx = 0 and bx = p.Step 2 Sketch the two vertical asymptotes found in Step 1.

Step 3 Divide the interval formed by the vertical asymptotes into four equal parts.

Step 4 Evaluate the function for the first-quarter point, midpoint, and third-quarter point, using the x-values found in Step 3.

Step 5 Join the points with a smooth curve, approaching the vertical asymp-totes. Indicate additional asymptotes and periods of the graph as necessary.

EXAMPLE 1 Graphing y = tan bx

Graph y = tan 2x.SOLUTION

Step 1 The period of this function is p2 . To locate two adjacent vertical asymp-totes, solve 2x = - p2 and 2x =

p2 (because this is a tangent function). The

two asymptotes have equations x = - p4 and x =p4 .

Step 2 Sketch the two vertical asymptotes x = {p4 , as shown in Figure 28.Step 3 Divide the interval A - p4 , p4 B into four equal parts to find key x-values.first-quarter value: - p

8 , middle value: 0, third-quarter value:

8 Key x-values

Step 4 Evaluate the function for the x-values found in Step 3.

Step 5 Join these points with a smooth curve, approaching the vertical asymp-totes. See Figure 28. ■✔ Now Try Exercise 13.

x - p8 0p8

2 x - p4 0p4

tan 2 x -1 0 1

Another period has been graphed, one half period to the left and one half period to the right.

Figure 28

y = tan 2x

–1

Period:p2

–p2

–p4

–p8

p8p4

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168 CHAPTER 4 Graphs of the Circular Functions

EXAMPLE 2 Graphing y = a tan bx

Graph y = -3 tan 12 x.

SOLUTION The period is p12

= p , 12 = p # 21 = 2p. Adjacent asymptotes are at x = -p and x = p. Dividing the interval 1-p, p2 into four equal parts gives key x-values of - p2 , 0, and

p2 . Evaluating the function at these x-values gives

the following key points.a - p2

, 3b , 10, 02, ap2

, -3b Key pointsBy plotting these points and joining them with a smooth curve, we obtain the graph shown in Figure 29. Because the coefficient -3 is negative, the graph is reflected across the x-axis compared to the graph of y = 3 tan 12 x.

■✔ Now Try Exercise 21.

Figure 29

–3

y = –3 tan xPeriod: 2p

–p2

p p–p2

NOTE The function y = -3 tan 12 x in Example 2, graphed in Figure 29, has a graph that compares to the graph of y = tan x as follows.

1. The period is larger because b = 12 , and 12 6 1.

2. The graph is stretched vertically because a = -3, and ' -3 ' 7 1.3. Each branch of the graph falls from left to right (that is, the function

decreases) between each pair of adjacent asymptotes because a = -3, and -3 6 0. When a 6 0, the graph is reflected across the x-axis com-pared to the graph of y = 'a ' tan bx.

EXAMPLE 3 Graphing y = a cot bx

Graph y = 12 cot 2x.

SOLUTION Because this function involves the cotangent, we can locate two adjacent asymptotes by solving the equations 2x = 0 and 2x = p. The lines x = 0 (the y-axis) and x = p2 are two such asymptotes. We divide the interval A0, p2 B into four equal parts, obtaining key x-values of p8 , p4 , and 3p8 . Evaluating the function at these x-values gives the key points Ap8 , 12 B , Ap4 , 0 B , A3p8 , - 12 B . We plot these points and join them with a smooth curve approaching the asymptotes to obtain the graph shown in Figure 30.

Figure 30

–1

Period:

y = cot 2x12p2

p8p4

3p8

■✔ Now Try Exercise 23.

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1694.3 Graphs of the Tangent and Cotangent Functions

Like the other circular functions, the graphs of the tangent and cotangent functions may be translated horizontally and vertically.

EXAMPLE 4 Graphing y = c + tan x

Graph y = 2 + tan x.ANALYTIC SOLUTION

Every value of y for this function will be 2 units more than the corresponding value of y in y = tan x, causing the graph of y = 2 + tan x to be translated 2 units up compared to the graph of y = tan x. See Figure 31.

GRAPHING CALCULATOR SOLUTION

Observe Figures 32 and 33. In these figures

y2 = tan x

is the red graph and

y1 = 2 + tan x

is the blue graph. Notice that for the arbitrarily-chosen value of p4 1approximately 0.785398162, the difference in the y-values is

y1 - y2 = 3 - 1 = 2.

This illustrates the vertical translation 2 units up.

Figure 31

–1

y = 2 + tan x

–2

– 3p2

3p2

–p–p p2

Figure 32

−4

−

Figure 33

−4

−Three periods of the function are shown in Figure 31. Because the period of y = 2 + tan x is p, additional asymptotes and periods of the function can be drawn by repeating the basic graph every p units on the x-axis to the left or to the right of the graph shown.

■✔ Now Try Exercise 29.

EXAMPLE 5 Graphing y = c + a cot 1x − d 2Graph y = -2 - cot Ax - p4 B .SOLUTION Here b = 1, so the period is p. The negative sign in front of the cotangent will cause the graph to be reflected across the x-axis, and the argument Ax - p4 B indicates a phase shift (horizontal shift) p4 unit to the right. Because c = -2, the graph will then be translated 2 units down. To locate adjacent asymp-totes, because this function involves the cotangent, we solve the following.

x - p4

= 0 and x - p4

= p

x = p4

and x = 5p4

Add p4 .

Dividing the interval Ap4 , 5p4 B into four equal parts and evaluating the function at the three key x-values within the interval give these points.ap

2 , -3b , a 3p

4 , -2b , 1p, -12 Key points

We join these points with a smooth curve. This period of the graph, along with the one in the domain interval A - 3p4 , p4 B , is shown in Figure 34 on the next page.

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170 CHAPTER 4 Graphs of the Circular Functions

Figure 34

–3

y = –2

y = –2 – cot(x – )4P

5p4

3p4

–p2

–3p4

– p4

■✔ Now Try Exercise 37.

EXAMPLE 6 Determining an Equation for a Graph

Determine an equation for each graph.

(a)

–1

–2

0 p2

–p2

(b)

–2

–1

–3

0 p2

SOLUTION

(a) This graph is that of y = tan x but reflected across the x-axis and stretched vertically by a factor of 2. Therefore, an equation for this graph is

y = -2 tan x. Vertical stretch

x-axis reflection

(b) This is the graph of a cotangent function, but the period is p2 rather than p. Therefore, the coefficient of x is 2. This graph is vertically translated 1 unit down compared to the graph of y = cot 2 x. An equation for this graph is

y = -1 + cot 2 x.Period is p2 .

Vertical translation 1 unit down

■✔ Now Try Exercises 39 and 43.

NOTE Because the circular functions are periodic, there are infinitely many equations that correspond to each graph in Example 6. Confirm that both

y = -1 - cot1-2x2 and y = -1 - tan a2x - p2b

are equations for the graph in Example 6(b). When writing the equation from a graph, it is practical to write the simplest form. Therefore, we choose values of b where b 7 0 and write the function without a phase shift when possible.

Connecting Graphs with Equations

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1714.3 Graphs of the Tangent and Cotangent Functions

3. Between any two successive vertical asymptotes, the graph of y = tan x .

(increases/decreases)

4. Between any two successive vertical asymptotes, the graph of y = cot x .

(increases/decreases)

5. The negative value k with the greatest value for which x = k is a vertical asymptote of the graph of y = tan x is .

6. The negative value k with the greatest value for which x = k is a vertical asymptote of the graph of y = cot x is .

CONCEPT PREVIEW Fill in the blank to correctly complete each sentence.

1. The least positive value x for which tan x = 0 is .

2. The least positive value x for which cot x = 0 is .

4.3 Exercises

Concept Check Match each function with its graph from choices A–F.

7. y = - tan x 8. y = -cot x 9. y = tan ax - p4b

10. y = cot ax - p4b 11. y = cot ax + p

4b 12. y = tan ax + p

0 p

4–p 3p

0–p2p2

0p4

5p4

0p4

3p4

–

4–p 3p

Graph each function over a one-period interval. See Examples 1–3.

13. y = tan 4 x 14. y = tan 12

x 15. y = 2 tan x

16. y = 2 cot x 17. y = 2 tan 14

x 18. y = 12

cot x

19. y = cot 3x 20. y = -cot 12

x 21. y = -2 tan 14

22. y = 3 tan 12

x 23. y = 12

cot 4 x 24. y = - 12

cot 2 x

Graph each function over a two-period interval. See Examples 4 and 5.

25. y = tan12 x - p2 26. y = tan a x2

+ pb 27. y = cot a3x + p4b

28. y = cot a2 x - 3p2b 29. y = 1 + tan x 30. y = 1 - tan x

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172 CHAPTER 4 Graphs of the Circular Functions

31. y = 1 - cot x 32. y = -2 - cot x

33. y = -1 + 2 tan x 34. y = 3 +12

tan x

35. y = -1 + 12

cot12 x - 3p2 36. y = -2 + 3 tan14 x + p237. y = 1 - 2 cot c 2 ax + p

2b d 38. y = -2 + 2

3 tan a3

4 x - pb

Connecting Graphs with Equations Determine the simplest form of an equation for each graph. Choose b 7 0, and include no phase shifts. (Midpoints and quarter-points are identified by dots.) See Example 6.

39.

3p22

–p

–2

40. y

x0 2pp

–2

41.

–1

2p3

42.

6–p p

6p2

43.

p–p2

–p

–2

44.

–2

–4

Concept Check Decide whether each statement is true or false. If false, explain why.

45. The least positive number k for which x = k is an asymptote for the tangent function is p2 .

46. The least positive number k for which x = k is an asymptote for the cotangent func-tion is p2 .

47. The graph of y = tan x in Figure 23 suggests that tan1-x2 = tan x for all x in the domain of tan x.

48. The graph of y = cot x in Figure 26 suggests that cot1-x2 = -cot x for all x in the domain of cot x.

Work each exercise.

49. Concept Check If c is any number, then how many solutions does the equation c = tan x have in the interval 1-2p, 2p4?

50. Concept Check Consider the function defined by ƒ1x2 = -4 tan12x + p2. What is the domain of ƒ? What is its range?

51. Show that tan1-x2 = - tan x by writing tan1-x2 as sin1- x2cos1- x2 and then using the rela-tionships for sin1-x2 and cos1-x2.

52. Show that cot1-x2 = -cot x by writing cot1-x2 as cos1- x2sin1- x2 and then using the rela-tionships for cos1-x2 and sin1-x2.

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1734.4 Graphs of the Secant and Cosecant Functions

(Modeling) Distance of a Rotating Beacon A rotating beacon is located at point A next to a long wall. The beacon is 4 m from the wall. The distance d is given by

d = 4 tan 2pt,

where t is time measured in seconds since the beacon started rotating. (When t = 0, the beacon is aimed at point R. When the beacon is aimed to the right of R, the value of d is positive; d is negative when the beacon is aimed to the left of R.) Find d for each time. Round to the nearest tenth if applicable.

53. t = 0

54. t = 0.4

55. t = 1.2

56. Why is 0.25 a meaningless value for t?

4 m

4.4 Graphs of the Secant and Cosecant Functions

Graph of the Secant Function Consider the table of selected points accompanying the graph of the secant function in Figure 35 on the next page. These points include special values from -p to p. The secant function is undefined for odd multiples of

2 and thus, like the tangent function, has vertical asymptotes for such values. Furthermore, because

sec1-x2 = sec x, Even functionthe graph of the secant function is symmetric with respect to the y-axis.

■ Graph of the Secant Function

■ Graph of the Cosecant Function

■ Techniques for Graphing

■ Connecting Graphs with Equations

■ Addition of Ordinates

Relating Concepts

For individual or collaborative investigation (Exercises 57–62)

Consider the following function from Example 5. Work these exercises in order.

y = -2 - cot ax - p4b

57. What is the least positive number for which y = cot x is undefined?

58. Let k represent the number found in Exercise 57. Set x - p4 equal to k, and solve to find a positive number for which cot Ax - p4 B is undefined.

59. Based on the answer in Exercise 58 and the fact that the cotangent function has period p, give the general form of the equations of the asymptotes of the graph

of y = -2 - cot Ax - p4 B . Let n represent any integer.60. Use the capabilities of a calculator to find the x-intercept with least positive

x-value of the graph of this function. Round to the nearest hundredth.

61. Use the fact that the period of this function is p to find the next positive x-intercept. Round to the nearest hundredth.

62. Give the solution set of the equation -2 - cot Ax - p4 B = 0 over all real num-bers. Let n represent any integer.

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174 CHAPTER 4 Graphs of the Circular Functions

Because secant values are reciprocals of corresponding cosine values, the period of the secant function is 2p, the same as for y = cos x. When cos x = 1, the value of sec x is also 1. Likewise, when cos x = -1, sec x = -1. For all x, -1 … cos x … 1, and thus, ' sec x ' Ú 1 for all x in its domain. Figure 36 shows how the graphs of y = cos x and y = sec x are related.

x y = sec x

0 1

{p6 2233 ≈ 1.2{p4 22 ≈ 1.4{p3 2

{2p3 -2

{3p4 -22 ≈ -1.4{5p6 - 2233 ≈ -1.2{p -1

Figure 35

–2

–1

y = sec x

pP2

–P2

–p

Figure 36

y = cos x

y = sec x Period: 2p

–2p 2p

–p p1

–1

Secant Function f 1x 2 = sec xDomain: E x ' x ≠ 12n + 12 p2 ,

where n is any integer F Range: 1-∞, -14 ´ 31, ∞2x y

- p2 undefined

- p4 220 1p4 22p2 undefined

3p4 -22p -1

3p2 undefined

The graph is discontinuous at values of x of the form x = 12n + 12 p2 and has vertical asymptotes at these values.

There are no x-intercepts.

Its period is 2p.

There are no minimum or maximum values, so its graph has no amplitude.

The graph is symmetric with respect to the y-axis, so the function is an even function. For all x in the domain, sec1-x2 = sec x.

Figure 37

–1

f(x) = sec x

–2p 2p–p p

−4

11p4

f(x) = sec x

11p4

−

As we shall see, locating the vertical asymptotes for the graph of a function involving the secant (as well as the cosecant) is helpful in sketching its graph.

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1754.4 Graphs of the Secant and Cosecant Functions

Graph of the Cosecant function A similar analysis for selected points between -p and p for the graph of the cosecant function yields the graph in Figure 38. The vertical asymptotes are at x-values that are integer multiples of p. This graph is symmetric with respect to the origin because

csc1-x2 = -csc x. Odd function

Because cosecant values are reciprocals of corresponding sine values, the period of the cosecant function is 2p, the same as for y = sin x. When sin x = 1, the value of csc x is also 1. Likewise, when sin x = -1, csc x = -1. For all x, -1 … sin x … 1, and thus ' csc x ' Ú 1 for all x in its domain. Figure 39 shows how the graphs of y = sin x and y = csc x are related.Figure 39

–1

1y = sin x

y = csc x Period: 2p

–2p 2p–p p Figure 38

y = csc x

–p2

–P p2

–1

x y = csc x x y = csc x

p6 2 -

p6 -2

p4 22 ≈ 1.4 - p4 -22 ≈ -1.4p3

2233 ≈ 1.2 -

p3 - 2233 ≈ -1.2

p2 1 -

p2 -1

2p3

2233 ≈ 1.2 -

2p3 - 2233 ≈ -1.2

3p4 22 ≈ 1.4 - 3p4 -22 ≈ -1.4

5p6 2 -

5p6 -2

Cosecant Function f 1x 2 = csc xDomain: 5x ' x ≠ np, Range: 1-∞, -14 ´ 31, ∞2

where n is any integer6x y

0 undefinedp6 2p3

2233

p2 1

2p3

2233

p undefined3p2 -1

2p undefined

The graph is discontinuous at values of x of the form x = np and has vertical asymptotes at these values.

There are no x-intercepts.

Its period is 2p.

There are no minimum or maximum values, so its graph has no amplitude.

The graph is symmetric with respect to the origin, so the function is an odd function. For all x in the domain, csc1-x2 = -csc x.

Figure 40

–10

f (x) = csc x

x–2p 2p–p p

−4

11p4

f(x) = csc x

11p4

−

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176 CHAPTER 4 Graphs of the Circular Functions

Calculators do not have keys for the cosecant and secant functions. To graph them with a graphing calculator, use

csc x = 1sin x

and sec x = 1cos x

. Reciprocal identities ■

Guidelines for Sketching Graphs of Cosecant and Secant Functions

To graph y = a csc bx or y = a sec bx, with b 7 0, follow these steps.Step 1 Graph the corresponding reciprocal function as a guide, using a

dashed curve.

To Graph Use as a Guide

y = a csc bx y = a sin bxy = a sec bx y = a cos bx

Step 2 Sketch the vertical asymptotes. They will have equations of the form x = k, where k corresponds to an x-intercept of the graph of the guide function.

Step 3 Sketch the graph of the desired function by drawing the typical U-shaped branches between the adjacent asymptotes. The branches will be above the graph of the guide function when the guide function values are positive and below the graph of the guide function when the guide function values are negative. The graph will resemble those in Figures 37 and 40 in the function boxes given earlier in this section.

Like graphs of the sine and cosine functions, graphs of the secant and cose-cant functions may be translated vertically and horizontally. The period of both basic functions is 2p.

EXAMPLE 1 Graphing y = a sec bx

Graph y = 2 sec 12 x.

SOLUTION

Step 1 This function involves the secant, so the corresponding reciprocal func-tion will involve the cosine. The guide function to graph is

y = 2 cos 12

Using the guidelines given earlier, we find that this guide function has amplitude 2 and that one period of the graph lies along the interval that satisfies the following inequality.

0 … 12

x … 2p

0 … x … 4p Multiply each part by 2.

Dividing the interval 30, 4p4 into four equal parts gives these key points.10, 22, 1p, 02, 12p, -22, 13p, 02, 14p, 22 Key points

Techniques for Graphing

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1774.4 Graphs of the Secant and Cosecant Functions

These key points are plotted and joined with a dashed red curve to indi-cate that this graph is only a guide. An additional period is graphed as shown in Figure 41(a).

–2

y = 2 cos xis used as a guide.

–4p –3p –2p 2p 3p 4p–p p–2

y = 2 sec x

–4p –3p –2p 2p 3p 4p–p p

(a) (b)

Figure 41

Step 2 Sketch the vertical asymptotes as shown in Figure 41(a). These occur at x-values for which the guide function equals 0, such as

x = -3p, x = -p, x = p, x = 3p.

Step 3 Sketch the graph of y = 2 sec 12 x by drawing typical U-shaped branches, approaching the asymptotes. See the solid blue graph in Fig ure 41(b).

■✔ Now Try Exercise 11.

EXAMPLE 2 Graphing y = a csc 1x − d 2Graph y = 32 csc Ax - p2 B .SOLUTION

Step 1 Graph the corresponding reciprocal function

y = 32

sin ax - p2b ,

shown as a red dashed curve in Figure 42.

Step 2 Sketch the vertical asymptotes through the x-intercepts of the graph of y = 32 sin Ax - p2 B . These x-values have the form 12n + 12 p2 , where n is any integer. See the black dashed lines in Figure 42.

Step 3 Sketch the graph of y = 32 csc Ax - p2 B by drawing the typical U-shaped branches between adjacent asymptotes. See the solid blue graph in Figure 42.

−4

4p−4p

This is a calculator graph of the function in Example 1.

Figure 42

–1

y = sin(x – )32 !2

y = csc(x – )32 !2

–! ! 2!– 3!2

3!2

5!2

– !2

■✔ Now Try Exercise 13.

−4

11p4

−

This is a calculator graph of the function in Example 2. (The use of decimal equivalents when defining y1 eliminates the need for some parentheses.)

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178 CHAPTER 4 Graphs of the Circular Functions

Connecting Graphs with Equations

EXAMPLE 3 Determining an Equation for a Graph

Determine an equation for each graph.

(a)

–3

–2

–1

4p3p2pp

(b)

03p2

–p2

SOLUTION

(a) This graph is that of a cosecant function that is stretched horizontally having period 4p. If y = csc bx, where b 7 0, then we must have b = 12 and

y = csc 12

x. 2p12

= 4p

Horizontal stretch

(b) This is the graph of y = sec x, translated 1 unit up. An equation is

y = 1 + sec x.

Vertical translation

■✔ Now Try Exercises 25 and 27.

EXAMPLE 4 Illustrating Addition of Ordinates

Use the functions y1 = cos x and y2 = sin x to illustrate addition of ordinates fory3 = cos x + sin x

with the value p6 for x.

SOLUTION In Figures 43–45, y1 = cos x is graphed in blue, y2 = sin x is graphed in red, and their sum, y1 + y2 = cos x + sin x, is graphed as y3 in green. If the ordinates ( y-values) for x = p6 (approximately 0.52359878) in Figures 43 and 44 are added, their sum is found in Figure 45. Verify that

0.8660254 + 0.5 = 1.3660254.(This would occur for any value of x.) ■✔ Now Try Exercise 43.

−2

11p4

−

Figure 43

−2

11p4

−

Figure 44

−2

11p4

−

Figure 45

Addition of Ordinates A function formed by combining two other func-tions, such as

y3 = y1 + y2,has historically been graphed using a method known as addition of ordinates. (The x-value of a point is sometimes called its abscissa, while its y-value is called its ordinate.)

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1794.4 Graphs of the Secant and Cosecant Functions

CONCEPT PREVIEW Match each description in Column I with the correct value in Column II. Refer to the basic graphs as needed.

4.4 Exercises

1. The least positive value k for which x = k is a vertical asymptote for y = sec x

2. The least positive value k for which x = k is a vertical asymptote for y = csc x

3. The least positive value that is in the range of y = sec x

4. The greatest negative value that is in the range of y = csc x

5. The greatest negative value of x for which sec x = -1

6. The least positive value of x for which csc x = -1

A. p

B. p

C. -p

D. 1

E. 3p2

F . -1

Concept Check Match each function with its graph from choices A–D.

7. y = -csc x 8. y = -sec x 9. y = sec ax - p2b 10. y = csc ax + p

3p2

–1–p2

y B.

–1 p 2p

–1–p2

–p p

–1

Graph each function over a one-period interval. See Examples 1 and 2.

11. y = 3 sec 14

x 12. y = -2 sec 12

x 13. y = - 12

csc ax + p2b

14. y = 12

csc ax - p2b 15. y = csc ax - p

4b 16. y = sec ax + 3p

17. y = sec ax + p4b 18. y = csc ax + p

19. y = csc a12

x - p4b 20. y = sec a1

2 x + p

21. y = 2 + 3 sec12x - p2 22. y = 1 - 2 csc ax + p2b

23. y = 1 - 12

csc ax - 3p4b 24. y = 2 + 1

4 sec a1

2 x - pb

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180 CHAPTER 4 Graphs of the Circular Functions

Connecting Graphs with Equations Determine an equation for each graph. See Exam-ple 3.

25.

–p8

–p4

–1

26.

–1

–p4

–p2

27.

–1

–3

p 3p 2p2

28.

–1–2

p 3p 2p2

29.

–2

0pp

2–p

30.

–2

–1

0p 3p2p 4p

Work each problem.

35. Concept Check If c is any number such that -1 6 c 6 1, then how many solutions does the equation c = sec x have over the entire domain of the secant function?

36. Concept Check Consider the function g1x2 = -2 csc14x + p2. What is the domain of g? What is its range?

37. Show that sec1-x2 = sec x by writing sec1-x2 as 1cos 1- x2 and then using the rela-tionship between cos1-x2 and cos x.

38. Show that csc1-x2 = -csc x by writing csc1-x2 as 1sin 1- x2 and then using the rela-tionship between sin1-x2 and sin x.

(Modeling) Distance of a Rotating Beacon The distance a in the figure (repeated from the exercise set in the previous section) is given by

a = 4 ' sec 2pt ' .

4 m

Find the value of a for each time t. Round to the nearest tenth if applicable.

39. t = 0 40. t = 0.86 41. t = 1.24 42. t = 0.25

Concept Check Decide whether each statement is true or false. If false, explain why.

31. The tangent and secant functions are undefined for the same values.

32. The secant and cosecant functions are undefined for the same values.

33. The graph of y = sec x in Figure 37 suggests that sec1-x2 = sec x for all x in the domain of sec x.

34. The graph of y = csc x in Figure 40 suggests that csc1-x2 = -csc x for all x in the domain of csc x.

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1814.5 Harmonic Motion

Given y1 and y2, define their sum to be

y3 = y1 + y2.

Evaluate y1 and y2 at the given value of x and show that their sum is equal to y3 evalu-ated at x. Use the method of addition of ordinates. See Example 4.

43. y1 = sin x, y2 = sin 2x; x =p6 44. y1 = cos x, y2 = cos 2x; x =

2p3

45. y1 = tan x, y2 = sec x; x =p4 46. y1 = cot x, y2 = csc x; x =

Summary Exercises on Graphing Circular Functions

These summary exercises provide practice with the various graphing techniques pre-sented in this chapter. Graph each function over a one-period interval.

1. y = 2 sin px 2. y = 4 cos 32

3. y = -2 + 12

cos p

4 x 4. y = 3 sec p

2 x

5. y = -4 csc 12

x 6. y = 3 tan ap2

x + pbGraph each function over a two-period interval.

7. y = -5 sin x3

8. y = 10 cos a x4

+ p2b

9. y = 3 - 4 sin a52

x + pb 10. y = 2 - sec3p1x - 324 4.5 Harmonic Motion

Simple Harmonic Motion In part A of Figure 46, a spring with a weight attached to its free end is in equilibrium (or rest) position. If the weight is pulled down a units and released (part B of the figure), the spring’s elasticity causes the weight to rise a units 1a 7 02 above the equilibrium position, as seen in part C, and then to oscillate about the equilibrium position.

■ Simple Harmonic Motion

■ Damped Oscillatory Motion

–aA. B. C.

Figure 46

If friction is neglected, this oscillatory motion is described mathematically by a sinusoid. Other applications of this type of motion include sound, electric current, and electromagnetic waves.

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182 CHAPTER 4 Graphs of the Circular Functions

P(x, y)

(a, 0)(–a, 0)

(0, a)

(0, –a)

R(x, 0)

Q(0, y)y

Figure 47

Simple Harmonic Motion

The position of a point oscillating about an equilibrium position at time t is modeled by either

s 1 t 2 = a cos V t or s 1 t 2 = a sin V t,where a and v are constants, with v 7 0. The amplitude of the motion is 'a ' , the period is

2pv

, and the frequency is v

2p oscillations per time unit.

EXAMPLE 1 Modeling the Motion of a Spring

Suppose that an object is attached to a coiled spring such as the one in Figure 46 on the preceding page. It is pulled down a distance of 5 in. from its equilibrium position and then released. The time for one complete oscillation is 4 sec.

(a) Give an equation that models the position of the object at time t.

(b) Determine the position at t = 1.5 sec.(c) Find the frequency.

SOLUTION

(a) When the object is released at t = 0, the distance of the object from the equi-librium position is 5 in. below equilibrium. If s1t2 is to model the motion, then s102 must equal -5. We use

s1t2 = a cos vt, with a = -5.We choose the cosine function here because cos v102 = cos 0 = 1, and -5 # 1 = -5. (Had we chosen the sine function, a phase shift would have been required.) Use the fact that the period is 4 to solve for v.

2pv

= 4 The period is 2pv .

v = p2

Solve for v.

Thus, the motion is modeled by s1t2 = -5 cos p2 t.

To develop a general equation for such motion, consider Figure 47. Sup-pose the point P1x, y2 moves around the circle counterclockwise at a uniform angular speed v. Assume that at time t = 0, P is at 1a, 02. The angle swept out by ray OP at time t is given by u = vt. The coordinates of point P at time t are

x = a cos u = a cos vt and y = a sin u = a sin vt.

As P moves around the circle from the point 1a, 02, the point Q10, y2 oscil-lates back and forth along the y-axis between the points 10, a2 and 10, -a2. Similarly, the point R1x, 02 oscillates back and forth between 1a, 02 and 1-a, 02. This oscillatory motion is simple harmonic motion.

The amplitude of the motion is 'a ' , and the period is 2pv

. The moving points P and Q or P and R complete one oscillation or cycle per period. The number of cycles per unit of time, called the frequency, is the reciprocal of the period, v2p , where v 7 0.

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1834.5 Harmonic Motion

(b) Substitute the given value of t in the equation found in part (a).

s1t2 = -5 cos p2

t Equation from part (a)

s11.52 = -5 cos cp2

11.52 d Let t = 1.5. s11.52 ≈ 3.54 in. Use a calculator.Because 3.54 7 0, the object is above the equilibrium position.

■✔ Now Try Exercise 7.

EXAMPLE 2 Analyzing Harmonic Motion

Suppose that an object oscillates according to the model

s1t2 = 8 sin 3t,where t is in seconds and s1t2 is in feet. Analyze the motion.SOLUTION The motion is harmonic because the model is

s1t2 = a sin vt. Because a = 8, the object oscillates 8 ft in either direction from its starting point. The period 2p3 ≈ 2.1 is the time, in seconds, it takes for one complete oscillation. The frequency is the reciprocal of the period, so the object completes 3

2p ≈ 0.48 oscillation per sec.

■✔ Now Try Exercise 17.

Damped Oscillatory Motion In the example of the stretched spring, we disregard the effect of friction. Friction causes the amplitude of the motion to diminish gradually until the weight comes to rest. In this situation, we say that the motion has been damped by the force of friction. Most oscillatory motions are damped. For instance, shock absorbers are put on an automobile in order to damp oscillatory motion. Instead of oscillating up and down for a long while after hitting a bump or pothole, the oscillations of the car are quickly damped out for a smoother ride.

The decrease in amplitude of a damped oscillatory motion usually follows the pattern of exponential decay.

EXAMPLE 3 Analyzing Damped Oscillatory Motion

A typical example of damped oscillatory motion is provided by the function

s1x2 = e-x cos 2px.(The number e ≈ 2.718 is the base of the natural logarithm function.) We use x rather than t to match the variable for graphing calculators.

(a) Provide a calculator graph of y3 = e-x cos 2px, along with the graphs of y1 = e-x and y2 = -e-x for 0 … x … 3.

(b) Describe the relationships among the three graphs drawn in part (a).

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184 CHAPTER 4 Graphs of the Circular Functions

y3 = e–x cos 2px

−1

y2 = −e–x

y1 = e–x

Figure 48

CONCEPT PREVIEW Refer to the equations in the definition of simple harmonic motion in this section, and consider the following equation.

s1t2 = 5 cos 2t, where t is time in secondsAnswer each question.

1. What is the amplitude of this motion?

2. What is the period of this motion?

3. What is the frequency?

4. What is s102? 5. What is s Ap2 B? 6. What is the range of the graph of this function? (Hint: See the answers to Exer-

cises 4 and 5.)

4.5 Exercises

(d) For what values of x does the graph of y3 intersect the x-axis?

SOLUTION

(a) Figure 48 shows a TI-84 Plus graph of y1, y2, and y3 in the window 30, 34 by 3-1, 14.

(b) The graph of y3 is bounded above by the graph of y1 and below by the graph of y2. (The graphs of y1 and y2 are referred to as envelopes for the graph of y3.)

(c) When 2px = 0, 2p, 4p, and 6p, cos 2px = 1. Thus, the value of e-x cos 2px is the same as the value of e-x when 2px = 0, 2p, 4p, and 6p—that is, when x = 0, 1, 2, and 3.

(d) When 2px = p2 , 3p2 ,

5p2 ,

7p2 ,

9p2 , and

11p2 , cos 2px = 0. Thus, the graph of y3

intersects the x-axis when x = 14 , 34 ,

54 ,

74 ,

94 , and

114 .

■✔ Now Try Exercise 33.

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1854.5 Harmonic Motion

(Modeling) Solve each problem. See Examples 1 and 2.

7. Spring Motion An object is attached to a coiled spring, as in Figure 46. It is pulled down a distance of 4 units from its equilibrium position and then released. The time for one complete oscillation is 3 sec.

(a) Give an equation that models the position of the object at time t.

(b) Determine the position at t = 1.25 sec to the nearest hundredth.(c) Find the frequency.

8. Spring Motion Repeat Exercise 7, but assume that the object is pulled down a dis-tance of 6 units and that the time for one complete oscillation is 4 sec.

9. Voltage of an Electrical Circuit The voltage E in an electrical circuit is modeled by

E = 5 cos 120pt,

where t is time measured in seconds.

(a) Find the amplitude and the period.

(b) Find the frequency.

10. Voltage of an Electrical Circuit For another electrical circuit, the voltage E is modeled by

E = 3.8 cos 40pt,

where t is time measured in seconds.

(a) Find the amplitude and the period.

(b) Find the frequency.

11. Particle Movement Write the equation and then determine the amplitude, period, and frequency of the simple harmonic motion of a particle moving uniformly around a circle of radius 2 units, with the given angular speed.

(a) 2 radians per sec (b) 4 radians per sec

12. Spring Motion The height attained by a weight attached to a spring set in motion is

s1t2 = -4 cos 8pt inches after t seconds.(a) Find the maximum height that the weight rises above the equilibrium position of

s1t2 = 0.(b) When does the weight first reach its maximum height if t Ú 0?(c) What are the frequency and the period?

13. Pendulum Motion What are the period P and frequency T of

oscillation of a pendulum of length 12 ft? 1Hint: P = 2p3 L32 ,

where L is the length of the pendulum in feet and the period P is in seconds.2

14. Pendulum Motion In Exercise 13, how long should the pen-dulum be to have a period of 1 sec?

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186 CHAPTER 4 Graphs of the Circular Functions

15. Spring Motion The formula for the up and down motion of a weight on a spring is given by

s1t2 = a sin A km t.If the spring constant k is 4, what mass m must be used to produce a period of 1 sec?

16. Spring Motion (See Exercise 15.) A spring with spring constant k = 2 and a 1-unit mass m attached to it is stretched and then allowed to come to rest.

(a) If the spring is stretched 12 ft and released, what are the amplitude, period, and frequency of the resulting oscillatory motion?

(b) What is the equation of the motion?

17. Spring Motion The position of a weight attached to a spring is

s1t2 = -5 cos 4pt inches after t seconds.(a) Find the maximum height that the weight rises above the equilibrium position of

s1t2 = 0. (b) What are the frequency and period?

(d) Calculate and interpret s11.32 to the nearest tenth.18. Spring Motion The position of a weight attached to a spring is

s1t2 = -4 cos 10t inches after t seconds.(a) Find the maximum height that the weight rises above the equilibrium position of

s1t2 = 0.(b) What are the frequency and period?

(d) Calculate and interpret s11.4662.19. Spring Motion A weight attached to a spring is pulled down 3 in. below the equi-

librium position.

(a) Assuming that the frequency is 6p

cycles per sec, determine a model that gives

the position of the weight at time t seconds.

(b) What is the period?

20. Spring Motion A weight attached to a spring is pulled down 2 in. below the equi-librium position.

(a) Assuming that the period is 13 sec, determine a model that gives the position of the weight at time t seconds.

(b) What is the frequency?

(Modeling) Springs A weight on a spring has initial position s102 and period P.(a) To model displacement of the weight, find a function s given by

s1t2 = a cos vt.(b) Evaluate s112. Is the weight moving upward, downward, or neither when t = 1?

Support the results graphically or numerically.

21. s102 = 2 in.; P = 0.5 sec 22. s102 = 5 in.; P = 1.5 sec23. s102 = -3 in.; P = 0.8 sec 24. s102 = -4 in.; P = 1.2 sec

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187

(Modeling) Music A note on a piano has given frequency F. Suppose the maximum displacement at the center of the piano wire is given by s102. Find constants a and v so that the equation

s1t2 = a cos vtmodels this displacement. Graph s in the viewing window 30, 0.054 by 3-0.3, 0.34.25. F = 27.5; s102 = 0.21 26. F = 110; s102 = 0.1127. F = 55; s102 = 0.14 28. F = 220; s102 = 0.06(Modeling) Spring Motion Consider the spring in Figure 46, but assume that because of friction and other resistive forces, the amplitude is decreasing over time, and that t seconds after the spring is released, its position in inches is given by the function

s1t2 = -11e-0.2t cos 0.5pt.29. How far was the weight pulled down from the equilibrium position before it was

released?

30. How far, to the nearest hundredth of an inch, is the weight from the equilibrium position after 6 sec?

31. Graph the function on the interval 30, 124 by 3-12, 124, and determine the values for which the graph intersects the horizontal axis.

32. How many complete oscillations will the graph make during 12 sec?

(Modeling) Damped Oscillatory Motion Work each exercise. See Example 3.

33. Consider the damped oscillatory function

s1x2 = 5e-0.3x cos px.(a) Graph the function y3 = 5e-0.3x cos px in the window 30, 34 by 3-5, 54.(b) The graph of which function is the upper envelope of the graph of y3?

34. Consider the damped oscillatory function

s1x2 = 10e-x sin 2px.(a) Graph the function y3 = 10e-x sin 2px in the window 30, 34 by 3-10, 104.(b) The graph of which function is the lower envelope of the graph of y3?

Key Terms

4.1 periodic function period sine wave (sinusoid) amplitude

4.2 phase shift argument

4.3 vertical asymptote

4.4 addition of ordinates 4.5 simple harmonic

motion

frequency damped oscillatory

motion envelope

Chapter 4 Test Prep

CHAPTER 4 Test Prep

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188 CHAPTER 4 Graphs of the Circular Functions

Quick Review

Concepts Examples

Graphs of the Sine and Cosine Functions

Translations of the Graphs of the Sine and Cosine Functions

Sine and Cosine Functions

4.1

4.2

–1

1 y = sin x

3p 2pp2

Domain: 1-∞, ∞2Range: 3-1, 14Amplitude: 1

Period: 2p

–1

1 y = cos x

3p 2pp2

Domain: 1-∞, ∞2Range: 3-1, 14Amplitude: 1

Period: 2p

The graph of

y = c + a sin 3b 1x − d 2 4 or y = c + a cos 3b 1x − d 2 4 ,with b 7 0, has the following characteristics.

1. amplitude $ a $

2. period 2pb3. vertical translation c units up if c 7 0 or $ c $ units down

if c 6 04. phase shift d units to the right if d 7 0 or $d $ units to the

left if d 6 0

Graphs of the Tangent and Cotangent Functions

Tangent and Cotangent Functions

4.3Graph y = 2 tan x over a one-period interval.

y = 2 tan x

–2–p

period: p

domain: E x $ x ≠ 12n + 12 p2 , where n is any integer F

range: 1-∞, ∞2

y = tan x

–p2

Do main: E x $ x ≠ 12n + 12 p2 , where n is any integer F

Range: 1-∞, ∞2Period: p

y = cot x

p2p

Do main: 5x $ x ≠ np, where n is any integer6

Range: 1-∞, ∞2Period: p

Graph y = 1 + sin 3x.

–1

2y = 1 + sin 3x

34ppp3

2p3

amplitude: 1 domain: 1-∞, ∞2period: 2p3 range: 30, 24vertical translation: 1 unit up

Graph y = -2 cos Ax + p2 B .x

–2

2y = –2 cos (x + )P2

5p2

3p2

7p2

–

amplitude: 2 domain: 1-∞, ∞2period: 2p range: 3-2, 24phase shift: p2 units left

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Concepts Examples

Graphs of the Secant and Cosecant Functions4.4

Secant and Cosecant Functions Graph y = sec Ax + p4 B over a one-period interval.x

y = sec x

–1–p2

p–p

Do main: E x ' x ≠ 12n + 12 p2 , where n is any integer F

Range: 1-∞, -14 ´ 31, ∞2Period: 2p

y = csc x1

–1

–p2

p p–p2

Do main: 5x ' x ≠ np, where n is any integer6

Range: 1-∞, -14 ´ 31, ∞2Period: 2p

Harmonic Motion

Simple Harmonic MotionThe position of a point oscillating about an equilibrium position at time t is modeled by either

s 1 t 2 = a cos V t or s 1 t 2 = a sin V t,where a and v are constants, with v 7 0. The amplitude of the motion is ' a ' , the period is 2pv , and the frequency is

v2p

oscillations per time unit.

4.5

5p4

7p4

3p4

y = sec(x + )P4

–1–

period: 2p

phase shift: p4 unit left

domain: E x ' x ≠ p4 + np, where n is any integer F

range: 1-∞, -14 ´ 31, ∞2A spring oscillates according to

s1t2 = -5 cos 6t,where t is in seconds and s1t2 is in inches. Find the amplitude, period, and frequency.

amplitude = ' -5 ' = 5 in. period = 2p6

= p3

sec

frequency = 3p

oscillation per sec

1. Concept Check Which one of the following statements is true about the graph of y = 4 sin 2x?

A. It has amplitude 2 and period p2 . B. It has amplitude 4 and period p.

C. Its range is 30, 44. D. Its range is 3-4, 04. 2. Concept Check Which one of the following statements is false about the graph of

y = -3 cos 12 x?A. Its range is 3-3, 34. B. Its domain is 1-∞, ∞2.C. Its amplitude is 3, and its period is 4p. D. Its amplitude is -3, and its period is p.

3. Concept Check Which of the basic circular functions can have y-value 12 ?

4. Concept Check Which of the basic circular functions can have y-value 2?

Chapter 4 Review Exercises

CHAPTER 4 Review Exercises 189

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190 CHAPTER 4 Graphs of the Circular Functions

For each function, give the amplitude, period, vertical translation, and phase shift, as applicable.

5. y = 2 sin x 6. y = tan 3x 7. y = - 12

cos 3x

8. y = 2 sin 5x 9. y = 1 + 2 sin 14

x 10. y = 3 - 14

cos 23

11. y = 3 cos ax + p2b 12. y = -sin ax - 3p

4b 13. y = 1

2 csc a2x - p

14. y = 2 sec1px - 2p2 15. y = 13

tan a3x - p3b 16. y = cot a x

2+ 3p

Concept Check Identify the circular function that satisfies each description.

17. period is p; x-intercepts have x-values of the form np, where n is any integer

18. period is 2p; graph passes through the origin

19. period is 2p; graph passes through the point Ap2 , 0 B20. period is 2p; domain is 5x ' x ≠ np, where n is any integer621. period is p; function is decreasing on the interval 10, p222. period is 2p; has vertical asymptotes of the form x = 12n + 12 p2 , where n is any

integer

Provide a short explanation.

23. Suppose that ƒ defines a sine function with period 10 and ƒ152 = 3. Explain why ƒ1252 = 3.

24. Suppose that ƒ defines a sine function with period p and ƒ A6p5 B = 1. Explain why ƒ A - 4p5 B = 1.

Graph each function over a one-period interval.

25. y = 3 sin x 26. y = 12

sec x 27. y = - tan x

28. y = -2 cos x 29. y = 2 + cot x 30. y = -1 + csc x

31. y = sin 2x 32. y = tan 3x 33. y = 3 cos 2x

34. y = 12

cot 3x 35. y = cos ax - p4b 36. y = tan ax - p

37. y = sec a2x + p3b 38. y = sin a3x + p

2b 39. y = 1 + 2 cos 3x

40. y = -1 - 3 sin 2x 41. y = 2 sin px 42. y = - 12

cos1px - p2(Modeling) Monthly Temperatures A set of temperature data (in °F) is given for a par-ticular location. (Source: www.weatherbase.com)

(a) Plot the data over a two-year interval.

(b) Use sine regression to determine a model for the two-year interval. Let x = 1 repre-sent January of the first year.

43. Average Monthly Temperature, Auckland, New Zealand

Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec

67.6 68.5 65.8 61.3 57.2 53.2 51.6 52.9 55.4 58.1 61.2 64.9

M04_LHSD7642_11_AIE_C04_pp139-194.indd 190 06/11/15 2:09 pm

http://www.weatherbase.com

44. Average Low Temperature, Auckland, New Zealand

Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec

60.8 61.7 58.8 54.9 51.1 47.1 45.5 46.8 49.5 52.2 55.0 58.8

Connecting Graphs with Equations Determine the simplest form of an equation for each graph. Choose b 7 0, and include no phase shifts.

45.

–1

–2

pp2

2p3p2

46.

–1

–2

pp2

3p4

47.

–3

–p pp2

–

48.

–1 pp2

2p3p2

–3

Solve each problem.

49. Viewing Angle to an Object Suppose that a person whose eyes are h1 feet from the ground is standing d feet from an object h2 feet tall, where h2 7 h1. Let u be the angle of ele-vation to the top of the object. See the figure.

(a) Show that d = 1h2 - h12 cot u.(b) Let h2 = 55 and h1 = 5. Graph d for the interval 0 6 u …

p2 .

h1d

50. (Modeling) Tides The figure shows a func-tion ƒ that models the tides in feet at Clearwater Beach, Florida, x hours after midnight. (Source: Pentcheff, D., WWW Tide and Current Predictor.)

(a) Find the time between high tides.

(b) What is the difference in water levels between high tide and low tide?

ƒ1x2 = 0.6 cos 30.5111x - 2.424 + 2.Estimate the tides, to the nearest hundredth, when x = 10.

40 8 12 16 20 24 28

Time (in hours)

Tide

s (i

n fe

et) (2.4, 2.6)

(14.7, 2.6)

(27, 2.6)

(21, 1.4)(8.7, 1.4)

51. (Modeling) Maximum Temperatures The maximum afternoon temperature (in °F) in a given city can be modeled by

t = 60 - 30 cos xp6

where t represents the maximum afternoon temperature in month x, with x = 0 representing January, x = 1 representing February, and so on. Find the maximum afternoon temperature, to the nearest degree, for each month.

(a) January (b) April (c) May

(d) June (e) August (f ) October

CHAPTER 4 Review Exercises 191

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192 CHAPTER 4 Graphs of the Circular Functions

53. (Modeling) Pollution Trends The amount of pollution in the air is lower after heavy spring rains and higher after periods of little rain. In addition to this seasonal fluctuation, the long-term trend is upward. An idealized graph of this situation is shown in the figure.

600

Circular functions can be used to model the fluctuating part of the pollution levels. Powers of the number e (e is the base of the natural logarithm; e ≈ 2.718282) can be used to model long-term growth. The pollution level in a certain area might be given by

y = 711 - cos 2px21x + 102 + 100e0.2x,where x is the time in years, with x = 0 representing January 1 of the base year. July 1 of the same year would be represented by x = 0.5, October 1 of the following year would be represented by x = 1.75, and so on. Find the pollution levels on each date.(a) January 1, base year (See the screen.) (b) July 1, base year

52. (Modeling) Average Monthly Temperature The average monthly temperature (in °F) in Chicago, Illinois, is shown in the table.

(a) Plot the average monthly temperature over a two-year period. Let x = 1 correspond to January of the first year.

(b) To model the data, determine a function of the form ƒ1x2 = a sin3b1x - d24 + c, where a, b, c, and d are constants.

(d) Use the sine regression capability of a graphing calculator to find the equation of a sine curve of the form y = a sin1bx + c) + d that fits these data.

Source: World Almanac and Book of Facts.

Month °F Month °FJan 22 July 73

Feb 27 Aug 72

Mar 37 Sept 64

Apr 48 Oct 52

May 59 Nov 39

June 68 Dec 27

54. (Modeling) Lynx and Hare Populations The figure shows the populations of lynx and hares in Canada for the years 1847–1903. The hares are food for the lynx. An increase in hare population causes an increase in lynx population some time later. The increasing lynx population then causes a decline in hare population. The two graphs have the same period.

Year

Canadian Lynx and Hare Populations

Num

ber

150,000

100,000

50,000

1850 1860 1870 1880 1890 1900

HareLynx

(a) Estimate the length of one period.

(b) Estimate the maximum and minimum hare populations.

M04_LHSD7642_11_AIE_C04_pp139-194.indd 192 06/11/15 2:09 pm

An object in simple harmonic motion has position function s1t2 inches from an equilib-rium point, where t is the time in seconds. Find the amplitude, period, and frequency.

55. s1t2 = 4 sin pt 56. s1t2 = 3 cos 2t57. In Exercise 55, what does the frequency represent? Find the position of the object

relative to the equilibrium point at 1.5 sec, 2 sec, and 3.25 sec.

58. In Exercise 56, what does the period represent? What does the amplitude represent?

CHAPTER 4 Test 193

1. Identify each of the following basic circular function graphs.

Chapter 4 Test

(a)

–1

–2p 2p–p p

(b)

0–1–2

3p2

2ppp2

–p–2p

(c)

0–1–2

3p2

–p

–2p

(d)

–2

x–p

2p4p2

(e)

–10

–2p 2p–p px

(f )

–1

xpp

4p2

2. Connecting Graphs with Equations Determine the simplest form of an equation for each graph. Choose b 7 0, and include no phase shifts.(a)

–1

–2

4px

3p2pp

(b)

–1

–2

3p4

3. Answer each question.

(a) What is the domain of the cosine function?

(b) What is the range of the sine function?

(d) What is the range of the secant function?

M04_LHSD7642_11_AIE_C04_pp139-194.indd 193 06/11/15 2:09 pm

194 CHAPTER 4 Graphs of the Circular Functions

4. Consider the function y = 3 - 6 sin A2x + p2 B.(a) What is its period?

(b) What is the amplitude of its graph?

(d) What is the y-intercept of its graph?

(e) What is its phase shift?

Graph each function over a two-period interval. Identify asymptotes when applicable.

5. y = sin12x + p2 6. y = -cos 2x 7. y = 2 + cos x 8. y = -1 + 2 sin1x + p2 9. y = tan ax - p

2b 10. y = -2 - cot ax - p

11. y = -csc 2x 12. y = 3 csc px

(Modeling) Solve each problem.

13. Average Monthly Temperature The average monthly temperature (in °F) in San Antonio, Texas, can be modeled by

ƒ 1x2 = 16.5 sin cp6

1x - 42 d + 67.5,where x is the month and x = 1 corresponds to January. (Source: World Almanac and Book of Facts.)

(a) Graph ƒ in the window 30, 254 by 340, 904. (b) Determine the amplitude, period, phase shift, and vertical translation of ƒ.

(d) Determine the minimum and maximum average monthly temperatures and the months when they occur.

(e) What would be an approximation for the average annual temperature in San Antonio? How is this related to the vertical translation of the sine function in the formula for ƒ?

14. Spring Motion The position of a weight attached to a spring is

s1t2 = -4 cos 8pt inches after t seconds. (a) Find the maximum height that the weight rises above the equilibrium position of

s1t2 = 0. (b) When does the weight first reach its maximum height if t Ú 0? (c) What are the frequency and period?

15. Explain why the domains of the tangent and secant functions are the same, and then give a similar explanation for the cotangent and cosecant functions.

M04_LHSD7642_11_AIE_C04_pp139-194.indd 194 06/11/15 2:09 pm

195

Electricity that passes through wires to homes and businesses alternates its direction on those wires and is modeled by sine and cosine functions.

Fundamental Identities

Verifying Trigonometric Identities

Sum and Difference Identities for Cosine

Sum and Difference Identities for Sine and Tangent

Chapter 5 Quiz

Double-Angle Identities

Half-Angle Identities

Summary Exercises on Verifying Trigonometric Identities

5.1

5.2

5.3

5.4

5.5

5.6

Trigonometric Identities5

M05_LHSD7642_11_AIE_C05_pp195-250.indd 195 16/11/15 3:18 pm

196 CHAPTER 5 Trigonometric Identities

5.1 Fundamental Identities

Fundamental Identities Recall that a function is even if ƒ 1−x 2 = ƒ 1x 2 for all x in the domain of ƒ, and a function is odd if ƒ 1−x 2 = −ƒ 1x 2 for all x in the domain of ƒ. We have used graphs to classify the trigono-metric functions as even or odd. We can also use Figure 1 to do this.

As suggested by the circle in Figure 1, an angle u having the point 1x, y2 on its terminal side has a corresponding angle -u with the point 1x, -y2 on its terminal side.

From the definition of sine, we see that sin1-u2 and sin u are negatives of each other. That is,

sin1-u2 = -yr and sin u =

yr ,

so sin 1−U 2 = −sin U Sine is an odd function.This is an example of an identity, an equation that is satisfied by every value in the domain of its variable. Some examples from algebra follow.

x2 - y2 = 1x + y21x - y2 x1x + y2 = x2 + xy Identities

x2 + 2xy + y2 = 1x + y22Figure 1 shows an angle u in quadrant II, but the same result holds for u in

any quadrant. The figure also suggests the following identity for cosine.

cos1-u2 = xr and cos u = x

cos 1−U 2 = cos U Cosine is an even function.We use the identities for sin1-u2 and cos1-u2 to find tan1-u2 in terms of tan u.

tan1-u2 = sin1-u2cos1-u2 = -sin ucos u = - sin ucos u

tan 1−U 2 = − tan U Tangent is an odd function.The reciprocal identities are used to determine that cosecant and cotangent are odd functions and secant is an even function. These even-odd identities together with the reciprocal, quotient, and Pythagorean identities make up the fundamental identities.

■ Fundamental Identities

■ Uses of the Fundamental Identities

sin(–U) = – = –sin Uyr

(x, y)

–yr

(x, –y)

–U

Figure 1

NOTE In trigonometric identities, u can represent an angle in degrees or radians, or a real number.

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1975.1 Fundamental Identities

Fundamental Identities

Reciprocal Identities

cot U =1

tan U sec U =

1cos U

csc U =1

sin U

Quotient Identities

tan U =sin Ucos U

cot U =cos Usin U

Pythagorean Identities

sin2 U + cos2 U = 1 tan2 U + 1 = sec2 U 1 + cot2 U = csc2 U

Even-Odd Identities

sin 1−U 2 = −sin U cos 1−U 2 = cos U tan 1−U 2 = − tan U csc 1−U 2 = −csc U sec 1−U 2 = sec U cot 1−U 2 = −cot U

EXAMPLE 1 Finding Trigonometric Function Values Given One Value and the Quadrant

If tan u = - 53 and u is in quadrant II, find each function value.(a) sec u (b) sin u (c) cot1-u2SOLUTION

(a) We use an identity that relates the tangent and secant functions.

tan2 u + 1 = sec2 u Pythagorean identity

a - 53b 2 + 1 = sec2 u tan u = - 53 259

+ 1 = sec2 u Square - 53 .

349

= sec2 u Add; 1 = 99

-B349 = sec u Take the negative square root because u is in quadrant II. sec u = - 234

Simplify the radical: -3349 = - 23429 = - 2343 , and rewrite.Choose the correct sign.

Uses of the Fundamental Identities We can use these identities to find the values of other trigonometric functions from the value of a given trigonomet-ric function.

NOTE We will also use alternative forms of the fundamental identities. For example, two other forms of sin2 U + cos2 U = 1 are

sin2 U = 1 − cos2 U and cos2 U = 1 − sin2 U.

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198 CHAPTER 5 Trigonometric Identities

(b) tan u = sin ucos u

Quotient identity

cos u tan u = sin u Multiply each side by cos u.

a 1sec u

b tan u = sin u Reciprocal identity ¢ - 3234

34≤ a - 5

3b = sin u tan u = - 53 , and from part (a), 1

sec u =1

- 2343 = - 3234 = - 3234 # 234234 = - 323434 . sin u = 5234

34 Multiply and rewrite.

cot1-u2 = 1- tan u Even-odd identity cot1-u2 = 1

- A - 53 B tan u = - 53 cot1-u2 = 3

- A - 53 B = 1 , 53 = 1 # 35 = 35■✔ Now Try Exercises 11, 19, and 31.

EXAMPLE 2 Writing One Trignometric Function in Terms of Another

Write cos x in terms of tan x.

SOLUTION By identities, sec x is related to both cos x and tan x.

1 + tan2 x = sec2 x Pythagorean identity

1 + tan2 x =1

sec2 x Take reciprocals.

1 + tan2 x = cos2 x The reciprocal of sec2 x is cos2 x.

{B 11 + tan2 x = cos x Take the square root of each side. cos x = {121 + tan2 x Quotient rule for radicals: 3n ab = 1na2n b ; rewrite. cos x = {21 + tan2 x

1 + tan2 x Rationalize the denominator.

Remember both the positive and negative roots.

The choice of the + sign or the - sign is made depending on the quadrant of x.■✔ Now Try Exercise 47.

Figure 2 supports the identity sin2 x + cos2 x = 1. ■

y1 = sin2 x + cos2 xy2 = 1

−4

11p4

−

With an identity, there should be no difference between the two graphs.

Figure 2

CAUTION When taking the square root, be sure to choose the sign based on the quadrant of U and the function being evaluated.

M05_LHSD7642_11_AIE_C05_pp195-250.indd 198 16/11/15 3:18 pm

1995.1 Fundamental Identities

The functions tan u, cot u, sec u, and csc u can easily be expressed in terms of sin u, cos u, or both. We make such substitutions in an expression to simplify it.

CONCEPT PREVIEW For each expression in Column I, choose the expression from Column II that completes an identity.

5.1 Exercises

1. cos xsin x

2. tan x =

3. cos1-x2 = 4. tan2 x + 1 =

5. 1 =

A. sin2 x + cos2 x

B. cot x

C. sec2 x

D. sin xcos x

E. cos x

EXAMPLE 3 Rewriting an Expression in Terms of Sine and Cosine

Write 1 + cot2 u

1 - csc2 u in terms of sin u and cos u, and then simplify the expression so that no quotients appear.

SOLUTION

1 + cot2 u1 - csc2 u Given expression

=1 + cos

2 usin2 u

1 - 1sin2 u

Quotient identities

=a1 + cos2 u

sin2 ubsin2 u

a1 - 1sin2 u

bsin2 u Simplify the complex fraction by multiplying both numerator and denominator by the LCD. = sin

2 u + cos2 usin2 u - 1

Distributive property: 1b + c2a = ba + ca = 1-cos2 u Pythagorean identities

= - sec2 u Reciprocal identity■✔ Now Try Exercise 59.

y2 = −sec2 x

−4

11p4

1 + cot2 x1 − csc2 x

y1 =

11p4

−

The graph supports the result in Example 3. The graphs of y1 and y2 coincide.

CAUTION When working with trigonometric expressions and identities, be sure to write the argument of the function. For example, we would not write sin2 + cos2 = 1. An argument such as u is necessary to write this cor-rectly as sin2 u + cos2 u = 1.

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200 CHAPTER 5 Trigonometric Identities

CONCEPT PREVIEW Use identities to correctly complete each sentence.

6. If tan u = 2.6, then tan1-u2 = . 7. If cos u = -0.65, then cos1-u2 = . 8. If tan u = 1.6, then cot u = .

9. If cos u = 0.8 and sin u = 0.6, then tan1-u2 = . 10. If sin u = 23 , then - sin1-u2 = .Find sin u. See Example 1.

11. cos u = 34 , u in quadrant I 12. cos u =56 , u in quadrant I

13. cot u = - 15 , u in quadrant IV 14. cot u = - 13 , u in quadrant IV

15. cos1-u2 = 255 , tan u 6 0 16. cos1-u2 = 236 , cot u 6 017. tan u = - 262 , cos u 7 0 18. tan u = - 272 , sec u 7 019. sec u = 114 , cot u 6 0 20. sec u =

72 , tan u 6 0

21. csc u = - 94 22. csc u = - 85

23. Why is it unnecessary to give the quadrant of u in Exercises 21 and 22?

24. Concept Check What is WRONG with the statement of this problem?

Find cos1-u2 if cos u = 3.Concept Check Find ƒ1-x2 to determine whether each function is even or odd.25. ƒ1x2 = sin x

x 26. ƒ1x2 = x cos x

Concept Check Identify the basic trigonometric function graphed and determine whether it is even or odd.

27.

−4

11p4

−

28.

−4

11p4

−

29.

−4

11p4

−

30.

−4

11p4

−

Find the remaining five trigonometric functions of u. See Example 1.

31. sin u = 23 , u in quadrant II 32. cos u =15 , u in quadrant I

33. tan u = - 14 , u in quadrant IV 34. csc u = - 52 , u in quadrant III

35. cot u = 43 , sin u 7 0 36. sin u = - 45 , cos u 6 0

37. sec u = 43 , sin u 6 0 38. cos u = - 14 , sin u 7 0

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2015.1 Fundamental Identities

Concept Check For each expression in Column I, choose the expression from Column II that completes an identity. One or both expressions may need to be rewritten.

39. - tan x cos x =

40. sec2 x - 1 =

41. sec xcsc x

42. 1 + sin2 x =

43. cos2 x =

A. sin2 xcos2 x

B. 1

sec2 x

C. sin1-x2 D. csc2 x - cot2 x + sin2 x

E. tan x

44. A student writes “1 + cot2 = csc2.” Comment on this student’s work.

45. Concept Check Suppose that cos u = xx + 1 . Find an expression in x for sin u.

46. Concept Check Suppose that sec u = x + 4x . Find an expression in x for tan u.

Perform each transformation. See Example 2.

47. Write sin x in terms of cos x. 48. Write cot x in terms of sin x.

49. Write tan x in terms of sec x. 50. Write cot x in terms of csc x.

51. Write csc x in terms of cos x. 52. Write sec x in terms of sin x.

Write each expression in terms of sine and cosine, and then simplify the expression so that no quotients appear and all functions are of u only. See Example 3.

53. cot u sin u 54. tan u cos u

55. sec u cot u sin u 56. csc u cos u tan u

57. cos u csc u 58. sin u sec u

59. sin2 u1csc2 u - 12 60. cot2 u11 + tan2 u2 61. 11 - cos u211 + sec u2 62. 1sec u - 121sec u + 12 63.

1 + tan1-u2tan1-u2 64. 1 + cot ucot u

65. 1 - cos21-u21 + tan21-u2 66. 1 - sin21-u21 + cot21-u2

67. sec u - cos u 68. csc u - sin u

69. 1sec u + csc u21cos u - sin u2 70. 1sin u - cos u21csc u + sec u271. sin u1csc u - sin u2 72. cos u1cos u - sec u273.

1 + tan2 u1 + cot2 u 74.

sec2 u - 1csc2 u - 1

75. csc u

cot1-u2 76. tan1-u2sec u77. sin21-u2 + tan21-u2 + cos21-u2 78. -sec21-u2 + sin21-u2 + cos21-u2

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202 CHAPTER 5 Trigonometric Identities

Use a graphing calculator to make a conjecture about whether each equation is an identity.

81. cos 2x = 1 - 2 sin2 x 82. 2 sin x = sin 2x

83. sin x = 21 - cos2 x 84. cos 2x = cos2 x - sin2 xRelating Concepts

For individual or collaborative investigation (Exercises 85–90)

Previously we graphed functions of the form

y = c + a # ƒ 3b1x - d24with the assumption that b 7 0. To see what happens when b 6 0, work Exer-cises 85–90 in order.

85. Use an even-odd identity to write y = sin1-2x2 as a function of 2x.86. How is the answer to Exercise 85 related to y = sin 2x?

87. Use an even-odd identity to write y = cos1-4x2 as a function of 4x.88. How is the answer to Exercise 87 related to y = cos 4x?

89. Use the results from Exercises 85 – 88 to rewrite the following with a positive value of b.

(a) y = sin1-4x2 (b) y = cos1-2x2 (c) y = -5 sin1-3x290. Write a short response to this statement, which is often used by one of the

authors of this text in trigonometry classes:

Students who tend to ignore negative signs should enjoy graphing functions involving the cosine and the secant.

5.2 Verifying Trigonometric Identities

Strategies One of the skills required for more advanced work in math-ematics, especially in calculus, is the ability to use identities to write expressions in alternative forms. We develop this skill by using the fundamental identities to verify that a trigonometric equation is an identity (for those values of the variable for which it is defined).

■ Strategies■ Verifying Identities by

Working with One Side■ Verifying Identities

by Working with Both Sides

CAUTION The procedure for verifying identities is not the same as that for solving equations. Techniques used in solving equations, such as add-ing the same term to each side, and multiplying each side by the same term, should not be used when working with identities.

Work each problem.

79. Let cos x = 15 . Find all possible values of sec x - tan x

sin x .

80. Let csc x = -3. Find all possible values of sin x + cos xsec x .

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2035.2 Verifying Trigonometric Identities

Verifying Identities by Working with One Side Avoid the temptation to use algebraic properties of equations to verify identities.

One strategy is to work with one side and rewrite it to match the other side.

Hints for Verifying Identities

1. Learn the fundamental identities. Whenever you see either side of a fundamental identity, the other side should come to mind. Also, be aware of equivalent forms of the fundamental identities. For example,

sin2 u = 1 - cos2 u is an alternative form of sin2 u + cos2 u = 1.2. Try to rewrite the more complicated side of the equation so that it is

identical to the simpler side.

3. It is sometimes helpful to express all trigonometric functions in the equation in terms of sine and cosine and then simplify the result.

4. Usually, any factoring or indicated algebraic operations should be performed. These algebraic identities are often used in verifying trigo-nometric identities.

x2 + 2xy + y2 = 1x + y22 x2 - 2xy + y2 = 1x - y22

x3 - y3 = 1x - y21x2 + xy + y22 x3 + y3 = 1x + y21x2 - xy + y22 x2 - y2 = 1x + y21x - y2

For example, the expression

sin2 x + 2 sin x + 1 can be factored as 1sin x + 122. The sum or difference of two trigonometric expressions can be found in

the same way as any other rational expression. For example,

1sin u

+ 1cos u

= 1# cos u

sin u cos u+ 1

# sin ucos u sin u

Write with the LCD.

= cos u + sin usin u cos u

. ac +bc =

a + bc

5. When selecting substitutions, keep in mind the side that is not changing, because it represents the goal. For example, to verify that the equation

tan2 x + 1 = 1cos2 x

is an identity, think of an identity that relates tan x to cos x. In this case, because sec x = 1cos x and sec2 x = tan2 x + 1, the secant function is the best link between the two sides.

6. If an expression contains 1 + sin x, multiplying both numerator and denominator by 1 - sin x would give 1 - sin2 x, which could be replaced with cos2 x. Similar procedures apply for 1 - sin x, 1 + cos x, and 1 - cos x.

LOOKING AHEAD TO CALCULUSTrigonometric identities are used in

calculus to simplify trigonometric

expressions, determine derivatives of

trigonometric functions, and change

the form of some integrals.

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204 CHAPTER 5 Trigonometric Identities

EXAMPLE 1 Verifying an Identity (Working with One Side)

Verify that the following equation is an identity.

cot u + 1 = csc u1cos u + sin u2SOLUTION We use the fundamental identities to rewrite one side of the equa-tion so that it is identical to the other side. The right side is more complicated, so we work with it, as suggested in Hint 2, and use Hint 3 to change all functions to expressions involving sine or cosine.

Steps Reasons Right side of given equation $11111111%11111111&

csc u1cos u + sin u2 = 1sin u

1cos u + sin u2 csc u = 1sin u= cos u

sin u+ sin u

sin uDistributive property: a1b + c2 = ab + ac

= cot u + 1 cos usin u = cot u; sin usin u = 1

(11)11* Left side of given equation

−4

11p4

For u = x, y1 = cot x + 1 y2 = csc x(cos x + sin x)

11p4

−

The graphs coincide, which supports the conclusion in Example 1. The given equation is an identity. The right side of the equation is identical to the

left side.■✔ Now Try Exercise 45.

EXAMPLE 2 Verifying an Identity (Working with One Side)

Verify that the following equation is an identity.

tan2 x11 + cot2 x2 = 11 - sin2 x

SOLUTION We work with the more complicated left side, as suggested in Hint 2. Again, we use the fundamental identities.

Left side of given equation$111111%111111&

tan2 x11 + cot2 x2 = tan2 x + tan2 x cot2 x Distributive property = tan2 x + tan2 x # 1

tan2 x cot2 x = 1tan2 x

= tan2 x + 1 tan2 x # 1tan2 x = 1 = sec2 x Pythagorean identity

= 1cos2 x

sec2 x = 1cos2 x

= 11 - sin2 x Pythagorean identity(111)111*

Right side of given equation

Because the left side of the equation is identical to the right side, the given equa-tion is an identity.

■✔ Now Try Exercise 49.

−4

11p4

y1 = tan2 x(1 + cot2 x)1

1 − sin2 xy2 =

(Video) An Exponential Equation | 2^{x^3}=3^{x^2}

11p4

−

The screen supports the conclusion in Example 2.

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2055.2 Verifying Trigonometric Identities

EXAMPLE 3 Verifying an Identity (Working with One Side)

Verify that the following equation is an identity.

tan t - cot tsin t cos t

= sec2 t - csc2 t

SOLUTION We transform the more complicated left side to match the right side.

tan t - cot tsin t cos t

= tan tsin t cos t

- cot tsin t cos t

a - bc =ac -

= tan t # 1sin t cos t

- cot t # 1sin t cos t

ab = a # 1b

= sin tcos t

# 1sin t cos t

- cos tsin t

# 1sin t cos t

tan t = sin tcos t ; cot t =cos tsin t

= 1cos2 t

- 1sin2 t

Multiply.

= sec2 t - csc2 t 1cos2 t = sec2 t; 1

sin2 t = csc2 t

Hint 3 about writing all trigonometric functions in terms of sine and cosine was used in the third line of the solution.

■✔ Now Try Exercise 53.

EXAMPLE 4 Verifying an Identity (Working with One Side)

Verify that the following equation is an identity.

cos x1 - sin x =

1 + sin xcos x

SOLUTION We work on the right side, using Hint 6 in the list given earlier to multiply the numerator and denominator on the right by 1 - sin x.

1 + sin x

cos x=11 + sin x211 - sin x2

cos x11 - sin x2 Multiply by 1 in the form 1 - sin x1 - sin x . = 1 - sin

2 xcos x11 - sin x2 1x + y21x - y2 = x2 - y2

= cos2 x

cos x11 - sin x2 1 - sin2 x = cos2 x = cos x

# cos xcos x11 - sin x2 a2 = a # a

= cos x1 - sin x Write in lowest terms.

■✔ Now Try Exercise 59.

Verifying Identities by Working with Both Sides If both sides of an identity appear to be equally complex, the identity can be verified by working independently on the left side and on the right side, until each side is changed into some common third result. Each step, on each side, must be reversible. With all steps reversible, the procedure is as shown in the margin. The left side leads to a common third expression, which leads back to the right side.

left = right

common third expression

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206 CHAPTER 5 Trigonometric Identities

NOTE Working with both sides is often a good alternative for identities that are difficult. In practice, if working with one side does not seem to be effective, switch to the other side. Somewhere along the way it may happen that the same expression occurs on both sides.

EXAMPLE 5 Verifying an Identity (Working with Both Sides)

Verify that the following equation is an identity.

sec a + tan asec a - tan a =

1 + 2 sin a + sin2 acos2 a

SOLUTION Both sides appear equally complex, so we verify the identity by changing each side into a common third expression. We work first on the left, multiplying the numerator and denominator by cos a.

sec a + tan asec a - tan a =

1sec a + tan a2 cos a1sec a - tan a2 cos a Multiply by 1 in the form cos acos a .(1111)1111* Left side of given equation =

sec a cos a + tan a cos asec a cos a - tan a cos a Distributive property

= 1 + tan a cos a1 - tan a cos a sec a cos a = 1

=1 + sin a

cos a# cos a

1 - sin acos a

# cos a tan a = sin acos a

= 1 + sin a1 - sin a Simplify.

On the right side of the original equation, we begin by factoring.

1 + 2 sin a + sin2 a

cos2 a=11 + sin a22

cos2 a

Factor the numerator; x2 + 2xy + y2 = 1x + y22.

(11111111)11111111* Right side of given equation =

11 + sin a221 - sin2 a cos

2 a = 1 - sin2 a

=11 + sin a2211 + sin a211 - sin a2 Factor the denominator; x2 - y2 = 1x + y21x - y2.

= 1 + sin a1 - sin a Write in lowest terms.

We have shown that

Left side of Common third Right side of given equation expression given equation$1111%1111& $11%11& $11111111%11111111&

sec a + tan asec a - tan a =

1 + sin a1 - sin a =

1 + 2 sin a + sin2 acos2 a

and thus have verified that the given equation is an identity.

■✔ Now Try Exercise 75.

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2075.2 Verifying Trigonometric Identities

CAUTION Use the method of Example 5 only if the steps are reversible.

An Inductor and a Capacitor

Figure 3

There are usually several ways to verify a given identity. Another way to begin verifying the identity in Example 5 is to work on the left as follows.

sec a + tan asec a - tan a =

1cos a

+ sin acos a

1cos a

- sin acos a

Fundamental identities

(11111)11111* Left side of given equation in Example 5

1 + sin acos a

1 - sin acos a

Add and subtract fractions.

= 1 + sin acos a

, 1 - sin acos a

Simplify the complex fraction. Use the definition of division.

= 1 + sin acos a

# cos a1 - sin a Multiply by the reciprocal.

= 1 + sin a1 - sin a Multiply and write in lowest terms.

Compare this with the result shown in Example 5 for the right side to see that the two sides indeed agree.

EXAMPLE 6 Applying a Pythagorean Identity to Electronics

Tuners in radios select a radio station by adjusting the frequency. A tuner may contain an inductor L and a capacitor C, as illustrated in Figure 3. The energy stored in the inductor at time t is given by

L1t2 = k sin2 2pFtand the energy stored in the capacitor is given by

C1t2 = k cos2 2pFt,where F is the frequency of the radio station and k is a constant. The total energy E in the circuit is given by

E1t2 = L1t2 + C1t2.Show that E is a constant function. (Source: Weidner, R. and R. Sells, Elemen-tary Classical Physics, Vol. 2, Allyn & Bacon.)

SOLUTION

E1t2 = L1t2 + C1t2 Given equation = k sin2 2pFt + k cos2 2pFt Substitute. = k3sin2 2pFt + cos2 2pFt4 Factor out k. = k112 sin2 u + cos2 u = 1 1Here u = 2pFt.2 = k Identity property

Because k is a constant, E1t2 is a constant function. ■✔ Now Try Exercise 105.

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208 CHAPTER 5 Trigonometric Identities

5.2 Exercises

To the student: Exercises 1–44 are designed for practice in using the fundamental iden-tities and applying algebraic techniques to trigonometric expressions. These skills are essential in verifying the identities that follow.

CONCEPT PREVIEW Match each expression in Column I with its correct factorization in Column II.

1. x2 - y2

2. x3 - y3

3. x3 + y3

4. x2 + 2xy + y2

A. 1x + y21x2 - xy + y22B. 1x + y21x - y2C. 1x + y22D. 1x - y21x2 + xy + y22

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each fundamental identity.

5. sin2 u + cos2 u = 6. tan2 u + 1 =

7. sin1-u2 = 8. sec1-u2 = 9. tan u = 1

= sin u

10. cot u = 1

= cos u

Perform each indicated operation and simplify the result so that there are no quotients.

11. cot u + 1cot u

12. sec xcsc x

+ csc xsec x

13. tan x1cot x + csc x214. cos b1sec b + csc b2 15. 1

csc2 u+ 1

sec2 u 16.

cos xsec x

+ sin xcsc x

17. 1sin a - cos a22 18. 1tan x + cot x22 19. 11 + sin t22 + cos2 t20. 11 + tan u22 - 2 tan u 21. 1

1 + cos x -1

1 - cos x 22. 1

sin a - 1 -1

sin a + 1

Factor each trigonometric expression.

23. sin2 u - 1 24. sec2 u - 1

25. 1sin x + 122 - 1sin x - 122 26. 1tan x + cot x22 - 1tan x - cot x2227. 2 sin2 x + 3 sin x + 1 28. 4 tan2 b + tan b - 3

29. cos4 x + 2 cos2 x + 1 30. cot4 x + 3 cot2 x + 2

31. sin3 x - cos3 x 32. sin3 a + cos3 a

Each expression simplifies to a constant, a single function, or a power of a function. Use fundamental identities to simplify each expression.

33. tan u cos u 34. cot a sin a 35. sec r cos r

36. cot t tan t 37. sin b tan b

cos b 38.

csc u sec ucot u

39. sec2 x - 1 40. csc2 t - 1 41. sin2 x

cos2 x+ sin x csc x

42. 1

tan2 a+ cot a tan a 43. 1 - 1

csc2 x 44. 1 - 1

sec2 x

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2095.2 Verifying Trigonometric Identities

Verify that each equation is an identity. See Examples 1–5.

45. cot ucsc u

= cos u 46. tan asec a

= sin a

47. 1 - sin2 b

cos b= cos b 48. tan

2 a + 1sec a

= sec a

49. cos2 u1tan2 u + 12 = 1 50. sin2 b11 + cot2 b2 = 151. cot u + tan u = sec u csc u 52. sin2 a + tan2 a + cos2 a = sec2 a

53. cos asec a

+ sin acsc a

= sec2 a - tan2 a 54. sin2 u

cos u= sec u - cos u

55. sin4 u - cos4 u = 2 sin2 u - 1 56. sec4 x - sec2 x = tan4 x + tan2 x

57. 1 - cos x1 + cos x = 1cot x - csc x22 58. 1sec a - tan a22 = 1 - sin a1 + sin a

59. cos u + 1

tan2 u= cos u

sec u - 1 60. 1sec u - tan u22 + 1

sec u csc u - tan u csc u = 2 tan u

61. 1

1 - sin u +1

1 + sin u = 2 sec2 u 62.

1sec a - tan a = sec a + tan a

63. cot a + 1cot a - 1 =

1 + tan a1 - tan a 64.

csc u + cot utan u + sin u = cot u csc u

65. cos u

sin u cot u= 1 66. sin2 u11 + cot2 u2 - 1 = 0

67. sec4 u - tan4 usec2 u + tan2 u = sec

2 u - tan2 u 68. sin4 a - cos4 a

sin2 a - cos2 a = 1

69. tan2 t - 1

sec2 t= tan t - cot t

tan t + cot t 70. cot2 t - 11 + cot2 t = 1 - 2 sin

2 t

71. sin2 a sec2 a + sin2 a csc2 a = sec2 a 72. tan2 a sin2 a = tan2 a + cos2 a - 1

73. tan x

1 + cos x +sin x

1 - cos x = cot x + sec x csc x

74. sin u

1 - cos u -sin u cos u1 + cos u = csc u11 + cos2 u2

75. 1 + cos x1 - cos x -

1 - cos x1 + cos x = 4 cot x csc x

76. 1 + sin u1 - sin u -

1 - sin u1 + sin u = 4 tan u sec u

77. 1 - sin u1 + sin u = sec

2 u - 2 sec u tan u + tan2 u

78. sin u + cos u = sin u1 - cot u +

cos u1 - tan u

79. -1

tan a - sec a +-1

tan a + sec a = 2 tan a

80. 11 + sin x + cos x22 = 211 + sin x211 + cos x281. 11 - cos2 a211 + cos2 a2 = 2 sin2 a - sin4 a82. 1sec a + csc a21cos a - sin a2 = cot a - tan a83.

1 - cos x1 + cos x = csc

2 x - 2 csc x cot x + cot2 x

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210 CHAPTER 5 Trigonometric Identities

84. 1 - cos u1 + cos u = 2 csc

2 u - 2 csc u cot u - 1

85. 12 sin x + cos x22 + 12 cos x - sin x22 = 5 86. sin2 x11 + cot x2 + cos2 x11 - tan x2 + cot2 x = csc2 x 87. sec x - cos x + csc x - sin x - sin x tan x = cos x cot x

88. sin3 u + cos3 u = 1cos u + sin u211 - cos u sin u2Graph each expression and use the graph to make a conjecture, predicting what might be an identity. Then verify your conjecture algebraically.

89. 1sec u + tan u211 - sin u2 90. 1csc u + cot u21sec u - 12 91.

cos u + 1sin u + tan u 92. tan u sin u + cos u

Graph the expressions on each side of the equals symbol to determine whether the equation might be an identity. (Note: Use a domain whose length is at least 2p.) If the equation looks like an identity, then verify it algebraically. See Example 1.

93. 2 + 5 cos x

sin x= 2 csc x + 5 cot x 94. 1 + cot2 x = sec

2 xsec2 x - 1

95. tan x - cot xtan x + cot x = 2 sin

2 x 96. 1

1 + sin x +1

1 - sin x = sec2 x

By substituting a number for t, show that the equation is not an identity.

97. sin1csc t2 = 1 98. 2cos2 t = cos t 99. csc t = 21 + cot2 t 100. cos t = 21 - sin2 t(Modeling) Work each problem.

101. Intensity of a Lamp According to Lambert’s law, the intensity of light from a single source on a flat surface at point P is given by

I = k cos2 u,

where k is a constant. (Source: Winter, C., Solar Power Plants, Springer-Verlag.)

(a) Write I in terms of the sine function.

(b) Why does the maximum value of I occur when u = 0?

102. Oscillating Spring The distance or displacement y of a weight attached to an oscillating spring from its natural posi-tion is modeled by

y = 4 cos 2pt,

where t is time in seconds. Potential energy is the energy of position and is given by

P = ky2,

where k is a constant. The weight has the greatest potential energy when the spring is stretched the most. (Source: Weidner, R. and R. Sells, Elementary Classical Physics, Vol. 2, Allyn & Bacon.)

(a) Write an expression for P that involves the cosine function.

(b) Use a fundamental identity to write P in terms of sin 2pt.

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2115.3 Sum and Difference Identities for Cosine

(Modeling) Radio Tuners See Example 6. Let the energy stored in the inductor be given by

L1t2 = 3 cos2 6,000,000tand let the energy stored in the capacitor be given by

C1t2 = 3 sin2 6,000,000t,where t is time in seconds. The total energy E in the circuit is given by

E1t2 = L1t2 + C1t2.103. Graph L, C, and E in the window 30, 10-64 by 3-1, 44, with Xscl = 10-7 and

Yscl = 1. Interpret the graph.

104. Make a table of values for L, C, and E starting at t = 0, incrementing by 10-7. Interpret the results.

105. Use a fundamental identity to derive a simplified expression for E1t2. 5.3 Sum and Difference Identities for Cosine

Difference Identity for Cosine Several examples presented earlier should have convinced you by now that

cos 1A − B 2 does not equal cos A − cos B.For example, if A = p2 and B = 0, then

cos1A - B2 = cos ap2

- 0b = cos p2

= 0,

while cos A - cos B = cos p2

- cos 0 = 0 - 1 = -1.

To derive a formula for cos1A - B2, we start by locating angles A and B in standard position on a unit circle, with B 6 A. Let S and Q be the points where the terminal sides of angles A and B, respectively, intersect the circle. Let P be the point 11, 02, and locate point R on the unit circle so that angle POR equals the difference A - B. See Figure 4.

■ Difference Identity for Cosine

■ Sum Identity for Cosine

■ Cofunction Identities■ Applications of the

Sum and Difference Identities

■ Verifying an Identity

(cos(A – B), sin(A – B)) R

BA A – B

(cos A, sin A) S

P (1, 0)

Q (cos B, sin B)

Figure 4

Because point Q is on the unit circle, the x-coordinate of Q is the cosine of angle B, while the y-coordinate of Q is the sine of angle B.

Q has coordinates 1cos B, sin B2.

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212 CHAPTER 5 Trigonometric Identities

In the same way,

S has coordinates 1cos A, sin A2,and R has coordinates 1cos1A - B2, sin1A - B22.

Angle SOQ also equals A - B. The central angles SOQ and POR are equal, so chords PR and SQ are equal. Because PR = SQ, by the distance formula,23cos1A - B2 - 142 + 3sin1A - B2 - 042

d = 21x2 - x122 + 1y2 - y122 = 21cos A - cos B22 + 1sin A - sin B22.Square each side of this equation. Then square each expression, remembering that for any values of x and y, 1x - y22 = x2 - 2xy + y2.3cos1A - B2 - 142 + 3sin1A - B2 - 042

= 1cos A - cos B22 + 1sin A - sin B22cos21A - B2 - 2 cos1A - B2 + 1 + sin21A - B2 = cos2 A - 2 cos A cos B + cos2 B + sin2 A - 2 sin A sin B + sin2 B

For any value of x, sin2 x + cos2 x = 1, so we can rewrite the equation.

2 - 2 cos1A - B2 = 2 - 2 cos A cos B - 2 sin A sin B Use sin2 x + cos2 x = 1 three times and add like terms.

cos 1A − B 2 = cos A cos B + sin A sin B Subtract 2, and then divide by -2.This is the identity for cos1A - B2. Although Figure 4 shows angles A and B in the second and first quadrants, respectively, this result is the same for any values of these angles.

Cosine of a Sum or Difference

cos 1A + B 2 = cos A cos B − sin A sin B cos 1A − B 2 = cos A cos B + sin A sin B

These identities are important in calculus and useful in certain applications. For example, the method shown in Example 1 can be applied to find an exact value for cos 15°.

Sum Identity for Cosine To find a similar expression for cos1A + B2, rewrite A + B as A - 1-B2 and use the identity for cos1A - B2.

cos1A + B2 = cos3A - 1-B24 Definition of subtraction = cos A cos1-B2 + sin A sin1-B2 Cosine difference identity = cos A cos B + sin A1-sin B2 Even-odd identities

cos 1A + B 2 = cos A cos B − sin A sin B Multiply.

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2135.3 Sum and Difference Identities for Cosine

EXAMPLE 1 Finding Exact Cosine Function Values

Find the exact value of each expression.

(a) cos 15° (b) cos 5p12

SOLUTION

(a) To find cos 15°, we write 15° as the sum or difference of two angles with known function values, such as 45° and 30°, because

15° = 45° - 30°. (We could also use 60° - 45°.)

Then we use the cosine difference identity.

cos 15°

= cos145° - 30°2 15° = 45° - 30° = cos 45° cos 30° + sin 45° sin 30° Cosine difference identity

= 222

# 232

+ 222

# 12

Substitute known values.

= 26 + 224

Multiply, and then add fractions.

(b) cos 5p12

= cos ap6

+ p4b p6 = 2p12 and p4 = 3p12

= cos p6

cos p

4- sin p

6 sin p

4 Cosine sum identity

= 232

# 222

- 12# 22

2 Substitute known values.

= 26 - 224

Multiply, and then subtract fractions.

(c) cos 87° cos 93° - sin 87° sin 93° = cos187° + 93°2 Cosine sum identity = cos 180° Add. = -1 cos 180° = -1 ■✔ Now Try Exercises 9, 13, and 17.

This screen supports the solution in Example 1(b) by showing that the decimal approximations for cos 5p12

and 26 - 224 agree.

Cofunction Identities We can use the identity for the cosine of the differ-ence of two angles and the fundamental identities to derive cofunction identities, presented previously for values of u in the interval 30°, 90°4. For example, sub-stituting 90° for A and u for B in the identity for cos1A - B2 gives the following.

cos 190° − U 2 = cos 90° cos u + sin 90° sin u Cosine difference identity = 0 # cos u + 1 # sin u cos 90° = 0 and sin 90° = 1 = sin U Simplify.

This result is true for any value of u because the identity for cos1A - B2 is true for any values of A and B.

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214 CHAPTER 5 Trigonometric Identities

NOTE Because trigonometric (circular) functions are periodic, the solu-tions in Example 2 are not unique. We give only one of infinitely many possibilities.

Cofunction Identities

The following identities hold for any angle u for which the functions are defined.

cos 190° − U 2 = sin U cot 190° − U 2 = tan U sin 190° − U 2 = cos U sec 190° − U 2 = csc U tan 190° − U 2 = cot U csc 190° − U 2 = sec U

The same identities can be obtained for a real number domain by replacing 90° with p2 .

EXAMPLE 2 Using Cofunction Identities to Find U

Find one value of u or x that satisfies each of the following.

(a) cot u = tan 25° (b) sin u = cos1-30°2 (c) csc 3p4

= sec x

SOLUTION

(a) Because tangent and cotangent are cofunctions, tan190° - u2 = cot u. cot u = tan 25°

tan190° - u2 = tan 25° Cofunction identity 90° - u = 25° Set angle measures equal.

u = 65° Solve for u.

(b) sin u = cos1-30°2 cos190° - u2 = cos1-30°2 Cofunction identity 90° - u = -30° Set angle measures equal. u = 120° Solve for u.

= sec x

csc 3p4

= cscap2

- xb Cofunction identity

3p4

= p2

- x Set angle measures equal.

x = - p4

Solve for x; p2 -3p4 =

2p4 -

3p4 = -

p4 .

■✔ Now Try Exercises 37 and 41.

Applications of the Sum and Difference Identities If either angle A or angle B in the identities for cos1A + B2 and cos1A - B2 is a quadrantal angle, then the identity allows us to write the expression in terms of a single function of A or B.

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2155.3 Sum and Difference Identities for Cosine

EXAMPLE 3 Reducing cos 1A − B 2 to a Function of a Single VariableWrite cos1180° - u2 as a trigonometric function of u alone.SOLUTION cos1180° - u2

= cos 180° cos u + sin 180° sin u Cosine difference identity

= 1-12 cos u + 102 sin u cos 180° = -1 and sin 180° = 0 = - cos u Simplify.

■✔ Now Try Exercise 49.

(a)

y13

–12

(b)

Figure 5

EXAMPLE 4 Finding cos 1s + t 2 Given Information about s and tSuppose that sin s = 35 , cos t = -

1213 , and both s and t are in quadrant II. Find

cos1s + t2.SOLUTION By the cosine sum identity,

cos1s + t2 = cos s cos t - sin s sin t.The values of sin s and cos t are given, so we can find cos1s + t2 if we know the values of cos s and sin t. There are two ways to do this.

Method 1 We use angles in standard position. To find cos s and sin t, we sketch two reference triangles in the second quadrant, one with sin s = 35 and the other with cos t = - 1213 . Notice that for angle t, we use -12 to denote the length of the side that lies along the x-axis. See Figure 5.

In Figure 5(a), y = 3 and r = 5. We must find x.

x2 + y2 = r2 Pythagorean theorem x2 + 32 = 52 Substitute.

x2 = 16 Isolate x2. x = -4 Choose the negative

square root here.

Choose the positive square root here.

Thus, cos s = xr = - 45 .

In Figure 5(b), x = -12 and r = 13. We must find y.

x2 + y2 = r2 Pythagorean theorem 1-1222 + y2 = 132 Substitute.

y2 = 25 Isolate y2. y = 5

Thus, sin t = yr =5

13 .

Now we can find cos1s + t2. cos1s + t2 = cos s cos t - sin s sin t Cosine sum identity (1)

= - 45

a - 1213b - 3

5# 5

13 Substitute.

= 4865

- 1565

Multiply.

cos1s + t2 = 3365

Subtract.

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216 CHAPTER 5 Trigonometric Identities

Method 2 We use Pythagorean identities here. To find cos s, recall that sin2 s + cos2 s = 1, where s is in quadrant II.

a 35b 2 + cos2 s = 1 sin s = 35 925

+ cos2 s = 1 Square 35 .

cos2 s = 1625

Subtract 925 .

cos s = - 45

To find sin t, we use sin2 t + cos2 t = 1, where t is in quadrant II.

sin2 t + a - 1213b 2 = 1 cos t = - 1213

sin2 t + 144169

= 1 Square - 1213 .

sin2 t = 25169

Subtract 144169 .

sin t = 513

From this point, the problem is solved using the same steps beginning with the equation marked (1) in Method 1 on the previous page. The result is

cos1s + t2 = 3365

. Same result as in Method 1

■✔ Now Try Exercise 51.

EXAMPLE 5 Applying the Cosine Difference Identity to Voltage

Common household electric current is called alternating current because the current alternates direction within the wires. The voltage V in a typical 115-volt outlet can be expressed by the function

V1t2 = 163 sin vt,where v is the angular speed (in radians per second) of the rotating generator at the electrical plant and t is time in seconds. (Source: Bell, D., Fundamentals of Electric Circuits, Fourth Edition, Prentice-Hall.)

(a) It is essential for electric generators to rotate at precisely 60 cycles per sec so household appliances and computers will function properly. Determine v for these electric generators.

(b) Graph V in the window 30, 0.054 by 3-200, 2004.(c) Determine a value of f so that the graph of

V1t2 = 163 cos1vt - f2 is the same as the graph of

V1t2 = 163 sin vt.

sin t 7 0 because t is in quadrant II.

cos s 6 0 because s is in quadrant II.

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2175.3 Sum and Difference Identities for Cosine

SOLUTION

(a) We convert 60 cycles per sec to radians per second as follows.

v =60 cycles

1 sec# 2p radians

1 cycle= 120p radians per sec.

(b) V1t2 = 163 sin vt V1t2 = 163 sin 120pt From part (a), v = 120p radians per sec. Because the amplitude of the function V1t2 is 163, an appropriate interval for

the range is 3-200, 2004, as shown in the graph in Figure 6.(c) Use the even-odd identity for cosine and a cofunction identity.

cos ax - p2b = cos c - ap

2- xb d = cos ap

2- xb = sin x

Therefore, if f = p2 , then

V1t2 = 163 cos1vt - f2 V1t2 = 163 cos avt - p

V1t2 = 163 sin vt.■✔ Now Try Exercise 75.

−200

200

0 0.05

For x = t,V(t) = 163 sin 120pt

Figure 6

Verifying an Identity

EXAMPLE 6 Verifying an Identity

Verify that the following equation is an identity.

sec a 3p2

- xb = -csc xSOLUTION We work with the more complicated left side.

sec a 3p2

- xb = 1cos A3p2 - x B Reciprocal identity

= 1cos 3p2 cos x + sin

3p2 sin x

Cosine difference identity

= 10 # cos x + 1-12sin x cos 3p2 = 0 and sin 3p2 = -1

= 1-sin x Simplify.

= -csc x Reciprocal identity

The left side is identical to the right side, so the given equation is an identity.

■✔ Now Try Exercise 67.

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218 CHAPTER 5 Trigonometric Identities

5.3 Exercises

CONCEPT PREVIEW Match each expression in Column I with the correct expression in Column II to form an identity. Choices may be used once, more than once, or not at all.

1. cos1x + y2 = 2. cos1x - y2 = 3. cos Ap2 - x B = 4. sin Ap2 - x B = 5. cos Ax - p2 B = 6. sin Ax - p2 B = 7. tan Ap2 - x B = 8. cot Ap2 - x B =

A. cos x cos y + sin x sin y

B. tan x

C. -cos x

D. -sin x

E. sin x

F . cos x cos y - sin x sin y

G. cos x

H. cot x

Find the exact value of each expression. (Do not use a calculator.) See Example 1.

9. cos 75° 10. cos1-15°211. cos1-105°2 1Hint: -105° = -60° + 1-45°22 12. cos 105° (Hint: 105° = 60° + 45°)13. cos

7p12

14. cos p

15. cos a - p12b 16. cos a - 7p

12b

17. cos 40° cos 50° - sin 40° sin 50° 18. cos 7p9

cos 2p9

- sin 7p9

sin 2p9

Write each function value in terms of the cofunction of a complementary angle. See Example 2.

19. tan 87° 20. sin 15° 21. cos p

12 22. sin

2p5

23. csc 14° 24′ 24. sin 142° 14′ 25. sin 5p8

26. cot 9p10

27. sec 146° 42′ 28. tan 174° 03′ 29. cot 176.9814° 30. sin 98.0142°

Use identities to fill in each blank with the appropriate trigonometric function name. See Example 2.

31. cot p

3= p

6 32. sin

2p3

= a - p6b

33. 33° = sin 57° 34. 72° = cot 18°

35. cos 70° = 1 20°

36. tan 24° = 1 66°

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2195.3 Sum and Difference Identities for Cosine

Find one value of u or x that satisfies each of the following. See Example 2.

37. tan u = cot145° + 2u2 38. sin u = cos12u + 30°239. sec x = csc 2p

3 40. cos x = sin p

41. sin13u - 15°2 = cos1u + 25°2 42. cot1u - 10°2 = tan12u - 20°2Use the identities for the cosine of a sum or difference to write each expression as a trigonometric function of u alone. See Example 3.

43. cos10° - u2 44. cos190° - u2 45. cos1u - 180°2 46. cos1u - 270°2 47. cos10° + u2 48. cos190° + u2 49. cos1180° + u2 50. cos1270° + u2Find cos1s + t2 and cos1s - t2. See Example 4.51. sin s = 35 and sin t = -

1213 , s in quadrant I and t in quadrant III

52. cos s = - 817 and cos t = - 35 , s and t in quadrant III

53. cos s = - 15 and sin t =35 , s and t in quadrant II

54. sin s = 23 and sin t = - 13 , s in quadrant II and t in quadrant IV

55. sin s = 257 and sin t = 268 , s and t in quadrant I56. cos s = 224 and sin t = - 256 , s and t in quadrant IVConcept Check Determine whether each statement is true or false.

57. cos 42° = cos130° + 12°2 58. cos1-24°2 = cos 16° - cos 40°59. cos 74° = cos 60° cos 14° + sin 60° sin 14°

60. cos 140° = cos 60° cos 80° - sin 60° sin 80°

61. cos p

3= cos p

12 cos p

4- sin p

12 sin p

62. cos 2p3

= cos 11p12

cos p

4+ sin 11p

12 sin p

63. cos 70° cos 20° - sin 70° sin 20° = 0

64. cos 85° cos 40° + sin 85° sin 40° = 222

65. tan ax - p2b = cot x 66. sin ax - p

2b = cos x

Verify that each equation is an identity. 1Hint: cos 2x = cos1x + x2.2 See Example 6.67. cos ap

2+ xb = -sin x 68. sec1p - x2 = -sec x

69. cos 2x = cos2 x - sin2 x 70. 1 + cos 2x - cos2 x = cos2 x

71. cos 2x = 1 - 2 sin2 x 72. cos 2x = 2 cos2 x - 1

73. cos 2x = cot2 x - 1

cot2 x + 1 74. sec 2x =cot2 x + 1cot2 x - 1

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220 CHAPTER 5 Trigonometric Identities

76. Sound Waves Sound is a result of waves applying pressure to a person’s eardrum. For a pure sound wave radiating outward in a spherical shape, the trig-onometric function

P = ar cos a2pr

l- ctb

can be used to model the sound pressure at a radius of r feet from the source, where t is time in seconds, l is length of the sound wave in feet, c is speed of sound in feet per second, and a is maximum sound pressure at the source measured in pounds per square foot. (Source: Beranek, L., Noise and Vibration Control, Institute of Noise Control Engineering, Washington, D.C.) Let l = 4.9 ft and c = 1026 ft per sec.(a) Let a = 0.4 lb per ft2. Graph the sound pressure at distance r = 10 ft from its

source in the window 30, 0.054 by 3-0.05, 0.054. Describe P at this distance.(b) Now let a = 3 and t = 10. Graph the sound pressure in the window 30, 204 by 3-2, 24. What happens to pressure P as radius r increases?(c) Suppose a person stands at a radius r so that r = nl, where n is a positive integer.

Use the difference identity for cosine to simplify P in this situation.

Relating Concepts

For individual or collaborative investigation (Exercises 77—82)

(This discussion applies to functions of both angles and real numbers.) The result of Example 3 in this section can be written as an identity.

cos 1180° − U 2 = −cos UThis is an example of a reduction formula, which is an identity that reduces a func-tion of a quadrantal angle plus or minus u to a function of u alone. Another example of a reduction formula is

cos 1270° + U 2 = sin U.Here is an interesting method for quickly determining a reduction formula for a

trigonometric function ƒ of the form

ƒ 1Q t U 2 , where Q is a quadrantal angle.There are two cases to consider, and in each case, think of U as a small positive angle in order to determine the quadrant in which Q{ u will lie.

(Modeling) Solve each problem. See Example 5.

75. Electric Current The voltage V in a typical 115-volt outlet can be expressed by the function

V1t2 = 163 sin 120pt, where 120p is the angular speed (in radians per second) of the rotating generator at

an electrical power plant, and t is time in seconds. (Source: Bell, D., Fundamentals of Electric Circuits, Fourth Edition, Prentice-Hall.)

(a) How many times does the current oscillate in 0.05 sec?

(b) What are the maximum and minimum voltages in this outlet?

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2215.4 Sum and Difference Identities for Sine and Tangent

Case 1 Q is a quadrantal angle whose terminal side lies along the x-axis.

Determine the quadrant in which Q{ u will lie for a small positive angle u. If the given function ƒ is positive in that quadrant, use a + sign on the reduced form. If ƒ is negative in that quadrant, use a - sign. The reduced form will have that sign, ƒ as the function, and u as the argument.

Example:

Cosine is negative in quadrant II.

Terminates on the x-axis

(1+)+1*This is in

quadrant II for small u.

Same function

cos1180° - u2 = -cos u

Case 2 Q is a quadrantal angle whose terminal side lies along the y-axis.

Example:

Terminates on the y-axis

Cosine is positive in quadrant IV.

cos1270° + u2 = + sin u (or sin u, as it is usually written)(1+)+1*This is in

quadrant IV for small u.

Cofunctions

5.4 Sum and Difference Identities for Sine and Tangent

Sum and Difference Identities for Sine We can use the cosine sum and difference identities from the previous section to derive similar identities for sine and tangent. In sin u = cos190° - u2, replace u with A + B.

sin1A + B2 = cos390° - 1A + B24 Cofunction identity = cos3190° - A2 - B4 Distribute negative sign and regroup. = cos190° - A2 cos B + sin190° - A2 sin B

Cosine difference identity

sin 1A + B 2 = sin A cos B + cos A sin B Cofunction identities

■ Sum and Difference Identities for Sine

■ Sum and Difference Identities for Tangent

■ Applications of the Sum and Difference Identities

■ Verifying an Identity

Use these ideas to write a reduction formula for each of the following.

77. cos190° + u2 78. cos1270° - u2 79. cos1180° + u280. cos1270° + u2 81. sin1180° + u2 82. tan1270° - u2

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222 CHAPTER 5 Trigonometric Identities

Sum and Difference Identities for Tangent We can derive the identity for tan1A + B2 as follows.

tan1A + B2 = sin1A + B2cos1A + B2 Fundamental identity

= sin A cos B + cos A sin Bcos A cos B - sin A sin B Sum identities

sin A cos B + cos A sin B1

cos A cos B - sin A sin B1

cos A cos B1

cos A cos B

Multiply by 1, where

1 =1

cos A cos B1

cos A cos B

sin A cos Bcos A cos B

+ cos A sin Bcos A cos B

cos A cos Bcos A cos B

- sin A sin Bcos A cos B

Multiply numerators. Multiply denominators.

sin Acos A

+ sin Bcos B

1 - sin Acos A

# sin Bcos B

Simplify.

tan 1A + B 2 = tan A + tan B1 − tan A tan B

sin ucos u = tan u

We express this result in terms of the

tangent function.

We can replace B with -B and use the fact that tan1-B2 = - tan B to obtain the identity for the tangent of the difference of two angles, as seen below.

Tangent of a Sum or Difference

tan 1A + B 2 = tan A + tan B1 − tan A tan B

tan 1A − B 2 = tan A − tan B1 + tan A tan B

Now we write sin1A - B2 as sin3A + 1-B24 and use the identity just found for sin1A + B2.

sin1A - B2 = sin3A + 1-B24 Definition of subtraction = sin A cos1-B2 + cos A sin1-B2 Sine sum identity

sin 1A − B 2 = sin A cos B − cos A sin B Even-odd identitiesSine of a Sum or Difference

sin 1A + B 2 = sin A cos B + cos A sin B sin 1A − B 2 = sin A cos B − cos A sin B

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2235.4 Sum and Difference Identities for Sine and Tangent

Applications of the Sum and Difference Identities

EXAMPLE 1 Finding Exact Sine and Tangent Function Values

Find the exact value of each expression.

(a) sin 75° (b) tan 7p12

SOLUTION

(a) sin 75°

= sin145° + 30°2 75° = 45° + 30° = sin 45° cos 30° + cos 45° sin 30° Sine sum identity

= 222

# 232

+ 222

# 12

Substitute known values.

= 26 + 224

Multiply, and then add fractions.

(c) sin 40° cos 160° - cos 40° sin 160° = sin140° - 160°2 Sine difference identity = sin1-120°2 Subtract. = - sin 120° Even-odd identity

= - 232

Substitute the known value.

■✔ Now Try Exercises 9, 15, and 25.

Factor first. Then divide out the

common factor.

(b) tan 7p12

= tan ap3

+ p4b p3 = 4p12 and p4 = 3p12

=tan p

3 + tan p

1 - tan p3 tan p

Tangent sum identity

= 23 + 11 - 23 # 1 Substitute known values.

= 23 + 11 - 23 # 1 + 231 + 23 Rationalize the denominator.

= 23 + 3 + 1 + 231 - 3

1a + b21c + d2 = ac + ad + bc + bd; 1x - y21x + y2 = x2 - y2 = 4 + 223-2 Combine like terms. =

212 + 23 221-12 Factor out 2.

= -2 - 23 Write in lowest terms.

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224 CHAPTER 5 Trigonometric Identities

(b) tan145° - u2 = tan 45° - tan u

1 + tan 45° tan u Tangent difference identity

= 1 - tan u1 + 1 # tan u tan 45° = 1

= 1 - tan u1 + tan u Multiply.

(c) sin1180° - u2 = sin 180° cos u - cos 180° sin u Sine difference identity = 0 # cos u - 1-12 sin u sin 180° = 0 and cos 180° = -1 = sin u Simplify.

■✔ Now Try Exercises 33, 39, and 43.

EXAMPLE 3 Finding Function Values and the Quadrant of A + B

Suppose that A and B are angles in standard position such that sin A = 45 , p2 6 A 6 p, and cos B = -

513 , p 6 B 6

3p2 . Find each of the following.

(a) sin1A + B2 (b) tan1A + B2 (c) the quadrant of A + BSOLUTION

(a) The identity for sin1A + B2 involves sin A, cos A, sin B, and cos B. We are given values of sin A and cos B. We must find values of cos A and sin B.

sin2 A + cos2 A = 1 Fundamental identity

a 45b 2 + cos2 A = 1 sin A = 45 1625

+ cos2 A = 1 Square 45 .

cos2 A = 925

Subtract 1625 .

cos A = - 35

Take square roots. Because A is in quadrant II, cos A 6 0.Pay attention

to signs.

EXAMPLE 2 Writing Functions as Expressions Involving Functions of U

Write each function as an expression involving functions of u alone.

(a) sin130° + u2 (b) tan145° - u2 (c) sin1180° - u2SOLUTION

(a) sin130° + u2 = sin 30° cos u + cos 30° sin u Sine sum identity

= 12

cos u + 232

sin u sin 30° = 12 and cos 30° =232

= cos u + 23 sin u2

ab # c = acb ; Add fractions.

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2255.4 Sum and Difference Identities for Sine and Tangent

In the same way, sin B = - 1213 . Now find sin1A + B2. sin1A + B2 = sin A cos B + cos A sin B Sine sum identity

= 45

a - 513b + a - 3

5b a - 12

13b Substitute the given values for sin A and cos B and the values

found for cos A and sin B.

= - 2065

+ 3665

Multiply.

sin1A + B2 = 1665

Add.

(b) To find tan1A + B2, use the values of sine and cosine from part (a), sin A = 45 , cos A = - 35 , sin B = -

1213 , and cos B = -

513 , to obtain tan A and tan B.

tan A = sin Acos A

tan B = sin Bcos B

=45

- 35 =

- 1213- 513

= 45

, a - 35b = - 12

13, a - 5

13b

= 45# a - 5

3b = - 12

13# a - 13

tan A = - 43

tan B = 125

tan1A + B2 = tan A + tan B1 - tan A tan B Tangent sum identity

=A - 43 B + 125

1 - A - 43 B A125 B Substitute. =

1615

1 + 4815 Perform the indicated operations.

=1615 6315

Add terms in the denominator.

= 1615

, 6315

Simplify the complex fraction.

= 1615

# 1563

Definition of division

tan1A + B2 = 1663

Multiply.

63 See parts (a) and (b).

Both are positive. Therefore, A + B must be in quadrant I, because it is the only quadrant in which both sine and tangent are positive.

■✔ Now Try Exercise 51.

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226 CHAPTER 5 Trigonometric Identities

CONCEPT PREVIEW Match each expression in Column I with the correct expression in Column II to form an identity.

5.4 Exercises

1. sin1A + B2 2. sin1A - B2 3. tan1A + B2 4. tan1A - B2

A. sin A cos B - cos A sin B

B. tan A + tan B1 - tan A tan B

C. tan A - tan B1 + tan A tan B

D. sin A cos B + cos A sin B

CONCEPT PREVIEW Match each expression in Column I with its equivalent expres-sion in Column II.

5. sin 60° cos 45° + cos 60° sin 45°

6. sin 60° cos 45° - cos 60° sin 45°

7. tan p3 + tan

1 - tan p3 tan p4

8. tan p3 - tan

1 + tan p3 tan p4

A. tan 7p12

B. sin 15°

C. sin 105°

D. tan p12

Find the exact value of each expression. See Example 1.

9. sin 165° 10. sin 255° 11. tan 165° 12. tan 285°

Verifying an Identity

EXAMPLE 4 Verifying an Identity

Verify that the equation is an identity.

sin ap6

+ ub + cos ap3

+ ub = cos uSOLUTION Work on the left side, using the sine and cosine sum identities.

sin ap6

+ ub + cos ap3

+ ub = asin p

6 cos u + cos p

6 sin ub + acos p

3 cos u - sin p

3 sin ub

Sine sum identity; cosine sum identity

= ¢ 12

cos u + 232

sin u≤ + ¢ 12

cos u - 232

sin u≤ sin p6 =

12 ; cos

p6 =

232 ; cos

p3 =

12 ; sin

p3 =

232

= 12

cos u + 12

cos u Simplify.

= cos u Add. ■✔ Now Try Exercise 63.

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2275.4 Sum and Difference Identities for Sine and Tangent

13. sin 5p12

14. sin 13p12

15. tan p

12 16. tan

5p12

17. sin 7p12

18. sin p

12 19. sin a - 7p

12b 20. sin a - 5p

12b

21. tan a - 5p12b 22. tan a - 7p

12b 23. tan 11p

12 24. sin a - 13p

12b

25. sin 76° cos 31° - cos 76° sin 31° 26. sin 40° cos 50° + cos 40° sin 50°

27. sin p

5 cos

3p10

+ cos p5

sin 3p10

28. sin 5p9

cos p

18- cos 5p

9 sin p

29. tan 80° + tan 55°

1 - tan 80° tan 55° 30. tan 80° - tan1-55°2

1 + tan 80° tan1-55°231.

tan 5p9 + tan 4p9

1 - tan 5p9 tan 4p9

32. tan 5p12 + tan

1 - tan 5p12 tan p4

Write each function as an expression involving functions of u or x alone. See Example 2.

33. cos130° + u2 34. cos1u - 30°2 35. cos160° + u236. cos145° - u2 37. cos a3p

4- xb 38. sin145° + u2

39. tan1u + 30°2 40. tan ap4

+ xb 41. sin ap4

+ xb42. sin a3p

4- xb 43. sin1270° - u2 44. tan1180° + u2

45. tan12p - x2 46. sin1p + x2 47. tan1p - x248. Why is it not possible to use the method of Example 2 to find a formula for

tan1270° - u2?49. Why is it that standard trigonometry texts usually do not develop formulas for the

cotangent, secant, and cosecant of the sum and difference of two numbers or angles?

50. Show that if A, B, and C are the angles of a triangle, then

sin1A + B + C2 = 0.Use the given information to find (a) sin1s + t2, (b) tan1s + t2, and (c) the quadrant of s + t. See Example 3.

51. cos s = 35 and sin t =5

13 , s and t in quadrant I

52. sin s = 35 and sin t = - 1213 , s in quadrant I and t in quadrant III

53. cos s = - 817 and cos t = - 35 , s and t in quadrant III

54. cos s = - 1517 and sin t =45 , s in quadrant II and t in quadrant I

55. sin s = 23 and sin t = - 13 , s in quadrant II and t in quadrant IV

56. cos s = - 15 and sin t =35 , s and t in quadrant II

Graph each expression and use the graph to make a conjecture, predicting what might be an identity. Then verify your conjecture algebraically.

57. sin ap2

+ ub 58. sin a3p2

+ ub 59. tan ap2

+ ub 60. tan ap2

- ub

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228 CHAPTER 5 Trigonometric Identities

Verify that each equation is an identity. See Example 4.

61. sin 2x = 2 sin x cos x 1Hint: sin 2 x = sin1x + x2262. sin1x + y2 + sin1x - y2 = 2 sin x cos y63. sin a7p

6+ xb - cos a2p

3+ xb = 0

64. tan1x - y2 - tan1y - x2 = 21tan x - tan y21 + tan x tan y

65. cos1a - b2cos a sin b

= tan a + cot b 66. sin1s + t2cos s cos t

= tan s + tan t

67. sin1x - y2sin1x + y2 = tan x - tan ytan x + tan y 68. sin1x + y2cos1x - y2 = cot x + cot y1 + cot x cot y

69. sin1s - t2

sin t+

cos1s - t2cos t

= sin ssin t cos t

70. tan1a + b2 - tan b

1 + tan1a + b2 tan b = tan a(Modeling) Solve each problem.

71. Back Stress If a person bends at the waist with a straight back making an angle of u degrees with the horizontal, then the force F exerted on the back muscles can be modeled by the equation

F =0.6W sin1u + 90°2

sin 12° ,

where W is the weight of the person. (Source: Metcalf, H., Topics in Classical Biophysics, Prentice-Hall.)

(a) Calculate force F, to the nearest pound, for W = 170 lb and u = 30°.(b) Use an identity to show that F is approximately equal to 2.9W cos u.

72. Back Stress Refer to Exercise 71.

(a) Suppose a 200-lb person bends at the waist so that u = 45°. Calculate the force, to the nearest pound, exerted on the person’s back muscles.

(b) Approximate graphically the value of u, to the nearest tenth, that results in the back muscles of a 200-lb person exerting a force of 400 lb.

73. Voltage A coil of wire rotating in a magnetic field induces a voltage

E = 20 sin apt4

- p2b .

Use an identity from this section to express this in terms of cos pt4 .

74. Voltage of a Circuit When the two voltages

V1 = 30 sin 120pt and V2 = 40 cos 120pt

are applied to the same circuit, the resulting voltage V will be equal to their sum. (Source: Bell, D., Fundamentals of Electric Circuits, Second Edition, Reston Publishing Company.)

(a) Graph the sum in the window 30, 0.054 by 3-60, 604.(b) Use the graph to estimate values for a and f so that V = a sin1120pt + f2.(c) Use identities to verify that the expression for V in part (b) is valid.

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2295.4 Sum and Difference Identities for Sine and Tangent

(Modeling) Roll of a Spacecraft The figure on the left below shows the three quantities that determine the motion of a spacecraft. A conventional three-dimensional spacecraft coordinate system is shown on the right.

PitchYaw

Roll

Y X

Q9 Z9

Angle YOQ = u and OQ = r. The coordinates of Q are 1x, y, z2, wherey = r cos u and z = r sin u.

When the spacecraft performs a rotation, it is necessary to find the coordinates in the spacecraft system after the rotation takes place. For example, suppose the spacecraft undergoes roll through angle R. The coordinates 1x, y, z2 of point Q become 1x′, y′, z′2, the coordinates of the corresponding point Q′. In the new reference system, OQ′ = r and, because the roll is around the x-axis and angle Y′OQ′ = YOQ = u,

x′ = x, y′ = r cos1u + R2, and z′ = r sin1u + R2.(Source: Kastner, B., Space Mathematics, NASA.)

75. Write y′ in terms of y, R, and z. 76. Write z′ in terms of y, R, and z.

Relating Concepts

For individual or collaborative investigation (Exercises 77—82)

Refer to the figure on the left below. By the definition of tan u,

m = tan u, where m is the slope and u is the angle of inclination of the line.

The following exercises, which depend on properties of triangles, refer to triangle ABC in the figure on the right below. Work Exercises 77– 82 in order. Assume that all angles are measured in degrees.

m = tan U

slope = m1

slope = m2

A B

77. In terms of b, what is the measure of angle ABC ?

78. Use the fact that the sum of the angles in a triangle is 180° to express u in terms of a and b.

79. Apply the formula for tan1A - B2 to obtain an expression for tan u in terms of tan a and tan b.

80. Replace tan a with m1 and tan b with m2 to obtain tan u =m2 - m11 + m1m2

Use the result from Exercise 80 to find the acute angle between each pair of lines. (Note that the tangent of the angle will be positive.) Use a calculator and round to the nearest tenth of a degree.

81. x + y = 9, 2x + y = -1 82. 5x - 2y + 4 = 0, 3x + 5y = 6

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230 CHAPTER 5 Trigonometric Identities

1. If sin u = - 725 and u is in quadrant IV, find the remaining five trigonometric func-tion values of u.

2. Express cot2 x + csc2 x in terms of sin x and cos x, and simplify.

3. Find the exact value of sin A - 7p12 B . 4. Express cos1180° - u2 as a function of u alone. 5. If cos A = 35 , sin B = -

513 , 0 6 A 6

p2 , and p 6 B 6

3p2 , find each of the following.

(a) cos1A + B2 (b) sin1A + B2 (c) the quadrant of A + B 6. Express tan A3p4 + x B as a function of x alone.Verify that each equation is an identity.

7. 1 + sin u

cot2 u= sin u

csc u - 1 8. sin ap3 + ub - sin ap3 - ub = sin u 9.

sin2 u - cos2 usin4 u - cos4 u = 1 10.

cos1x + y2 + cos1x - y2sin1x - y2 + sin1x + y2 = cot x

Chapter 5 Quiz (Sections 5.1—5.4)

5.5 Double-Angle Identities

Double-Angle Identities When A = B in the identities for the sum of two angles, the double-angle identities result. To derive an expression for cos 2A, we let B = A in the identity cos1A + B2 = cos A cos B - sin A sin B.

cos 2 A = cos1A + A2 2A = A + A = cos A cos A - sin A sin A Cosine sum identity

cos 2 A = cos2 A − sin2 A a # a = a2

Two other useful forms of this identity can be obtained by substituting

cos2 A = 1 - sin2 A or sin2 A = 1 - cos2 A.

Replacing cos2 A with the expression 1 - sin2 A gives the following.

cos 2 A = cos2 A - sin2 A Double-angle identity from above = 11 - sin2 A2 - sin2 A Fundamental identity

cos 2 A = 1 − 2 sin2 A Subtract.

Replacing sin2 A with 1 - cos2 A gives a third form.

cos 2 A = cos2 A - sin2 A Double-angle identity from above = cos2 A - 11 - cos2 A2 Fundamental identity = cos2 A - 1 + cos2 A Distributive property

cos 2 A = 2 cos2 A − 1 Add.

■ Double-Angle Identities

■ An Application■ Product-to-Sum and

Sum-to-Product Identities

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2315.5 Double-Angle Identities

We find sin 2A using sin1A + B2 = sin A cos B + cos A sin B, with B = A. sin 2 A = sin1A + A2 2A = A + A

= sin A cos A + cos A sin A Sine sum identity sin 2 A = 2 sin A cos A Add.

Using the identity for tan1A + B2, we find tan 2A. tan 2 A = tan1A + A2 2A = A + A

= tan A + tan A1 - tan A tan A Tangent sum identity

tan 2 A =2 tan A

1 − tan2 A Simplify.

LOOKING AHEAD TO CALCULUSThe identities

cos 2 A = 1 - 2 sin2 A

and cos 2 A = 2 cos2 A - 1

can be rewritten as

sin2 A = 12

11 - cos 2 A2and cos2 A = 1

2 11 + cos 2 A2.

These identities are used to integrate

the functions

ƒ1A2 = sin2 A and g1A2 = cos2 A.

NOTE In general, for a trigonometric function ƒ,

ƒ12 A2 ≠ 2ƒ1A2.Double-Angle Identities

cos 2 A = cos2 A − sin2 A cos 2 A = 1 − 2 sin2 A

cos 2 A = 2 cos2 A − 1 sin 2 A = 2 sin A cos A

tan 2 A =2 tan A

1 − tan2 A

EXAMPLE 1 Finding Function Values of 2U Given Information about U

Given cos u = 35 and sin u 6 0, find sin 2u, cos 2u, and tan 2u.

SOLUTION To find sin 2u, we must first find the value of sin u.

sin2 u + cos2 u = 1 Pythagorean identity

sin2 u + a 35b 2 = 1 cos u = 35

sin2 u = 1625

A35 B2 = 925 ; Subtract 925 . sin u = - 4

Take square roots. Choose the negative square root because sin u 6 0.

Pay attention to signs here.

Now use the double-angle identity for sine.

sin 2u = 2 sin u cos u = 2a - 45b a 3

5b = - 24

25 sin u = - 45 and cos u =

Now we find cos 2u, using the first of the double-angle identities for cosine.

cos 2u = cos2 u - sin2 u = 925

- 1625

= - 725

cos u = 35 and A35 B2 = 925 ; sin u = - 45 and A - 45 B2 = 1625Any of the three forms may be used.

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232 CHAPTER 5 Trigonometric Identities

The value of tan 2u can be found in either of two ways. We can use the double-

angle identity and the fact that tan u = sin ucos u =- 4535

= - 45 ,35 = -

45# 5

3 = - 43 .

tan 2u = 2 tan u1 - tan2 u =

2 A - 43 B1 - A - 43 B2 = - 83- 79 = 247

Alternatively, we can find tan 2u by finding the quotient of sin 2u and cos 2u.

tan 2u = sin 2ucos 2u

=- 2425- 725

= 247

Same result as above

■✔ Now Try Exercise 11.

–3

√10

y1sin U =

√10

Figure 7

EXAMPLE 2 Finding Function Values of U Given Information about 2U

Find the values of the six trigonometric functions of u given cos 2u = 45 and 90° 6 u 6 180°.

SOLUTION We must obtain a trigonometric function value of u alone.

cos 2u = 1 - 2 sin2 u Double-angle identity

= 1 - 2 sin2 u cos 2u = 45

- 15

= -2 sin2 u Subtract 1 from each side.

110

= sin2 u Multiply by - 12 .

sin u = B 110 Take square roots. Choose the positive square root because u terminates in quadrant II. sin u = 1210 # 210210 Quotient rule for radicals; rationalize the denominator. sin u = 210

10 2a # 2a = a

Now find values of cos u and tan u by sketching and labeling a right triangle in

quadrant II. Because sin u = 1110 , the triangle in Figure 7 is labeled accordingly. The Pythagorean theorem is used to find the remaining leg.

cos u = -3210 = - 321010 and tan u = 1-3 = - 13 cos u = xr and tan u = yxWe find the other three functions using reciprocals.

csc u = 1sin u

= 210 , sec u = 1cos u

= - 2103

, cot u = 1tan u

= -3

■✔ Now Try Exercise 15.

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2335.5 Double-Angle Identities

EXAMPLE 3 Verifying an Identity

Verify that the following equation is an identity.

cot x sin 2x = 1 + cos 2xSOLUTION We start by working on the left side, writing all functions in terms of sine and cosine and then simplifying the result.

cot x sin 2x = cos xsin x

# sin 2x Quotient identity

= cos xsin x

12 sin x cos x2 Double-angle identity= 2 cos2 x Multiply.

= 1 + cos 2x cos 2x = 2 cos2 x - 1, so

2 cos2 x = 1 + cos 2x.

■✔ Now Try Exercise 17.

Be able to recognize alternative forms

of identities.

This is not an obvious way to begin, but it is

indeed valid.

EXAMPLE 4 Simplifying Expressions Using Double-Angle Identities

Simplify each expression.

(a) cos2 7x - sin2 7x (b) sin 15° cos 15°SOLUTION

(a) This expression suggests one of the double-angle identities for cosine: cos 2 A = cos2 A - sin2 A. Substitute 7x for A.

cos2 7x - sin2 7x = cos 217x2 = cos 14x(b) If the expression sin 15° cos 15° were

2 sin 15° cos 15°,

we could apply the identity for sin 2 A directly because sin 2 A = 2 sin A cos A.

sin 15° cos 15°

= 12

122 sin 15° cos 15° Multiply by 1 in the form 12 122. = 1

2 12 sin 15° cos 15°2 Associative property

= 12

sin12 # 15°2 2 sin A cos A = sin 2 A, with A = 15° = 1

2 sin 30° Multiply.

= 12# 1

2 sin 30° = 12

= 14

Multiply.

■✔ Now Try Exercises 37 and 39.

Identities involving larger multiples of the variable can be derived by repeated use of the double-angle identities and other identities.

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234 CHAPTER 5 Trigonometric Identities

EXAMPLE 5 Deriving a Multiple-Angle Identity

Write sin 3x in terms of sin x.

SOLUTION

sin 3x

= sin12x + x2 3x = 2x + x = sin 2x cos x + cos 2x sin x Sine sum identity = 12 sin x cos x2cos x + 1cos2 x - sin2 x2sin x Double-angle identities = 2 sin x cos2 x + cos2 x sin x - sin3 x Multiply. = 2 sin x11 - sin2 x2 + 11 - sin2 x2sin x - sin3 x cos2 x = 1 - sin2 x = 2 sin x - 2 sin3 x + sin x - sin3 x - sin3 x Distributive property = 3 sin x - 4 sin3 x Combine like terms.

■✔ Now Try Exercise 49.

Use the simple fact that 3 = 2 + 1 here.

An Application

EXAMPLE 6 Determining Wattage Consumption

If a toaster is plugged into a common household outlet, the wattage consumed is not constant. Instead, it varies at a high frequency according to the model

W = V2

R ,

where V is the voltage and R is a constant that measures the resistance of the toaster in ohms. (Source: Bell, D., Fundamentals of Electric Circuits, Fourth Edition, Prentice-Hall.)

Graph the wattage W consumed by a toaster with R = 15 and V = 163 sin 120pt in the window 30, 0.054 by 3-500, 20004. How many oscillations are there?

SOLUTION Substituting the given values into the wattage equation gives

W = V2

R=1163 sin 120pt22

15 .

To determine the range of W, we note that sin 120pt has maximum value 1, so

the expression for W has maximum value 1632

15 ≈ 1771. The minimum value is 0. The graph in Figure 8 shows that there are six oscillations.

■✔ Now Try Exercise 69.

−500

2000

0 0.05

For x = t,W(t) = (163 sin 120pt)

Figure 8

Product-to-Sum and Sum-to-Product Identities We can add the corresponding sides of the identities for cos1A + B2 and cos1A - B2 to derive a product-to-sum identity that is useful in calculus.

cos1A + B2 = cos A cos B - sin A sin B cos1A - B2 = cos A cos B + sin A sin B cos1A + B2 + cos1A - B2 = 2 cos A cos B Add. cos A cos B =

3cos 1A + B 2 + cos 1A − B 2 4M05_LHSD7642_11_AIE_C05_pp195-250.indd 234 16/11/15 3:20 pm

2355.5 Double-Angle Identities

Similarly, subtracting cos1A + B2 from cos1A - B2 givessin A sin B =

3cos 1A − B 2 − cos 1A + B 2 4 .Using the identities for sin1A + B2 and sin1A - B2 in the same way, we

obtain two more identities. Those and the previous ones are now summarized.

LOOKING AHEAD TO CALCULUSThe product-to-sum identities are used

in calculus to find integrals of functions that are products of trigonometric

functions. The classic calculus text by

Earl Swokowski includes the following

example:

Evaluate Lcos 5x cos 3x dx.The first solution line reads:

“We may write

cos 5x cos 3x = 12

3cos 8x + cos 2x4.”

Product-to-Sum Identities

cos A cos B =12

3cos 1A + B 2 + cos 1A − B 2 4 sin A sin B =

3cos 1A − B 2 − cos 1A + B 2 4 sin A cos B =

3sin 1A + B 2 + sin 1A − B 2 4 cos A sin B =

3sin 1A + B 2 − sin 1A − B 2 4EXAMPLE 7 Using a Product-to-Sum Identity

Write 4 cos 75° sin 25° as the sum or difference of two functions.

SOLUTION

4 cos 75° sin 25°

= 4 c 12

1sin175° + 25°2 - sin175° - 25°22 d Use the identity for cos A sin B, with A = 75° and B = 25°. = 2 sin 100° - 2 sin 50° Simplify.

■✔ Now Try Exercise 57.

We can transform the product-to-sum identities into equivalent useful forms—the sum-to-product identities—using substitution. Consider the product-to-sum identity for sin A cos B.

sin A cos B = 12

3sin 1A + B2 + sin 1A - B24 Product-to-sum identityLet u = A + B and v = A - B.Then u + v = 2A and u - v = 2B,

so A = u + v2

and B = u - v2

sin a u + v2b cos a u - v

2b = 1

2 1sin u + sin v2 Substitute.

sin u + sin v = 2 sin a u + v2b cos a u - v

2b Multiply by 2. Interchange sides.

The other three sum-to-product identities are derived using the same substitu-tions into the other three product-to-sum formulas.

Use substitution variables to write the product-to-sum identity in terms of u and v.

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236 CHAPTER 5 Trigonometric Identities

Sum-to-Product Identities

sin A + sin B = 2 sin aA + B2

b cos aA − B2

b sin A − sin B = 2 cos aA + B

2b sin aA − B

cos A + cos B = 2 cos aA + B2

b cos aA − B2

b cos A − cos B = −2 sin aA + B

2b sin aA − B

EXAMPLE 8 Using a Sum-to-Product Identity

Write sin 2u - sin 4u as a product of two functions.

SOLUTION sin 2u - sin 4u

= 2 cos a 2u + 4u2

b sin a 2u - 4u2

b Use the identity for sin A - sin B, with A = 2u and B = 4u.

= 2 cos 6u2

sin a -2u2b Simplify the numerators.

= 2 cos 3u sin1-u2 Divide. = -2 cos 3u sin u sin1-u2 = - sin u

■✔ Now Try Exercise 63.

CONCEPT PREVIEW Match each expression in Column I with its value in Column II.

5.5 Exercises

1. 2 cos2 15° - 1 2. 2 tan 15°1 - tan2 15°

3. 2 sin 22.5° cos 22.5° 4. cos2 p

6- sin2 p

5. 4 sin p

3 cos p

3 6.

2 tan p3 1 - tan2 p3

A. 12

B. 222

C. 232

D. - 23 E. 233

F. 23Find values of the sine and cosine functions for each angle measure. See Examples 1 and 2.

7. 2u, given sin u = 25 and cos u 6 0 8. 2u, given cos u = - 1213 and sin u 7 0

9. 2x, given tan x = 2 and cos x 7 0 10. 2x, given tan x = 53 and sin x 6 0

11. 2u, given sin u = - 257 and cos u 7 0 12. 2u, given cos u = 235 and sin u 7 013. u, given cos 2u = 35 and u terminates

in quadrant I14. u, given cos 2u = 34 and u terminates

in quadrant III

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2375.5 Double-Angle Identities

15. u, given cos 2u = - 512 and 90° 6 u 6 180°

16. u, given cos 2u = 23 and 90° 6 u 6 180°

Verify that each equation is an identity. See Example 3.

17. 1sin x + cos x22 = sin 2x + 1 18. sec 2x = sec2 x + sec4 x2 + sec2 x - sec4 x

19. 1cos 2x + sin 2x22 = 1 + sin 4x 20. 1cos 2x - sin 2x22 = 1 - sin 4x21. tan 8u - tan 8u tan2 4u = 2 tan 4u 22. sin 2x = 2 tan x

1 + tan2 x

23. cos 2u = 2 - sec2 u

sec2 u 24. tan 2u = -2 tan u

sec2 u - 2

25. sin 4x = 4 sin x cos x cos 2x 26. 1 + cos 2xsin 2x

= cot x

27. 2 cos 2usin 2u

= cot u - tan u 28. cot 4u = 1 - tan2 2u

2 tan 2u

29. tan x + cot x = 2 csc 2x 30. cos 2x = 1 - tan2 x

1 + tan2 x

31. 1 + tan x tan 2x = sec 2x 32. cot A - tan Acot A + tan A = cos 2A

33. sin 2 A cos 2 A = sin 2 A - 4 sin3 A cos A34. sin 4x = 4 sin x cos x - 8 sin3 x cos x35. tan1u - 45°2 + tan1u + 45°2 = 2 tan 2u36. cot u tan1u + p2 - sin1p - u2 cos ap

2- ub = cos2 u

Simplify each expression. See Example 4.

37. cos2 15° - sin2 15° 38. 2 tan 15°1 - tan2 15°

39. 1 - 2 sin2 15°

40. 1 - 2 sin2 22 12°

41. 2 cos2 67 12°- 1 42. cos2

8- 1

43. tan 51°

1 - tan2 51° 44. tan 34°

211 - tan2 34°2 45. 14 - 12 sin2 47.1°46.

sin 29.5° cos 29.5° 47. sin2 2p5

- cos2 2p5

48. cos2 2x - sin2 2x

Express each function as a trigonometric function of x. See Example 5.

49. sin 4x 50. cos 3x 51. tan 3x 52. cos 4x

Graph each expression and use the graph to make a conjecture, predicting what might be an identity. Then verify your conjecture algebraically.

53. cos4 x - sin4 x 54. 4 tan x cos2 x - 2 tan x

1 - tan2 x

55. 2 tan x

2 - sec2 x 56. cot2 x - 1

2 cot x

Write each expression as a sum or difference of trigonometric functions. See Example 7.

57. 2 sin 58° cos 102° 58. 2 cos 85° sin 140° 59. 2 sin p

6 cos p

60. 5 cos 3x cos 2x 61. 6 sin 4x sin 5x 62. 8 sin 7x sin 9x

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238 CHAPTER 5 Trigonometric Identities

Write each expression as a product of trigonometric functions. See Example 8.

63. cos 4x - cos 2x 64. cos 5x + cos 8x 65. sin 25° + sin1-48°266. sin 102° - sin 95° 67. cos 4x + cos 8x 68. sin 9x - sin 3x

(Modeling) Solve each problem. See Example 6.

69. Wattage Consumption Use the identity cos 2u = 1 - 2 sin2 u to determine values of a, c, and v so that the equation

W =1163 sin 120pt22

15 becomes W = a cos1vt2 + c.

Round to the nearest tenth as necessary. Check by graphing both expressions for W on the same coordinate axes.

70. Amperage, Wattage, and Voltage Amperage is a measure of the amount of electricity that is moving through a circuit, whereas voltage is a measure of the force pushing the electricity. The wattage W consumed by an electrical device can be determined by calculating the product of the amperage I and voltage V. (Source: Wilcox, G. and C. Hesselberth, Electricity for Engineering Technology, Allyn & Bacon.)

(a) A household circuit has voltage

V = 163 sin 120pt

when an incandescent light bulb is turned on with amperage

I = 1.23 sin 120pt.

Graph the wattage W = VI consumed by the light bulb in the window 30, 0.054 by 3-50, 3004.

(b) Determine the maximum and minimum wattages used by the light bulb.

(c) Use identities to determine values for a, c, and v so that W = a cos1vt2 + c.(d) Check by graphing both expressions for W on the same coordinate axes.

(e) Use the graph to estimate the average wattage used by the light. For how many watts (to the nearest integer) would this incandescent light bulb be rated?

5.6 Half-Angle Identities

Half-Angle Identities From alternative forms of the identity for cos 2A,

we derive identities for sin A2 , cos A2 , and tan

A2 , known as half-angle identities.

We derive the identity for sin A2 as follows.

cos 2x = 1 - 2 sin2 x Cosine double-angle identity 2 sin2 x = 1 - cos 2x Add 2 sin2 x and subtract cos 2x.

sin x = tB1 - cos 2 x2 Divide by 2 and take square roots. sin

A2= tÅ1 − cos A2 Let 2x = A, so x = A2 . Substitute.

■ Half-Angle Identities■ Applications of the

Half-Angle Identities■ Verifying an Identity

Remember both the positive and negative

square roots.

The { symbol indicates that the appropriate sign is chosen depending on the quadrant of A2 . For example, if

A2 is a quadrant III angle, we choose the negative

sign because the sine function is negative in quadrant III.

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2395.6 Half-Angle Identities

We derive the identity for cos A2 using another double-angle identity.

cos 2x = 2 cos2 x - 1 Cosine double-angle identity 1 + cos 2x = 2 cos2 x Add 1.

cos2 x = 1 + cos 2x2

Rewrite and divide by 2.

cos x = {B1 + cos 2x2 Take square roots. cos

A2= tÅ1 + cos A2 Replace x with A2 .

An identity for tan A2 comes from the identities for sin A2 and cos

A2 .

tan A2

=sin A2cos A2

={B1 - cos A2{B1 + cos A2 = tÅ1 − cos A1 + cos A

We derive an alternative identity for tan A2 using double-angle identities.

tan A2

=sin A2cos A2

=2 sin A2 cos

2 cos2 A2

Multiply by 2 cos A2 in numerator and denominator.

=sin 2 AA2 B

1 + cos 2 AA2 B Double-angle identities tan

A2=

sin A1 + cos A

Simplify.

From the identity tan A2 =sin A

1 + cos A , we can also derive an equivalent identity.

tan A2=

1 − cos Asin A

Half-Angle Identities

In the following identities, the { symbol indicates that the sign is chosen based on the function under consideration and the quadrant of

A2 .

cos A2= tÅ1 + cos A2 sin A2 = tÅ1 − cos A2

tan A2= tÅ1 − cos A1 + cos A tan A2 = sin A1 + cos A tan A2 = 1 − cos Asin A

Three of these identities require a sign choice. When using these identities,

select the plus or minus sign according to the quadrant in which A2 terminates. For

example, if an angle A = 324°, then A2 = 162°, which lies in quadrant II. So when A = 324°, cos A2 and tan

A2 are negative, and sin

A2 is positive.

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240 CHAPTER 5 Trigonometric Identities

Applications of the Half-Angle Identities

EXAMPLE 1 Using a Half-Angle Identity to Find an Exact Value

Find the exact value of cos 15° using the half-angle identity for cosine.

SOLUTION cos 15° = cos 30°2

= B1 + cos 30°2Choose the positive square root.

= F1 + 2322 = FQ 1 + 232 R # 22 # 2 = 32 + 232Simplify the radicals.

■✔ Now Try Exercise 11.

EXAMPLE 2 Using a Half-Angle Identity to Find an Exact Value

Find the exact value of tan 22.5° using the identity tan A2 =sin A

1 + cos A .

SOLUTION Because 22.5° = 45°2 , replace A with 45°.

tan 22.5° = tan 45°2

= sin 45°1 + cos 45° =

222

1 + 222 =222

1 + 222 # 22 = 22

2 + 22 = 222 + 22 # 2 - 222 - 22 = 222 - 22 Rationalize the denominator. =

2 A22 - 1 B2

= 22 - 1 ■✔ Now Try Exercise 13.Factor first, and then divide out the common factor.

3p2

Figure 9

Notice that it is not necessary to use a half-angle identity for tan s2 once we find

sin s2 and cos

s2 . However, using this identity provides an excellent check.

■✔ Now Try Exercise 19.

EXAMPLE 3 Finding Function Values of s2 Given Information about s

Given cos s = 23 , with 3p2 6 s 6 2p, find sin

s2 , cos

s2 , and tan

s2 .

SOLUTION The angle associated with s2 terminates in quadrant II because

3p2

6 s 6 2p and 3p4

6 s2

6 p. Divide by 2.

See Figure 9. In quadrant II, the values of cos s2 and tan s2 are negative and the

value of sin s2 is positive. Use the appropriate half-angle identities and simplify.

sin s2

= D1 - 232 = B16 = 2126 # 2626 = 266 cos

= - D1 + 232 = - B56 = - 2526 # 2626 = - 2306 Rationalize all denominators. tan

=sin s2cos s2

=266

- 2306 = 26- 230 = - 26230 # 230230 = - 218030 = - 6256 # 5 = - 255

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2415.6 Half-Angle Identities

EXAMPLE 4 Simplifying Expressions Using Half-Angle Identities

Simplify each expression.

(a) {B1 + cos 12x2 (b) 1 - cos 5asin 5aSOLUTION

(a) This matches part of the identity for cos A2 . Replace A with 12x.

cos A2

= {B1 + cos A2 = {B1 + cos 12x2 = cos 12x2 = cos 6x(b) Use the identity tan A2 =

1 - cos Asin A with A = 5a.

1 - cos 5asin 5a

= tan 5a2

■✔ Now Try Exercises 37 and 39.

Verifying an Identity

EXAMPLE 5 Verifying an Identity

Verify that the following equation is an identity.asin x2

+ cos x2b 2 = 1 + sin x

SOLUTION We work on the more complicated left side.asin x2

+ cos x2b 2

= sin2 x2

+ 2 sin x2

cos x2

+ cos2 x2

1x + y22 = x2 + 2 xy + y2 = 1 + 2 sin x

2 cos

sin2 x2 + cos2 x2 = 1

= 1 + sin 2 a x2b 2 sin x2 cos x2 = sin 2 A x2 B

= 1 + sin x Multiply.■✔ Now Try Exercise 47.

Remember the middle term when

squaring a binomial.

CONCEPT PREVIEW Determine whether the positive or negative square root should be selected.

1. sin 195° = {B1 - cos 390°2 2. cos 58° = {B1 + cos 116°2 3. tan 225° = {B1 - cos 450°1 + cos 450° 4. sin1-10°2 = {B1 - cos1-20°22

5.6 Exercises

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242 CHAPTER 5 Trigonometric Identities

CONCEPT PREVIEW Match each expression in Column I with its value in Column II.

5. sin 15° 6. tan 15°

7. cos p

8 8. tan a - p

9. tan 67.5° 10. cos 67.5°

A. 2 - 23 B. 32 - 222

C. 32 - 23

2 D.

32 + 222

E. 1 - 22 F . 1 + 22Use a half-angle identity to find each exact value. See Examples 1 and 2.

11. sin 67.5° 12. sin 195° 13. tan 195°

14. cos 195° 15. cos 165° 16. sin 165°

17. Explain how to use identities from this section to find the exact value of sin 7.5°.

18. The half-angle identity

tan A2

= {B1 - cos A1 + cos Acan be used to find tan 22.5° = 33 - 222, and the half-angle identity

tan A2

= sin A1 + cos A

can be used to find tan 22.5° = 22 - 1. Show that these answers are the same, without using a calculator. (Hint: If a 7 0 and b 7 0 and a2 = b2 , then a = b.)

Use the given information to find each of the following. See Example 3.

19. cos x2 , given cos x =14 , with 0 6 x 6

20. sin x2 , given cos x = - 58 , with

p2 6 x 6 p

21. tan u2 , given sin u =35 , with 90° 6 u 6 180°

22. cos u2 , given sin u = - 45 , with 180° 6 u 6 270°

23. sin x2 , given tan x = 2, with 0 6 x 6p2

24. cos x2 , given cot x = -3, with p2 6 x 6 p

25. tan u2 , given tan u =273 , with 180° 6 u 6 270°

26. cot u2 , given tan u = - 252 , with 90° 6 u 6 180°

27. sin u, given cos 2u = 35 and u terminates in quadrant I

28. cos u, given cos 2u = 12 and u terminates in quadrant II

29. cos x, given cos 2x = - 512 , with p2 6 x 6 p

30. sin x, given cos 2x = 23 , with p 6 x 63p2

31. Concept Check If cos x ≈ 0.9682 and sin x = 0.250, then tan x2 ≈ .

32. Concept Check If cos x = -0.750 and sin x ≈ 0.6614, then tan x2 ≈ .

Simplify each expression. See Example 4.

33. B1 - cos 40°2 34. B1 + cos 76°2 35. B1 - cos 147°1 + cos 147° 36. B1 + cos 165°1 - cos 165° 37. 1 - cos 59.74°sin 59.74° 38. sin 158.2°1 + cos 158.2°

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2435.6 Half-Angle Identities

39. {B1 + cos 18x2 40. {B1 + cos 20a2 41. {B1 - cos 8 u1 + cos 8 u42. {B1 - cos 5A1 + cos 5A 43. {B1 + cos x4 2 44. {D1 - cos 3 u5 2Verify that each equation is an identity. See Example 5.

45. sec2 x2

= 21 + cos x 46. cot

2 x2

=11 + cos x22

sin2 x

47. sin2 x2

= tan x - sin x2 tan x

48. sin 2x2 sin x

= cos2 x2

- sin2 x2

49. 2

1 + cos x - tan2

= 1 50. tan u

2= csc u - cot u

51. 1 - tan2 u2

= 2 cos u1 + cos u 52. cos x =

1 - tan2 x2 1 + tan2 x2

53. Use the half-angle identity

tan A2

= sin A1 + cos A

to derive the equivalent identity

tan A2

= 1 - cos Asin A

by multiplying both the numerator and the denominator by 1 - cos A.

54. Use the identity tan A2 =sin A

1 + cos A to determine an identity for cot A2 .

Graph each expression and use the graph to make a conjecture, predicting what might be an identity. Then verify your conjecture algebraically.

55. sin x

1 + cos x56.

1 - cos xsin x

57. tan x2 + cot

cot x2 - tan x2

58. 1 - 8 sin2 x2

cos2 x2

(Modeling) Mach Number An airplane flying faster than the speed of sound sends out sound waves that form a cone, as shown in the figure. The cone intersects the ground to form a hyperbola. As this hyperbola passes over a particular point on the ground, a sonic boom is heard at that point. If u is the angle at the vertex of the cone, then

sin u

2= 1

m ,

where m is the Mach number for the speed of the plane. (We assume m 7 1.) The Mach number is the ratio of the speed of the plane to the speed of sound. Thus, a speed of Mach 1.4 means that the plane is flying at 1.4 times the speed of sound.

In each of the following exercises, u or m is given. Find the other value (u to the nearest degree and m to the nearest tenth as applicable).

59. m = 54

60. m = 32

61. u = 60° 62. u = 30°

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244 CHAPTER 5 Trigonometric Identities

Solve each problem.

63. (Modeling) Railroad Curves In the United States, circular railroad curves are desig-nated by the degree of curvature, the central angle subtended by a chord of 100 ft. See the figure. (Source: Hay, W. W., Railroad Engineering, John Wiley and Sons.)

50 50

Ru2

(a) Use the figure to write an expres-sion for cos u2 .

(b) Use the result of part (a) and the half-angle identity tan A2 =1 - cos A

sin A to write an expression for tan u4 .

64. In Exercise 63, if b = 12, what is the measure of angle u to the nearest degree?

Advanced methods of trigonometry can be used to find the following exact value.

sin 18° = 25 - 14

(See Hobson’s A Treatise on Plane Trigonometry.) Use this value and identities to find each exact value. Support answers with calculator approximations if desired.

65. cos 18° 66. tan 18° 67. cot 18° 68. sec 18°

69. csc 18° 70. cos 72° 71. sin 72° 72. tan 72°

73. cot 72° 74. csc 72° 75. sec 72° 76. sin 162°

Relating Concepts

For individual or collaborative investigation (Exercises 77–84)

These exercises use results from plane geom-etry to obtain exact values of the trigonometric functions of 15°.

Start with a right triangle ACB having a 60° angle at A and a 30° angle at B. Let the hypotenuse of this triangle have length 2. Extend side BC and draw a semicircle with diameter along BC extended, center at B, and radius AB. Draw segment AE. (See the figure.) Any angle inscribed in a semicircle is a right angle, so triangle EAD is a right triangle. Work Exercises 77– 84 in order.

77. Why is AB = BD true? Conclude that triangle ABD is isosceles.

78. Why does angle ABD have measure 150°?

79. Why do angles DAB and ADB both have measures of 15°?

80. What is the length DC?

81. Use the Pythagorean theorem to show that the length AD is 26 + 22.82. Use angle ADB of triangle EAD to find cos 15°.

83. Show that AE has length 26 - 22 and find sin 15°.84. Use triangle ACD to find tan 15°.

2 DBCE

90°160°

30°2

15°

√3

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245 Summary Exercises on Verifying Trigonometric Identities

These summary exercises provide practice with the various types of trigonometric identi-ties presented in this chapter. Verify that each equation is an identity.

Summary Exercises on Verifying Trigonometric Identities

1. tan u + cot u = sec u csc u 2. csc u cos2 u + sin u = csc u

3. tan x2

= csc x - cot x 4. sec1p - x2 = - sec x 5.

sin t1 + cos t =

1 - cos tsin t

6. 1 - sin t

cos t= 1

sec t + tan t

7. sin 2u = 2 tan u1 + tan2 u 8.

21 + cos x - tan

2 x2

= 1

9. cot u - tan u = 2 cos2 u - 1

sin u cos u 10.

1sec t - 1 +

1sec t + 1 = 2 cot t csc t

11. sin1x + y2cos1x - y2 = cot x + cot y1 + cot x cot y 12. 1 - tan2 u2 = 2 cos u1 + cos u

13. sin u + tan u

1 + cos u = tan u 14. csc4 x - cot4 x = 1 + cos

2 x1 - cos2 x

15. cos x =1 - tan2 x21 + tan2 x2

16. cos 2x = 2 - sec2 x

sec2 x

17. tan2 t + 1tan t csc2 t

= tan t 18. sin s1 + cos s +

1 + cos ssin s

= 2 csc s

19. tan 4u = 2 tan 2u2 - sec2 2u 20. tan a x2 + p4 b = sec x + tan x

21. cot s - tan scos s + sin s =

cos s - sin ssin s cos s

22. tan u - cot utan u + cot u = 1 - 2 cos

2 u

23. tan1x + y2 - tan y

1 + tan1x + y2 tan y = tan x 24. 2 cos2 x2 tan x = tan x + sin x 25.

cos4 x - sin4 xcos2 x

= 1 - tan2 x 26. csc t + 1csc t - 1 = 1sec t + tan t22

27. 21sin x - sin3 x2

cos x= sin 2x 28.

cot x2

- 12

tan x2

= cot x

29. cos1x + y2 + cos1y - x2sin1x + y2 - sin1y - x2 = cot x

30. sin160° - x2 - sin160° + x2 = - sin x 31. sin160° + x2 + sin160° - x2 = 23 cos x 32. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

33. sin3 u + cos3 u + sin u cos2 u + sin2 u cos u = sin u + cos u

34. cos x + sin xcos x - sin x -

cos x - sin xcos x + sin x = 2 tan 2x

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246 CHAPTER 5 Trigonometric Identities

Quick Review

Concepts Examples

If u is in quadrant IV and sin u = - 35 , find csc u, cos u, and sin1-u2.

csc u = 1sin u

= 1- 35

= - 53 Reciprocal

identity

sin2 u + cos2 u = 1 Pythagorean identity

a - 35b2 + cos2 u = 1 Substitute.

cos2 u = 1625

A - 35 B2 = 925 ; Subtract 925 . cos u = +B1625 cos u is positive in quadrant IV. cos u = 4

sin1-u2 = - sin u = - a - 35b = 3

5 Even-odd identity

Fundamental Identities

Reciprocal Identities

cot u = 1tan u

sec u = 1cos u

csc u = 1sin u

Quotient Identities

tan u = sin ucos u

cot u = cos usin u

Pythagorean Identities

sin2 u + cos2 u = 1 tan2 u + 1 = sec2 u

1 + cot2 u = csc2 u

Even-Odd Identities

sin1-u2 = - sin u cos1-u2 = cos u tan1-u2 = - tan u csc1-u2 = - csc u sec1-u2 = sec u cot1-u2 = - cot u

5.1

Chapter 5 Test Prep

Verifying Trigonometric Identities

See the box titled Hints for Verifying Identities in Section 5.2.

5.2

Sum and Difference Identities for Cosine

Sum and Difference Identities for Sine and Tangent

5.3

5.4

Cofunction Identities

cos190° - u2 = sin u cot190° - u2 = tan u sin190° - u2 = cos u sec190° - u2 = csc u tan190° - u2 = cot u csc190° - u2 = sec u

Sum and Difference Identities

cos1A - B2 = cos A cos B + sin A sin B cos1A + B2 = cos A cos B - sin A sin B sin1A + B2 = sin A cos B + cos A sin B sin1A - B2 = sin A cos B - cos A sin B tan1A + B2 = tan A + tan B

1 - tan A tan B

tan1A - B2 = tan A - tan B1 + tan A tan B

Find one value of u such that tan u = cot 78°.

tan u = cot 78°cot190° - u2 = cot 78° Cofunction identity

90° - u = 78° Set angles equal.u = 12° Solve for u.

Find the exact value of cos1-15°2.cos1-15°2

= cos130° - 45°2 -15° = 30° - 45° = cos 30° cos 45° + sin 30° sin 45°

Cosine difference identity

= 232

# 222

+ 12# 22

2 Substitute

known values.

= 26 + 224

Simplify.

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247CHAPTER 5 Test Prep

Concepts Examples

Double-Angle Identities

cos 2A = cos2 A - sin2 A cos 2A = 1 - 2 sin2 A

cos 2A = 2 cos2 A - 1 sin 2A = 2 sin A cos A

tan 2A = 2 tan A1 - tan2 A

Product-to-Sum Identities

cos A cos B = 12

3cos1A + B2 + cos1A - B24 sin A sin B = 1

2 3cos1A - B2 - cos1A + B24

sin A cos B = 12

3sin1A + B2 + sin1A - B24 cos A sin B = 1

2 3sin1A + B2 - sin1A - B24

Sum-to-Product Identities

sin A + sin B = 2 sin aA + B2b cos aA - B

sin A - sin B = 2 cos aA + B2b sin aA - B

cos A + cos B = 2 cos aA + B2b cos aA - B

cos A - cos B = -2 sin aA + B2b sin aA - B

5.5

Half-Angle Identities

cos A2

= {B1 + cos A2 sin A2 = {B1 - cos A2 tan

= {B1 - cos A1 + cos A tan A2 = sin A1 + cos Atan

= 1 - cos Asin AA In the identities involving radicals, the sign is chosen based

on the function under consideration and the quadrant of A2 . B

5.6

Given cos u = - 513 and sin u 7 0, find sin 2u.Sketch a triangle in quadrant II because cos u 6 0 and sin u 7 0. Use it to find that sin u = 1213 .

sin 2u = 2 sin u cos u

= 2 a1213b a - 5

13b

= - 120169

Write sin1-u2 sin 2u as the difference of two functions.sin1-u2 sin 2u

= 12

3cos1-u - 2u2 - cos1-u + 2u24 = 1

2 3cos1-3u2 - cos u4

= 12

cos1-3u2 - 12

cos u

= 12

cos 3u - 12

cos u

Write cos u + cos 3u as a product of two functions.

cos u + cos 3u

= 2 cos a u + 3u2b cos a u - 3u

= 2 cos a4u2b cos a -2u

= 2 cos 2u cos1-u2 = 2 cos 2u cos u

–5

13U

Find the exact value of tan 67.5°.We choose the last form with A = 135°.

tan 67.5° = tan 135°2

= 1 - cos 135°sin 135°

=1 - Q- 222 R22

=1 + 22222

# 22

= 2 + 2222 = 22 + 1Rationalize the denominator and simplify.

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248 CHAPTER 5 Trigonometric Identities

Chapter 5 Review ExercisesConcept Check For each expression in Column I, choose the expression from Column II that completes an identity.

1. sec x = 2. csc x =

3. tan x = 4. cot x =

5. tan2 x = 6. sec2 x =

A. 1

sin x B.

1cos x

C. sin xcos x

D. 1

cot2 x

E. 1

cos2 x F .

cos xsin x

Use identities to write each expression in terms of sin u and cos u, and then simplify so that no quotients appear and all functions are of u only.

7. sec2 u - tan2 u 8. cot 1-u2sec 1-u2 9. tan2 u11 + cot2 u2

10. csc u - sin u 11. tan u - sec u csc u 12. csc2 u + sec2 u

Work each problem.

13. Use the trigonometric identities to find sin x, tan x, and cot1-x2, given cos x = 35 and x in quadrant IV.

14. Given tan x = - 54 , where p2 6 x 6 p, use the trigonometric identities to find cot x,

csc x, and sec x.

15. Find the exact values of the six trigonometric functions of 165°.

16. Find the exact values of sin x, cos x, and tan x, for x = p12 , using (a) difference identities (b) half-angle identities.

Concept Check For each expression in Column I, use an identity to choose an expression from Column II with the same value. Choices may be used once, more than once, or not at all.

17. cos 210° 18. sin 35°

19. tan1-35°2 20. -sin 35° 21. cos 35° 22. cos 75°

23. sin 75° 24. sin 300°

25. cos 300° 26. cos1-55°2

A. sin1-35°2 B. cos 55°C. B1 + cos 150°2 D. 2 sin 150° cos 150° E. cot1-35°2 F . cos2 150° - sin2 150° G. cos1-35°2 H. cot 125°I . cos 150° cos 60° - sin 150° sin 60°

J . sin 15° cos 60° + cos 15° sin 60°

Use the given information to find sin1x + y2, cos1x - y2, tan1x + y2, and the quadrant of x + y.

27. sin x = - 35 , cos y = - 7

25 , x and y in quadrant III

28. sin x = 35 , cos y =2425 , x in quadrant I, y in quadrant IV

29. sin x = - 12 , cos y = - 25 , x and y in quadrant III

30. sin y = - 23 , cos x = - 15 , x in quadrant II, y in quadrant III

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249

Find values of the sine and cosine functions for each angle measure.

33. u, given cos 2u = - 34 , 90° 6 2u 6 180°

34. B, given cos 2B = 18 , 540° 6 2B 6 720°

35. 2x, given tan x = 3, sin x 6 0 36. 2y, given sec y = - 53 , sin y 7 0

Use the given information to find each of the following.

37. cos u2 , given cos u = - 12 , 90° 6 u 6 180°

38. sin A2 , given cos A = - 34 , 90° 6 A 6 180°

39. tan x, given tan 2x = 2, p 6 x 6 3p2 40. sin y, given cos 2y = - 13 , p2 6 y 6 p

41. tan x2 , given sin x = 0.8, 0 6 x 6p2 42. sin 2x, given sin x = 0.6,

p2 6 x 6 p

Graph each expression and use the graph to make a conjecture, predicting what might be an identity. Then verify your conjecture algebraically.

43. sin 2x + sin xcos x - cos 2x

44. 1 - cos 2x

sin 2x45.

sin x1 - cos x

46. cos x sin 2x1 + cos 2x 47.

21sin x - sin3 x2cos x

48. csc x - cot x

Verify that each equation is an identity.

49. sin2 x - sin2 y = cos2 y - cos2 x 50. 2 cos3 x - cos x =cos2 x - sin2 x

sec x

51. sin2 x

2 - 2 cos x = cos2

52. sin 2xsin x

= 2sec x

53. 2 cos A - sec A = cos A - tan Acsc A

54. 2 tan Bsin 2B

= sec2 B

55. 1 + tan2 a = 2 tan a csc 2a 56. 2 cot xtan 2x

= csc2 x - 2

57. tan u sin 2 u = 2 - 2 cos2 u 58. csc A sin 2 A - sec A = cos 2 A sec A

59. 2 tan x csc 2x - tan2 x = 1 60. 2 cos2 u - 1 =1 - tan2 u1 + tan2 u

61. tan u cos2 u = 2 tan u cos2 u - tan u

1 - tan2 u 62. sec2 a - 1 = sec 2a - 1

sec 2a + 1

63. sin2 x - cos2 x

csc x= 2 sin3 x - sin x 64. sin3 u = sin u - cos2 u sin u

65. tan 4u = 2 tan 2u2 - sec2 2u

66. 2 cos2 x2

tan x = tan x + sin x

67. tan a x2

+ p4b = sec x + tan x 68. 1

2 cot

- 12

tan x2

= cot x

69. - cot x2

= sin 2x + sin xcos 2x - cos x 70.

sin 3t + sin 2tsin 3t - sin 2t =

tan 5t2tan t2

CHAPTER 5 Review Exercises

31. sin x = 110 , cos y =45 , x in quadrant I, y in quadrant IV

32. cos x = 29 , sin y = - 12 , x in quadrant IV, y in quadrant III

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250 CHAPTER 5 Trigonometric Identities

(Modeling) Solve each problem.

71. Distance Traveled by an Object The distance D of an object thrown (or projected) from height h (in feet) at angle u with initial velocity v is illus-trated in the figure and modeled by the formula

D =v2 sin u cos u + v cos u 21v sin u22 + 64h

32 .

(Source: Kreighbaum, E. and K. Barthels, Biomechanics, Allyn & Bacon.)

(a) Find D when h = 0 — that is, when the object is projected from the ground. (b) Suppose a car driving over loose gravel kicks up a small stone at a velocity of

36 ft per sec (about 25 mph) and an angle u = 30°. How far, to the nearest foot, will the stone travel?

72. Amperage, Wattage, and Voltage Suppose that for an electric heater, voltage is given by V = a sin 2pvt and amperage by I = b sin 2pvt, where t is time in seconds.

(a) Find the period of the graph for the voltage.

(b) Show that the graph of the wattage W = VI has half the period of the voltage.

1. If cos u = 2425 and u is in quadrant IV, find the other five trigonometric functions of u.

2. Express sec u - sin u tan u as a single function of u.

3. Express tan2 x - sec2 x in terms of sin x and cos x, and simplify.

4. Find the exact value of cos 5p12 .

5. Express (a) cos1270° - u2 and (b) tan1p + x2 as functions of u or x alone. 6. Use a half-angle identity to find the exact value of sin1-22.5°2. 7. Graph y = cot 12 x - cot x and use the graph to make a conjecture, predicting what

might be an identity. Then verify your conjecture algebraically.

8. Given that sin A = 513 , cos B = - 35 , A is a quadrant I angle, and B is a quadrant II

angle, find each of the following.

(a) sin1A + B2 (b) cos1A + B2 (c) tan1A - B2 (d) the quadrant of A + B 9. Given that cos u = - 35 and 90° 6 u 6 180°, find each of the following.

(a) cos 2u (b) sin 2u (c) tan 2u (d) cos u

2 (e) tan

Chapter 5 Test

Verify that each equation is an identity.

13. tan2 x - sin2 x = 1tan x sin x22 14. tan x - cot xtan x + cot x = 2 sin

2 x - 1

15. (Modeling) Voltage The voltage in common household current is expressed as V = 163 sin vt, where v is the angular speed (in radians per second) of the genera-tor at an electrical plant and t is time (in seconds).

(a) Use an identity to express V in terms of cosine.

(b) If v = 120p, what is the maximum voltage? Give the least positive value of t when the maximum voltage occurs.

10. sec2 B = 11 - sin2 B

11. cos 2A = cot A - tan Acsc A sec A

12. sin 2x

cos 2x + 1 = tan x

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251

Sound waves, such as those initiated by musical instruments, travel in sinusoidal patterns that can be graphed as sine or cosine functions and described by trigonometric equations.

Inverse Circular Functions

Trigonometric Equations I

Trigonometric Equations II

Chapter 6 Quiz

Equations Involving Inverse Trigonometric Functions

6.1

6.2

6.3

6.4

Inverse Circular Functions and Trigonometric Equations6

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252 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Inverse Functions We now review some basic concepts from algebra. For a function ƒ, every element x in the domain corresponds to one and only one element y, or ƒ 1x 2 , in the range. This means the following:

1. If point 1a, b2 lies on the graph of ƒ, then there is no other point on the graph that has a as first coordinate.

2. Other points may have b as second coordinate, however, because the defi-nition of function allows range elements to be used more than once.

If a function is defined so that each range element is used only once, then it is a one-to-one function. For example, the function

ƒ1x2 = x3 is a [emailprotected]@one functionbecause every real number has exactly one real cube root. However,

g1x2 = x2 is not a [emailprotected]@one functionbecause g122 = 4 and g1-22 = 4. There are two domain elements, 2 and -2, that correspond to the range element 4.

The horizontal line test helps determine graphically whether a function is one-to-one.

6.1 Inverse Circular Functions■ Inverse Functions■ Inverse Sine Function■ Inverse Cosine Function■ Inverse Tangent

Function■ Other Inverse Circular

Functions■ Inverse Function Values

This test is applied to the graphs of ƒ1x2 = x3 and g1x2 = x2 in Figure 1.By interchanging the components of the ordered pairs of a one-to-one

function ƒ, we obtain a new set of ordered pairs that satisfies the definition of a function. This new function is the inverse function, or inverse, of ƒ.

The special notation used for inverse functions is ƒ−1 (read “ƒ-inverse”). It represents the function created by interchanging the input (domain) and the output (range) of a one-to-one function.

Horizontal Line Test

A function is one-to-one if every horizontal line intersects the graph of the function at most once.

Figure 1

f (x) = x3

f (x) = x3 is a one-to-one func-tion. It satisfies the conditionsof the horizontal line test.

g(x) = x2

g(x) = x2 is not one-to-one.It does not satisfy the conditionsof the horizontal line test.

Inverse Function

The inverse function of a one-to-one function ƒ is defined as follows.

ƒ −1 = 5 1 y, x 2 ∣ 1x, y 2 belongs to ƒ 6CAUTION Do not confuse the −1 in ƒ−1 with a negative exponent. The symbol ƒ-11x2 represents the inverse function of ƒ, not 1ƒ1x2 .

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2536.1 Inverse Circular Functions

The following statements summarize the concepts of inverse functions.

Summary of Inverse Functions

1. In a one-to-one function, each x-value corresponds to only one y-value and each y-value corresponds to only one x-value.

2. If a function ƒ is one-to-one, then ƒ has an inverse function ƒ-1.

3. The domain of ƒ is the range of ƒ-1, and the range of ƒ is the domain of ƒ -1. That is, if the point 1a, b2 lies on the graph of ƒ, then the point 1b, a2 lies on the graph of ƒ-1.

4. The graphs of ƒ and ƒ-1 are reflections of each other across the line y = x.

5. To find ƒ-11x2 for ƒ1x2, follow these steps.Step 1 Replace ƒ1x2 with y and interchange x and y.Step 2 Solve for y.

Step 3 Replace y with ƒ-11x2.Figure 2 illustrates some of these concepts.

Figure 2

(b, a) is the reflection of (a, b) acrossthe line y = x.

(b, a)

(a, b)y = xx

The graph of f –1 is the reflection ofthe graph of f across the line y = x.

y = x

f(x) = x3 – 1

f –1(x) = √x + 12

We often restrict the domain of a function that is not one-to-one to make it one-to-one without changing the range. We saw in Figure 1 that g1x2 = x2, with its natural domain 1-∞, ∞2, is not one-to-one. However, if we restrict its domain to the set of nonnegative numbers 30, ∞2, we obtain a new function ƒ that is one-to-one and has the same range as g, 30, ∞2. See Figure 3.

If the domain of g(x) = x2 isrestricted so that x ≥ 0, thenit is a one-to-one function.

g(x) = x2, x ≥ 0

x20

Figure 3

NOTE We could have restricted the domain of g1x2 = x2 to 1-∞, 04 to obtain a different one-to-one function. For the trigonometric functions, such choices are based on general agreement by mathematicians.

LOOKING AHEAD TO CALCULUSThe inverse circular functions are used in calculus to solve certain types

of related-rates problems and to

integrate certain rational functions.

Inverse Sine Function Refer to the graph of the sine function in Figure 4 on the next page. Applying the horizontal line test, we see that y = sin x does not define a one-to-one function. If we restrict the domain to the interval C - p2 , p2 D , which is the part of the graph in Figure 4 shown in color, this restricted function is one-to-one and has an inverse function. The range of y = sin x is 3-1, 14 , so the domain of the inverse function will be 3-1, 14, and its range will be C - p2 , p2 D .

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254 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Reflecting the graph of y = sin x on the restricted domain, shown in Fig-ure 5(a), across the line y = x gives the graph of the inverse function, shown in Figure 5(b). Some key points are labeled on the graph. The equation of the inverse of y = sin x is found by interchanging x and y to obtain

x = sin y.

This equation is solved for y by writing

y = sin−1 x (read “inverse sine of x”).

As Figure 5(b) shows, the domain of y = sin-1 x is 3-1, 14, while the restricted domain of y = sin x, C - p2 , p2 D , is the range of y = sin-1 x. An alternative nota-tion for sin−1 x is arcsin x.

Figure 4

(0, 0)–1

–2 y = sin x

Restricted domain

x–2p 2p–p p– 3p

3p2

–p2

[– , ]2P 2P

(– , –1)2P

( , 1)2P

(a) (b)

Figure 5

Restricted domain

[– , ]2P 2P

–1

–p2

(0, 0)

( , 1)P2

( , –1)P2–

y = sin x

y = sin–1 x or y = arcsin x

–1

(0, 0)

(1, )2P

(–1, – ) –

p3( , )√32

p4( , )√22

p4(– , – )√22

p3(– , – )√32

12( , )

p6(– , – )12

Inverse Sine Function

y = sin−1 x or y = arcsin x means that x = sin y, for - p2 … y …p2 .

We can think of y = sin−1 x or y = arcsin x as “y is the number (angle) in the interval C−P2 , P2 D whose sine is x.”

Thus, we can write y = sin-1 x as sin y = x to evaluate it. We must pay close attention to the domain and range intervals.

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2556.1 Inverse Circular Functions

CAUTION In Example 1(b), it is tempting to give the value of sin-11-12 as 3p2 because sin

3p2 = -1. However,

3p2 is not in the range of the inverse

sine function. Be certain that the number given for an inverse function value is in the range of the particular inverse function being considered.

Inverse Sine Function y = sin−1 x or y = arcsin x

Domain: 3-1, 14 Range: C - p2 , p2 Dx y

-1 - p2

- 122 - p40 0

122

1 p2

The inverse sine function is increasing on the open interval 1-1, 12 and continuous on its domain 3-1, 14.

Its x- and y-intercepts are both 10, 02. Its graph is symmetric with respect to the origin, so the function is an odd function. For all x in the domain, sin-11-x2 = -sin-1 x.

y = sin–1 x

–1 0 1

–

−1 1

−

y = sin–1x

Figure 7

EXAMPLE 1 Finding Inverse Sine Values

Find the value of each real number y if it exists.

(a) y = arcsin 12

(b) y = sin-11-12 (c) y = sin-11-22ALGEBRAIC SOLUTION

(a) The graph of the function defined by y = arcsin x (Figure 5(b)) includes the point A12 , p6 B . Therefore, arcsin

12 =

6 .

Alternatively, we can think of y = arcsin 12 as “y is the number in C - p2 , p2 D whose sine is 12 .” Then we can write the given equation as sin y = 12 . Because sin p6 =

12 and

p6 is in the range of the arc-

sine function, y = p6 .

(b) Writing the equation y = sin-11-12 in the form sin y = -1 shows that y = - p2 . Notice that the

point A -1, - p2 B is on the graph of y = sin-1 x.(c) Because -2 is not in the domain of the inverse sine

function, sin-11-22 does not exist.

GRAPHING CALCULATOR SOLUTION

(a)–(c) We graph the equation y1 = sin-1 x and find the points with x-values 12 = 0.5 and -1. For these two x-values, Figure 6 indicates that y = p6 ≈ 0.52359878 and y = - p2 ≈ -1.570796.

Figure 6

−1 1

−

−1 1

−

Because sin-11-22 does not exist, a calculator will give an error message for this input.

■✔ Now Try Exercises 13, 21, and 25.

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256 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Inverse Cosine Function The function

y = cos−1 x or y = arccos x

is defined by restricting the domain of the function y = cos x to the interval 30, p4 as in Figure 8. This restricted function, which is the part of the graph in Figure 8 shown in color, is one-to-one and has an inverse function. The inverse function, y = cos-1 x, is found by interchanging the roles of x and y. Reflecting the graph of y = cos x across the line y = x gives the graph of the inverse func-tion shown in Figure 9. Some key points are shown on the graph.

Figure 8

(0, 1)

( , 0)P2(P, –1)

Restricted domain[0, P]

–1

0–p

2p2

y = cos x

Figure 9

x10

y = cos–1 x or y = arccos x

(1, 0)–1

(–1, P)p

6( , )√32

4( , )√22

312( , )

2p3

(– , )√22 3p4(– , )√32 5p6

12(– , )

(0, )P2

We can think of y = cos−1 x or y = arccos x as

“y is the number (angle) in the interval 30, P 4 whose cosine is x.”Inverse Cosine Function

y = cos−1 x or y = arccos x means that x = cos y, for 0 … y … p.

EXAMPLE 2 Finding Inverse Cosine Values

Find the value of each real number y if it exists.

(a) y = arccos 1 (b) y = cos-1 ¢ - 222

≤SOLUTION

(a) Because the point 11, 02 lies on the graph of y = arccos x in Figure 9, the value of y, or arccos 1, is 0. Alternatively, we can think of y = arccos 1 as

“y is the number in 30, p4 whose cosine is 1,” or cos y = 1.Thus, y = 0, since cos 0 = 1 and 0 is in the range of the arccosine function.

(b) We must find the value of y that satisfies

cos y = - 222 , where y is in the interval 30, p4,which is the range of the function y = cos-1 x. The only value for y that satisfies these conditions is 3p4 . Again, this can be verified from the graph in Figure 9.

■✔ Now Try Exercises 15 and 23.

−1 10

These screens support the results of Example 2 because

- 222 ≈ -0.7071068and 3p4 ≈ 2.3561945.

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2576.1 Inverse Circular Functions

Inverse Cosine Function y = cos −1x or y = arccos x

Domain: 3-1, 14 Range: 30, p4x y

-1 p

- 122 3p40 p2

122

1 0

The inverse cosine function is decreasing on the open interval 1-1, 12 and continuous on its domain 3-1, 14.

Its x-intercept is 11, 02 and its y-intercept is A0, p2 B . Its graph is not symmetric with respect to either the y-axis or the origin.

y = cos–1 x

–1 0 1

−1 10

y = cos–1x

Figure 10

Figure 11 Figure 12

y = tan x

(0, 0)

Restricted domain

( , 1)4P

(– , –1)4P

–p2

(– , )2P 2P

–1

–2

x–1 (0, 0)

y = tan–1 x or y = arctan x

1 2

–2p

6( , )√33(1, )

(–1, – )p3(–√3 , – )

6(– , – )√33–

p2p4

–p4

3( √3 , )

Inverse Tangent Function

y = tan−1 x or y = arctan x means that x = tan y, for - p2 6 y 6p2 .

We can think of y = tan−1 x or y = arctan x as

“y is the number (angle) in the interval A−P2 , P2 B whose tangent is x.”

Inverse Tangent Function Restricting the domain of the function y = tan x to the open interval A - p2 , p2 B yields a one-to-one function. By interchanging the roles of x and y, we obtain the inverse tangent function given by

y = tan−1 x or y = arctan x.

Figure 11 shows the graph of the restricted tangent function. Figure 12 gives the graph of y = tan-1 x.

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258 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

We summarize this discussion about the inverse tangent function as follows.

Inverse Tangent Function y = tan−1 x or y = arctan x

Domain: 1-∞, ∞2 Range: A - p2 , p2 Bx y

-1 - p4

- 133 - p60 0

133

1 p4

The inverse tangent function is increasing on 1-∞, ∞2 and continuous on its domain 1-∞, ∞2.

Its x- and y-intercepts are both 10, 02. Its graph is symmetric with respect to the origin, so the function is an odd function. For all x in the domain, tan-11-x2 = - tan-1 x.

The lines y = p2 and y = - p2 are horizontal asymptotes.

y = tan–1 x

–1–2 0 1 2

–

−4 4

−

y = tan–1x

Figure 13

–2 2–1 0 1

y = cot–1 xp

x–2 2–1 0 1

y = sec–1 xp

y = csc–1 x

–2

–1 0

1 2

–p2

Figure 14

Inverse Cotangent, Secant, and Cosecant Functions*

y = cot−1 x or y = arccot x means that x = cot y, for 0 6 y 6 p.y = sec−1 x or y = arcsec x means that x = sec y, for 0 … y … p, y ≠ p2 .y = csc−1 x or y = arccsc x means that x = csc y, for - p2 … y …

2 ,y ≠ 0.

Other Inverse Circular Functions The other three inverse trigonometric functions are defined similarly. Their graphs are shown in Figure 14.

*The inverse secant and inverse cosecant functions are sometimes defined with different ranges. We use intervals that match those of the inverse cosine and inverse sine functions, respectively (except for one missing point).

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2596.1 Inverse Circular Functions

The table gives all six inverse circular functions with their domains and ranges.

Summary of Inverse Circular Functions

Range

Inverse Function Domain Interval

Quadrants of the Unit Circle

y = sin-1 x 3-1, 14 C - p2 , p2 D I and IVy = cos-1 x 3-1, 14 30, p4 I and IIy = tan-1 x 1-∞, ∞2 A - p2 , p2 B I and IVy = cot-1 x 1-∞, ∞2 10, p2 I and IIy = sec-1 x 1-∞, -14 ´ 31, ∞2 C 0, p2 B ´ Ap2 , p D I and IIy = csc-1 x 1-∞, -14 ´ 31, ∞2 C - p2 , 0 B ´ A0, p2 D I and IV

EXAMPLE 3 Finding Inverse Function Values (Degree-Measured Angles)

Find the degree measure of u if it exists.

(a) u = arctan 1 (b) u = sec-1 2SOLUTION

(a) Here u must be in 1-90°, 90°2, but because 1 is positive, u must be in quad-rant I. The alternative statement, tan u = 1, leads to u = 45°.

(b) Write the equation as sec u = 2. For sec-1 x, u is in quadrant I or II. Because 2 is positive, u is in quadrant I and u = 60°, since sec 60° = 2. Note that 60° 1the degree equivalent of p3 B is in the range of the inverse secant function.

■✔ Now Try Exercises 37 and 45.

The inverse trigonometric function keys on a calculator give correct results for the inverse sine, inverse cosine, and inverse tangent functions.

sin-1 0.5 = 30°, sin-1 1-0.52 = -30°, Degree mode

tan-11-12 = -45°, and cos-11-0.52 = 120°However, finding cot-1 x, sec-1 x, and csc-1 x with a calculator is not as

straightforward because these functions must first be expressed in terms of tan-1 x, cos-1 x, and sin-1 x, respectively. If y = sec-1 x, for example, then we have sec y = x, which must be written in terms of cosine as follows.

If sec y = x, then 1cos y = x, or cos y =1x , and y = cos-1

1x .

Inverse Function Values The inverse circular functions are formally defined with real number ranges. However, there are times when it may be con-venient to find degree-measured angles equivalent to these real number values. It is also often convenient to think in terms of the unit circle and choose the inverse function values on the basis of the quadrants given in the preceding table.

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260 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Use the following to evaluate these inverse functions on a calculator.

sec−1 x is evaluated as cos−1 1x ; csc−1 x is evaluated as sin−1 1x ;

cot−1 x is evaluated as e tan−1 1x if x + 0180° + tan−1 1x if x * 0.

Degree mode

EXAMPLE 4 Finding Inverse Function Values with a Calculator

Use a calculator to approximate each value.

(a) Find y in radians if y = csc-11-32.(b) Find u in degrees if u = arccot1-0.35412.SOLUTION

(a) With the calculator in radian mode, enter csc-1 1-32 as sin-1 A 1-3 B to obtain y ≈ -0.3398369095. See Figure 15(a).

(b) A calculator in degree mode gives the inverse tangent value of a negative number as a quadrant IV angle. The restriction on the range of arccotangent implies that u must be in quadrant II.

arccot1-0.35412 is entered as tan-1 A 1-0.3541 B + 180°.As shown in Figure 15(b),

u ≈ 109.4990544°.✔ Now Try Exercises 53 and 65.

CAUTION Be careful when using a calculator to evaluate the inverse cotangent of a negative quantity. Enter the inverse tangent of the recipro-cal of the negative quantity, which returns an angle in quadrant IV. Because inverse cotangent is negative in quadrant II, adjust the calculator result by adding p or 180° accordingly. (Note that cot-1 0 = p2 or 90°.)

EXAMPLE 5 Finding Function Values Using Definitions of the Trigonometric Functions

Evaluate each expression without using a calculator.

(a) sin a tan-1 32b (b) tan acos-1 a - 5

13b b

SOLUTION

(a) Let u = tan-1 32 , and thus tan u =32 . The inverse tangent function yields val-

ues only in quadrants I and IV, and because 32 is positive, u is in quadrant I. Sketch u in quadrant I, and label a triangle, as shown in Figure 16 on the next page. By the Pythagorean theorem, the hypotenuse is 213. The value of sine is the quotient of the side opposite and the hypotenuse.

sin a tan-1 32b = sin u = 3213 = 3213 # 213213 = 321313

Rationalize the denominator.

(b)

Figure 15

(a)

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2616.1 Inverse Circular Functions

(b) Let A = cos-1A - 513 B . Then, cos A = - 513 . Because cos-1 x for a negative value of x is in quadrant II, sketch A in quadrant II. See Figure 17.

tan acos-1 a- 513b b = tan A = - 12

✔ Now Try Exercises 75 and 77.

Figure 17

–5

A = cos–1(– )513Figure 16

U = tan–1 23

√13

EXAMPLE 6 Finding Function Values Using Identities

Evaluate each expression without using a calculator.

(a) cos aarctan 23 + arcsin 13b (b) tan a2 arcsin 2

SOLUTION

(a) Let A = arctan 23 and B = arcsin 13 . Therefore, tan A = 23 and sin B = 13 . Sketch both A and B in quadrant I, as shown in Figure 18, and use the Pythagorean theorem to find the unknown side in each triangle.

Figure 18

2√3

2√2

cos aarctan23 + arcsin 13b Given expression

= cos1A + B2 Let A = arctan23 and B = arcsin 13 . = cos A cos B - sin A sin B Cosine sum identity

= 12# 222

3- 23

2# 1

3 Substitute values using Figure 18.

= 222 - 236

Multiply and write as a single fraction.

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262 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

(b) Let B = arcsin 25 , so that sin B =25 . Sketch angle B in quadrant I, and use

the Pythagorean theorem to find the length of the third side of the triangle.

See Figure 19.

tan a2 arcsin 25b Given expression

=2 A 2121 B

1 - A 2121 B 2 Use tan 2B = 2 tan B1 - tan2 B with tan B = 2121 from Figure 19.=

41211 - 421

Multiply and apply the exponent.

=4121 # 121121

1721

Rationalize in the numerator. Subtract in the denominator.

=4121

211721

Multiply in the numerator.

= 422117

Divide; abcd

= ab ,cd =

ab# d

c .

✔ Now Try Exercises 79 and 87.

√21

Figure 19

While the work shown in Examples 5 and 6 does not rely on a calculator, we

can use one to support our algebraic work. By entering cos Aarctan 23 + arcsin 13 B from Example 6(a) into a calculator, we find the approximation 0.1827293862,

the same approximation as when we enter 222 - 236 (the exact value we obtained algebraically). Similarly, we obtain the same approximation when we evaluate

tan A2 arcsin 25 B and 422117 , supporting our answer in Example 6(b).EXAMPLE 7 Writing Function Values in Terms of u

Write each trigonometric expression as an algebraic expression in u.

(a) sin1tan-1 u2 (b) cos12 sin-1 u2SOLUTION(a) Let u = tan-1 u, so tan u = u. Because

- p2

6 tan-1 u 6 p2

sketch u in quadrants I and IV and label two triangles, as shown in Figure 20. Sine is given by the quotient of the side opposite and the hypotenuse, so we have the following.

sin1tan-1 u2 = sin u = u2u2 + 1 = u2u2 + 1 # 2u2 + 12u2 + 1 = u2u2 + 1u2 + 1Rationalize the denominator.

The result is positive when u is positive and negative when u is negative.

u, u > 0

u, u < 0

1UU

√u2 + 1

Figure 20

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2636.1 Inverse Circular Functions

An athlete can consistently put the shot with h = 6.6 ft and v = 42 ft per sec. At what angle should he release the ball to maximize distance?

SOLUTION To find this angle, substitute and use a calculator in degree mode.

u = arcsin aA 422214222 + 6416.62 b ≈ 42° Use h = 6.6, v = 42, and a calculator.✔ Now Try Exercise 105.

EXAMPLE 8 Finding Optimal Angle of Elevation of a Shot Put

The optimal angle of elevation u for a shot-putter to achieve the greatest dis-tance depends on the velocity v of the throw and the initial height h of the shot. See Figure 21. One model for u that attains this greatest distance is

u = arcsin aA v22v2 + 64h b .(Source: Townend, M. S., Mathematics in Sport, Chichester, Ellis Horwood Ltd.)

Figure 21

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. For a function to have an inverse, it must be -to- .

2. The domain of y = arcsin x equals the of y = sin x.

3. y = cos-1 x means that x = for 0 … y … p.

4. The point Ap4 , 1 B lies on the graph of y = tan x. Therefore, the point lies on the graph of y = tan-1 x.

5. If a function ƒ has an inverse and ƒ1p2 = -1, then ƒ-11-12 = . 6. To evaluate sec-1 x, use the value of cos-1 1 .

6.1 Exercises

CONCEPT PREVIEW Write a short answer for each of the following.

7. Consider the inverse sine function y = sin-1 x, or y = arcsin x.(a) What is its domain? (b) What is its range?

(d) Why is arcsin1-22 not defined?

(b) Let u = sin-1 u, so that sin u = u. To find cos 2u, use the double-angle identity cos 2u = 1 - 2 sin2 u.

cos12 sin-1 u2 = cos 2u = 1 - 2 sin2 u = 1 - 2u2✔ Now Try Exercises 95 and 99.

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264 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

8. Consider the inverse cosine function y = cos-1 x, or y = arccos x.(a) What is its domain? (b) What is its range?

(d) arccos A - 12 B = 2p3 . Why is arccos A - 12 B not equal to - 4p3 ? 9. Consider the inverse tangent function y = tan-1 x, or y = arctan x.

(a) What is its domain? (b) What is its range?

(d) Is there any real number x for which arctan x is not defined? If so, what is it (or what are they)?

10. Give the domain and range of each inverse trigonometric function, as defined in this section.

(a) inverse cosecant function (b) inverse secant function

11. Concept Check Why are different intervals used when restricting the domains of the sine and cosine functions in the process of defining their inverse functions?

12. Concept Check For positive values of a, cot-1 a is calculated as tan-1 1a . How is cot-1 a calculated for negative values of a?

Find the exact value of each real number y if it exists. Do not use a calculator. See Exam ples 1 and 2.

13. y = sin-1 0 14. y = sin-11-12 15. y = cos-11-1216. y = arccos 0 17. y = tan-1 1 18. y = arctan1-1219. y = arctan 0 20. y = tan-11-12 21. y = arcsin ¢- 23

2≤

22. y = sin-1 222

23. y = arccos ¢- 232

≤ 24. y = cos-1 a- 12b

25. y = sin-1 23 26. y = arcsin A -22 B 27. y = cot-11-1228. y = arccot A -23 B 29. y = csc-11-22 30. y = csc-1 2231. y = arcsec 223

332. y = sec-1 A -22 B 33. y = sec-1 1

34. y = sec-1 0 35. y = csc-1 222

36. y = arccsc a- 12b

Give the degree measure of u if it exists. Do not use a calculator. See Example 3.

37. u = arctan1-12 38. u = tan-1 23 39. u = arcsin ¢- 232

≤40. u = arcsin ¢- 22

2 ≤ 41. u = arccos a- 1

2b 42. u = sec-11-22

43. u = cot-1 a- 233b 44. u = cot-1 23

345. u = csc-11-22

46. u = csc-11-12 47. u = sin-1 2 48. u = cos-11-22

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2656.1 Inverse Circular Functions

Use a calculator to approximate each value in decimal degrees. See Example 4.

49. u = sin-11-0.133491222 50. u = arcsin 0.7790001651. u = arccos1-0.398764592 52. u = cos-11-0.13348816253. u = csc-1 1.9422833 54. u = cot-1 1.7670492

55. u = cot-11-0.607242262 56. u = cot-11-2.7733744257. u = tan-11-7.78286412 58. u = sec-11-5.11803782Use a calculator to approximate each real number value. (Be sure the calculator is in radian mode.) See Example 4.

59. y = arcsin 0.92837781 60. y = arcsin 0.81926439

67. y = sec-11-1.28716842 68. y = sec-1 4.796382563. y = arctan 1.1111111 64. y = cot-1 1.0036571

61. y = cos-11-0.326478912 62. y = arccos 0.4462459365. y = cot-11-0.921701282 66. y = cot-11-36.8746102The screen here shows how to define the inverse secant, cosecant, and cotangent functions in order to graph them using a TI-84 Plus graphing calculator.

Use this information to graph each inverse circular function and compare the graphs to those in Figure 14.

69. y = sec-1 x 70. y = csc-1 x 71. y = cot-1 x

Graph each inverse circular function by hand.

72. y = arccsc 2x 73. y = arcsec 12

x 74. y = 2 cot-1 x

Evaluate each expression without using a calculator. See Examples 5 and 6.

75. tan aarccos 34b 76. sin aarccos 1

4b 77. cos1tan-11-222

78. sec asin-1 a- 15b b 79. sin a2 tan-1 12

5b 80. cos a2 sin-1 1

81. cos a2 arctan 43b 82. tan a2 cos-1 1

4b 83. sin a2 cos-1 1

84. cos12 tan-11-222 85. sec1sec-1 22 86. csc Acsc-122 B87. cos atan-1 5

12- tan-1 3

4b 88. cos asin-1 3

5+ cos-1 5

13b

89. sin asin-1 12

+ tan-11-32b 90. tan acos-1 232

- sin-1 a- 35b b

Use a calculator to find each value. Give answers as real numbers.

91. cos1tan-1 0.52 92. sin1cos-1 0.25293. tan1arcsin 0.122510142 94. cot1arccos 0.582368412

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266 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

108. Landscaping Formula A shrub is planted in a 100-ft-wide space between buildings measuring 75 ft and 150 ft tall. The loca-tion of the shrub determines how much sun it receives each day. Show that if u is the angle in the figure and x is the distance of the shrub from the taller building, then the value of u (in radians) is given by

u = p - arctan a 75100 - x b - arctan a 150x b . 100 ft

75 ft

150 ft

Write each trigonometric expression as an algebraic expression in u, for u 7 0. See Example 7.

95. sin1arccos u2 96. tan1arccos u2 97. cos1arcsin u2 98. cot1arcsin u2 99. sin a2 sec-1 u

2b 100. cos a2 tan-1 3

101. tan asin-1 u2u2 + 2 b 102. sec acos-1 u2u2 + 5 b103. sec ¢arccot 24 - u2

u ≤ 104. csc ¢arctan 29 - u2

u ≤

(Modeling) Solve each problem.

105. Angle of Elevation of a Shot Put Refer to Example 8. Suppose a shot-putter can consistently release the steel ball with velocity v of 32 ft per sec from an initial height h of 5.0 ft. What angle, to the nearest degree, will maximize the distance?

106. Angle of Elevation of a Shot Put Refer to Example 8.

(a) What is the optimal angle, to the nearest degree, when h = 0?(b) Fix h at 6 ft and regard u as a function of v. As v increases without bound, the

graph approaches an asymptote. Find the equation of that asymptote.

107. Observation of a Painting A painting 1 m high and 3 m from the floor will cut off an angle u to an observer, where

u = tan-1 a xx2 + 2 b ,

assuming that the observer is x meters from the wall where the painting is displayed and that the eyes of the observer are 2 m above the ground. (See the figure.) Find the value of u for the following values of x. Round to the nearest degree.

(a) 1 (b) 2 (c) 3

(d) Derive the formula given above. (Hint: Use the identity for tan1u + a2. Use right triangles.)

(e) Graph the function for u with a graphing calcu-lator, and determine the distance that maximizes the angle.

(f ) The concept in part (e) was first investigated in 1471 by the astronomer Regiomontanus. (Source:

Maor, E., Trigonometric Delights, Princeton University Press.) If the bottom of the picture is a meters above eye level and the top of the picture is b meters

above eye level, then the optimum value of x is 2ab meters. Use this result to find the exact answer to part (e).

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2676.1 Inverse Circular Functions

109. Communications Satellite Coverage The figure shows a stationary communi-cations satellite positioned 20,000 mi above the equator. What percent, to the nearest tenth, of the equator can be seen from the satellite? The diameter of Earth is 7927 mi at the equator.

110. Oil in a Storage Tank The level of oil in a storage tank buried in the ground can be found in much the same way as a dipstick is used to determine the oil level in an automobile crankcase. Suppose the ends of the cylindrical storage tank in the figure are circles of radius 3 ft and the cylinder is 20 ft long. Determine the volume of oil in the tank to the nearest cubic foot if the rod shows a depth of 2 ft. (Hint: The volume will be 20 times the area of the shaded segment of the circle shown in the figure on the right.)

2 ft

3 ft

Relating Concepts

For individual or collaborative investigation (Exercises 111–114)*

111. Consider the function

ƒ1x2 = 3x - 2 and its inverse ƒ -11x2 = 13

x + 23

Simplify ƒ1ƒ -11x22 and ƒ -11ƒ1x22. What do you notice in each case?112. Now consider the general linear functions

ƒ1x2 = ax + b and ƒ -11x2 = 1a

x - ba

, for a≠ 0.

Simplify ƒ1ƒ-11x22 and ƒ-11ƒ1x22. What do you notice in each case? What is the graph in each case?

113. Use a graphing calculator to graph y = tan1tan-1 x2 in the standard viewing window, using radian mode. How does this compare to the graph you described in Exercise 112?

114. Use a graphing calculator to graph y = tan-11tan x2 in the standard viewing window, using radian and dot modes. Why does this graph not agree with the graph you found in Exercise 113?

*The authors wish to thank Carol Walker of Hinds Community College for making a suggestion on which these exercises are based.

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268 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

6.2 Trigonometric Equations I

Earlier we studied trigonometric equations that were identities. We now consider trigonometric equations that are conditional. These equations are satisfied by some values but not others.

■ Linear Methods■ Zero-Factor Property

Method■ Quadratic Methods■ Trigonometric Identity

Substitutions■ An Application

Linear Methods The most basic trigonometric equations are solved by first using properties of equality to isolate a trigonometric expression on one side of the equation.

EXAMPLE 1 Solving a Trigonometric Equation (Linear Methods)

Solve the equation 2 sin u + 1 = 0(a) over the interval 30°, 360°2 (b) for all solutions.

ALGEBRAIC SOLUTION

(a) Because sin u is to the first power, we use the same method as we would to solve the linear equation 2x + 1 = 0.

2 sin u + 1 = 0 Original equation 2 sin u = -1 Subtract 1.

sin u = - 12 Divide by 2.

To find values of u that satisfy sin u = - 12 , we observe that u must be in either quadrant III or quadrant IV because the sine function is negative only in these two quadrants. Furthermore, the ref-erence angle must be 30°. The graph of the unit circle in Figure 22 shows the two possible values of u. The solution set is 5210°, 330°6.

GRAPHING CALCULATOR SOLUTION

(a) Consider the original equation.

2 sin u + 1 = 0

We can find the solution set of this equation by graphing the function

y1 = 2 sin x + 1

and then determining its zeros. Because we are find-ing solutions over the interval 30°, 360°2, we use degree mode and choose this interval of values for the input x on the graph.

The screen in Figure 23(a) indicates that one solution is 210°, and the screen in Figure 23(b) indicates that the other solution is 330°. The solution set is 5210°, 330°6, which agrees with the algebraic solution.

330°210°30° 30° 1–1

–1

–12(– , – )√32 12( , – )√32

Figure 22

(b) To find all solutions, we add integer multiples of the period of the sine function, 360°, to each solu-tion found in part (a). The solution set is written as follows. 5210° + 360°n , 330° + 360°n ,

where n is any integer6(b) Because the graph of

y1 = 2 sin x + 1repeats the same y-values every 360°, all solutions are found by adding integer multiples of 360° to the solutions found in part (a). See the algebraic solution.

✔ Now Try Exercises 15 and 47.

−4

0° 360°

−4

0° 360°

Degree mode Degree mode

(a) (b)

Figure 23

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2696.2 Trigonometric Equations I

Zero-Factor Property Method

EXAMPLE 2 Solving a Trigonometric Equation (Zero-Factor Property)

Solve sin u tan u = sin u over the interval 30°, 360°2.SOLUTION sin u tan u = sin u Original equation sin u tan u - sin u = 0 Subtract sin u. sin u 1tan u - 12 = 0 Factor out sin u.

sin u = 0 or tan u - 1 = 0 Zero-factor property tan u = 1

u = 0° or u = 180° u = 45° or u = 225° Apply the inverse function.

See Figure 24. The solution set is 50°, 45°, 180°, 225°6.✔ Now Try Exercise 35.

45°

225°

0°180°(1, 0)(–1, 0)

–1

(– , – )√22 √22

( , )√22 √22

Figure 24

CAUTION Trying to solve the equation in Example 2 by dividing each side by sin u would lead to tan u = 1, which would give u = 45° or u = 225°. The missing two solutions are the ones that make the divisor, sin u, equal 0. For this reason, we avoid dividing by a variable expression.

Quadratic Methods The equation au2 + bu + c = 0, where u is an alge-braic expression, is solved by quadratic methods. The expression u may be a trigonometric function.

EXAMPLE 3 Solving a Trigonometric Equation (Zero-Factor Property)

Solve tan2 x + tan x - 2 = 0 over the interval 30, 2p2.SOLUTION tan2 x + tan x - 2 = 0 This equation is quadratic in form. 1tan x - 12 1tan x + 22 = 0 Factor. tan x - 1 = 0 or tan x + 2 = 0 Zero-factor property tan x = 1 or tan x = -2 Solve each equation.

The solutions for tan x = 1 over the interval 30, 2p2 are x = p4 and x = 5p4 .To solve tan x = -2 over that interval, we use a calculator set in radian mode.

We find that tan-11-22 ≈ -1.1071487. This is a quadrant IV number, based on the range of the inverse tangent function. However, because we want solutions over the interval 30, 2p2, we must first add p to -1.1071487, and then add 2p. See Figure 25.

x ≈ -1.1071487 + p ≈ 2.0344439 x ≈ -1.1071487 + 2p ≈ 5.1760366

The solutions over the required interval form the following solution set.Ep4 , 5p4 , 2.0344, 5.1760 F(1111)1111* (111111111)111111111* Exact Approximate values to values four decimal places

✔ Now Try Exercise 25.

–1.1071

–1.1071 + 2P = 5.1760

–1.1071 + P = 2.0344

1–1

–1

The solutions shown in bluerepresent angle measures, inradians, and their interceptedarc lengths on the unit circle.

5P4

Figure 25

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270 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

To find all solutions, we add integer multiples of the period of the tangent function, which is p, to each solution found previously. Although not unique, a common form of the solution set of the equation, written using the least possible nonnegative angle measures, is given as follows.52.8476 + np, 1.2768 + np, where n is any integer6

Round to four decimal places.

✔ Now Try Exercise 57.

EXAMPLE 4 Solving a Trigonometric Equation (Quadratic Formula)

Find all solutions of cot x1cot x + 32 = 1.SOLUTION We multiply the factors on the left and subtract 1 to write the equation in standard quadratic form.

cot x1cot x + 32 = 1 Original equation cot2 x + 3 cot x - 1 = 0 Distributive property; Subtract 1.

This equation is quadratic in form, but cannot be solved using the zero-factor property. Therefore, we use the quadratic formula, with a = 1, b = 3, c = -1, and cot x as the variable.

LOOKING AHEAD TO CALCULUSThere are many instances in calculus

where it is necessary to solve

trigonometric equations. Examples

include solving related-rates problems

and optimization problems.

Trigonometric Identity Substitutions Recall that squaring each side of an equation, such as 2x + 4 = x + 2,will yield all solutions but may also give extraneous solutions—solutions that satisfy the final equation but not the original equation. As a result, all proposed solutions must be checked in the original equation as shown in Example 5.

cot x = -b { 2b2 - 4ac2a

Quadratic formula

=-3 { 232 - 41121-12

2112 a = 1, b = 3, c = -1= -3 { 29 + 4

2 Simplify.

= -3 { 2132

Add under the radical.

cot x ≈ -3.302775638 or cot x ≈ 0.3027756377 Use a calculator.

x ≈ cot-11-3.3027756382 or x ≈ cot-110.30277563772Definition of inverse cotangent

x ≈ tan-1 a 1-3.302775638 b + p or x ≈ tan-1 a 10.3027756377 bWrite inverse cotangent in terms of inverse tangent.

x ≈ -0.2940013018 + p or x ≈ 1.276795025Use a calculator in radian mode.

x ≈ 2.847591352

Be careful with signs.

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2716.2 Trigonometric Equations I

EXAMPLE 5 Solving a Trigonometric Equation (Squaring)

Solve tan x + 23 = sec x over the interval 30, 2p2.SOLUTION We must rewrite the equation in terms of a single trigonometric function. Because the tangent and secant functions are related by the identity 1 + tan2 x = sec2 x, square each side and express sec2 x in terms of tan2 x.

A tan x + 23 B2 = 1sec x22 Square each side. tan2 x + 223 tan x + 3 = sec2 x 1x + y22 = x2 + 2x y + y2 tan2 x + 223 tan x + 3 = 1 + tan2 x Pythagorean identity

223 tan x = -2 Subtract 3 + tan2 x. tan x = - 123 Divide by 223. tan x = - 23

3 Rationalize the denominator.

Solutions of tan x = - 233 over 30, 2p2 are 5p6 and 11p6 . These possible, or pro-posed, solutions must be checked to determine whether they are also solutions of the original equation.

Don’t forget the middle term.

CHECK tan x + 23 = sec x Original equationtan a 5p

6b + 23≟ sec a 5p

- 233

+ 3233≟ - 223

2233

= - 2233

False

tan a 11p6b + 23≟ sec a 11p

- 233

+ 3233≟ 223

2233

= 2233

✓ True

As the check shows, only 11p6 is a solution, so the solution set is E 11p6 F .✔ Now Try Exercise 45.

−4

0 2p

Radian mode

The graph shows that on the interval 30, 2p2, the only zero of the function y = tan x + 23 - sec x is 5.7595865, which is an approximation for 11p6 , the solution found in Example 5.

Solving a Trigonometric Equation

1. Decide whether the equation is linear or quadratic in form in order to determine the solution method.

2. If only one trigonometric function is present, solve the equation for that function.

3. If more than one trigonometric function is present, rewrite the equation so that one side equals 0. Then try to factor and apply the zero-factor property.

4. If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval.

5. Try using identities to change the form of the equation. It may be helpful to square each side of the equation first. In this case, check for extraneous solutions.

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272 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

An Application

EXAMPLE 6 Describing a Musical Tone from a Graph

A basic component of music is a pure tone. The graph in Figure 26 models the sinusoidal pressure y = P in pounds per square foot from a pure tone at time x = t in seconds.(a) The frequency of a pure tone is often measured in hertz. One hertz is equal

to one cycle per second and is abbreviated Hz. What is the frequency ƒ, in hertz, of the pure tone shown in the graph?

(b) The time for the tone to produce one complete cycle is the period. Approxi-mate the period T, in seconds, of the pure tone.

(c) An equation for the graph is y = 0.004 sin 300px. Use a calculator to estimate all solutions that make y = 0.004 over the interval 30, 0.024.

SOLUTION

(a) From Figure 26, we see that there are 6 cycles in 0.04 sec. This is equivalent

to 6

0.04 = 150 cycles per sec. The pure tone has a frequency of ƒ = 150 Hz.

(b) Six periods cover a time interval of 0.04 sec. One period would be equal to T = 0.046 =

1150 , or 0.006 sec.

(c) If we reproduce the graph in Figure 26 on a calculator as y1 and also graph a second function as y2 = 0.004, we can determine that the approximate values of x at the points of intersection of the graphs over the interval 30, 0.024 are

0.0017, 0.0083, and 0.015.

The first value is shown in Figure 27. These values represent time in seconds.

✔ Now Try Exercise 65.

−0.006

0.006

0 0.04

Figure 26

−0.006

0.006

0 0.02

y1 = 0.004 sin 300px

y2 = 0.004

Figure 27

CONCEPT PREVIEW Use the unit circle shown here to solve each simple trigno-metric equation. If the variable is x, then solve over 30, 2p2. If the variable is u, then solve over 30°, 360°2.

6.2 Exercises

1. cos x = 12

2. cos x = 232

3. sin x = - 12

4. sin x = - 232

5. cos x = -1 6. cos x = 0

7. sin u = 0 8. sin u = -1 9. cos u = - 12

(0, 1)

(1, 0)

(0, –1)

(–1, 0) 00°180°

60°

90°

150°

210°

300°

360°

315°330°

2PP

135°120°

225°240°

270°

45°30°

12( , )√32

12( , – )√3212(– , – )√32

( , )√22√22

12( , – )√32

12( , )√32

12(– , )√32

(– , )√22√22

( , – )√22√22

12(– , – )√32

(– , – )√22√22

12(– , )√32

3P2

5P3

7P44P

5P4

7P6

5P6

3P4

2P3

11P6

10. cos u = - 222

11. sin u = - 222

12. sin u = - 232

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2736.2 Trigonometric Equations I

13. Concept Check Suppose that in solving an equation over the interval 30°, 360°2, we reach the step sin u = - 12 . Why is -30° not a correct answer?

14. Concept Check Lindsay solved the equation sin x = 1 - cos x by squaring each side to obtain

sin2 x = 1 - 2 cos x + cos2 x.

Several steps later, using correct algebra, she concluded that the solution set for solu-

tions over the interval 30, 2p2 is E 0, p2 , 3p2 F . Explain why this is not correct.

Solve each equation for solutions over the interval 30°, 360°2. Give solutions to the near-est tenth as appropriate. See Examples 2 – 5.

27. Acot u - 23 B A2 sin u + 23 B = 0 28. 1tan u - 121cos u - 12 = 029. 2 sin u - 1 = csc u 30. tan u + 1 = 23 + 23 cot u

45. cot u + 2 csc u = 3 46. 2 sin u = 1 - 2 cos u

37. sec2 u tan u = 2 tan u 38. cos2 u - sin2 u = 0

33. csc2 u - 2 cot u = 0 34. sin2 u cos u = cos u

41. tan2 u + 4 tan u + 2 = 0 42. 3 cot2 u - 3 cot u - 1 = 0

31. tan u - cot u = 0 32. cos2 u = sin2 u + 1

39. 9 sin2 u - 6 sin u = 1 40. 4 cos2 u + 4 cos u = 1

35. 2 tan2 u sin u - tan2 u = 0 36. sin2 u cos2 u = 0

43. sin2 u - 2 sin u + 3 = 0 44. 2 cos2 u + 2 cos u + 1 = 0

53. 2 cos2 x + cos x - 1 = 0 54. 4 cos2 x - 1 = 0

49. 3 csc x - 223 = 0 50. cot x + 23 = 057. sin x 13 sin x - 12 = 1 58. tan x 1tan x - 22 = 5

47. cos u + 1 = 0 48. tan u + 1 = 0

55. sin u cos u - sin u = 0 56. tan u csc u - 23 csc u = 051. 6 sin2 u + sin u = 1 52. 3 sin2 u - sin u = 2

Solve each equation (x in radians and u in degrees) for all exact solutions where appro-priate. Round approximate answers in radians to four decimal places and approximate answers in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. See Examples 1 – 5.

Solve each equation for exact solutions over the interval 30, 2p2. See Examples 1– 3.15. 2 cot x + 1 = -1 16. sin x + 2 = 3

17. 2 sin x + 3 = 4 18. 2 sec x + 1 = sec x + 3

19. tan2 x + 3 = 0 20. sec2 x + 2 = -1

21. 1cot x - 12A23 cot x + 1 B = 0 22. 1csc x + 22Acsc x - 22 B = 023. cos2 x + 2 cos x + 1 = 0 24. 2 cos2 x - 23 cos x = 025. -2 sin2 x = 3 sin x + 1 26. 2 cos2 x - cos x = 1

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274 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

The following equations cannot be solved by algebraic methods. Use a graphing calcula-tor to find all solutions over the interval 30, 2p2. Express solutions to four decimal places.63. x2 + sin x - x3 - cos x = 0 64. x3 - cos2 x = 1

2 x - 1

(Modeling) Solve each problem.

65. Pressure on the Eardrum See Example 6. No musical instrument can generate a true pure tone. A pure tone has a unique, constant frequency and amplitude that sounds rather dull and uninteresting. The pressures caused by pure tones on the eardrum are sinusoidal. The change in pressure P in pounds per square foot on a person’s eardrum from a pure tone at time t in seconds can be modeled using the equation

P = A sin12pƒt + f2,where ƒ is the frequency in cycles per second, and f is the phase angle. When P is positive, there is an increase in pressure and the eardrum is pushed inward. When P is negative, there is a decrease in pressure and the eardrum is pushed outward. (Source: Roederer, J., Introduction to the Physics and Psychophysics of Music, Second Edition, Springer-Verlag.)

(a) Determine algebraically the values of t for which P = 0 over 30, 0.0054.(b) From a graph and the answer in part (a), determine the interval for which P … 0

over 30, 0.0054.(c) Would an eardrum hearing this tone be vibrating outward or inward when P 6 0?

66. Accident Reconstruction To reconstruct accidents in which a vehicle vaults into the air after hitting an obstruction, the model

0.342D cos u + h cos2 u = 16D2

V0 2

can be used. V0 is velocity in feet per second of the vehicle when it hits the obstruction, D is distance (in feet) from the obstruction to the landing point, and h is the dif-ference in height (in feet) between landing point and takeoff point. Angle u is the takeoff angle, the angle between the horizontal and the path of the vehicle. Find u to the nearest degree if V0 = 60, D = 80, and h = 2.

67. Electromotive Force In an electric circuit, suppose that the electromotive force in volts at t seconds can be modeled by

V = cos 2pt.

Find the least value of t where 0 … t … 12 for each value of V.

(a) V = 0 (b) V = 0.5 (c) V = 0.25

68. Voltage Induced by a Coil of Wire A coil of wire rotating in a magnetic field induces a voltage modeled by

E = 20 sin apt4

- p2b ,

where t is time in seconds. Find the least positive time to produce each voltage.

(a) 0 (b) 1023

59. 5 + 5 tan2 u = 6 sec u 60. sec2 u = 2 tan u + 4

61. 2 tan u

3 - tan2 u = 1 62. 2 cot2 u

cot u + 3 = 1

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2756.3 Trigonometric Equations II

6.3 Trigonometric Equations II

In this section, we discuss trigonometric equations that involve functions of half-angles and multiple angles. Solving these equations often requires adjusting solution intervals to fit given domains.

■ Equations with Half-Angles

■ Equations with Multiple Angles

■ An Application

Equations with Half-Angles

−2

0 2p

The x-intercepts correspond to the solutions found in Example 1(a). Using Xscl = p3 makes it possible to support the exact solutions by counting the tick marks from 0 on the graph.

(b) The argument x2 in the expression sin x2 can also be written

12 x to see that the

value of b in sin bx is 12 . From earlier work we know that the period is 2pb , so

we replace b with 12 in this expression and perform the calculation. Here the period is

2p12

= 2p , 12

= 2p # 2 = 4p.

All solutions are found by adding integer multiples of 4p.Ep3 + 4np, 5p3 + 4np, where n is any integer F✔ Now Try Exercises 25 and 39.

EXAMPLE 1 Solving an Equation with a Half-Angle

Solve 2 sin x2 = 1

(a) over the interval 30, 2p2 (b) for all solutions.SOLUTION

(a) To solve over the interval 30, 2p2, we must have0 … x 6 2p.

The corresponding inequality for x2 is

0 … x2

6 p. Divide by 2.

To find all values of x2 over the interval 30, p2 that satisfy the given equation,

first solve for sin x2 .

2 sin x2

= 1 Original equation

sin x2

= 12 Divide by 2.

The two numbers over the interval 30, p2 with sine value 12 are p6 and 5p6 . x2

= p6 or

= 5p6

Definition of inverse sine

x = p3 or x = 5p

3 Multiply by 2.

The solution set over the given interval is Ep3 , 5p3 F .

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276 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Equations with Multiple Angles

EXAMPLE 2 Solving an Equation Using a Double-Angle Identity

Solve cos 2x = cos x over the interval 30, 2p2.SOLUTION First convert cos 2x to a function of x alone. Use the identity cos 2x = 2 cos2 x - 1 so that the equation involves only cos x. Then factor.

cos 2x = cos x Original equation 2 cos2 x - 1 = cos x Cosine double-angle identity

2 cos2 x - cos x - 1 = 0 Subtract cos x. 12 cos x + 12 1cos x - 12 = 0 Factor.

2 cos x + 1 = 0 or cos x - 1 = 0 Zero-factor property

cos x = - 12

or cos x = 1 Solve each equation for cos x.

If we use the unit circle to analyze these results, we recognize that a radian-measured angle having cosine - 12 must be in quadrant II or III with reference angle p3 . Another possibility is that it has a value of 1 at 0 radians. We can use

Figure 28 to determine that solutions over the required interval are as follows.

x = 2p3 or x = 4p

3 or x = 0

The solution set is E0, 2p3 , 4p3 F . ✔ Now Try Exercise 27.

x–1

–1

4P3

2P3(– , )

√32

(– , – )12 √32

0 (1, 0)

–

Figure 28

EXAMPLE 3 Solving an Equation Using a Double-Angle Identity

Solve 4 sin u cos u = 23(a) over the interval 30°, 360°2 (b) for all solutions.SOLUTION

(a) 4 sin u cos u = 23 Original equation 212 sin u cos u2 = 23 4 = 2 # 2

2 sin 2u = 23 Sine double-angle identity sin 2u = 23

2 Divide by 2.

From the given interval 0° … u 6 360°, the corresponding interval for 2u is 0° … 2u 6 720°. Because the sine is positive in quadrants I and II, solutions over this interval are as follows.

2u = 60°, 120°, 420°, 480°, Reference angle is 60°.

or u = 30°, 60°, 210°, 240° Divide by 2.

The final two solutions for 2u were found by adding 360° to 60° and 120°, respectively, which gives the solution set 530°, 60°, 210°, 240°6.

CAUTION Because 2 is not a factor of cos 2x, cos 2x2 ≠ cos x. In Exam-ple 2, we changed cos 2x to a function of x alone using an identity.

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2776.3 Trigonometric Equations II

(b) All angles 2u that are solutions of the equation sin 2u = 232 are found by adding integer multiples of 360° to the basic solution angles, 60° and 120°.

2u = 60° + 360°n and 2u = 120° + 360°n Add integer multiples of 360°. u = 30° + 180°n and u = 60° + 180°n Divide by 2.

All solutions are given by the following set, where 180° represents the period of sin 2u. 530° + 180°n, 60° + 180°n, where n is any integer6

✔ Now Try Exercises 23 and 47.

NOTE Solving an equation by squaring both sides may introduce extra-neous values. We use this method in Example 4, and all proposed solutions must be checked.

EXAMPLE 4 Solving an Equation with a Multiple Angle

Solve tan 3x + sec 3x = 2 over the interval 30, 2p2.SOLUTION The tangent and secant functions are related by the identity 1 + tan2 x = sec2 x. One way to begin is to express the left side in terms of secant.

tan 3x + sec 3x = 2

tan 3x = 2 - sec 3x Subtract sec 3x.

1tan 3x22 = 12 - sec 3x22 Square each side. tan2 3x = 4 - 4 sec 3x + sec2 3x 1x - y22 = x2 - 2xy + y2

sec2 3x - 1 = 4 - 4 sec 3x + sec2 3x Replace tan2 3x with sec2 3x - 1.

4 sec 3x = 5 Simplify.

sec 3x = 54

Divide by 4.

cos 3x= 5

4 sec u = 1cos u

cos 3x = 45

Use reciprocals.

Multiply each term of the inequality 0 … x 6 2p by 3 to find the interval for 3x: 30, 6p2. Use a calculator and the fact that cosine is positive in quadrants I and IV.3x ≈ 0.6435, 5.6397, 6.9267, 11.9229, 13.2099, 18.2061 These numbers have cosine value equal to 45 .

x ≈ 0.2145, 1.8799, 2.3089, 3.9743, 4.4033, 6.0687 Divide by 3.

Both sides of the equation were squared, so each proposed solution must be checked. Verify by substitution in the given equation that the solution set is50.2145, 2.3089, 4.40336.

✔ Now Try Exercise 53.

−3

0 2p

The screen shows the graphs of

y1 = tan 3x + sec 3xand

y2 = 2.One solution is approximately 2.3089. An advantage of using a graphing calculator is that extraneous values do not appear.

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278 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

An Application A piano string can vibrate at more than one frequency when it is struck. It produces a complex wave that can mathematically be modeled by a sum of several pure tones. When a piano key with a frequency of ƒ1 is played, the corresponding string vibrates not only at ƒ1 but also at the higher frequencies of 2ƒ1 , 3ƒ1 , 4ƒ1 , . . . , nƒ1. ƒ1 is the fundamental frequency of the string, and higher frequencies are the upper harmonics. The human ear will hear the sum of these frequencies as one complex tone. (Source: Roederer, J., Introduction to the Physics and Psychophysics of Music, Second Edition, Springer-Verlag.)

EXAMPLE 5 Analyzing Pressures of Upper Harmonics

Suppose that the A key above middle C is played on a piano. Its fundamental frequency is ƒ1 = 440 Hz, and its associated pressure is expressed as

P1 = 0.002 sin 880pt.

The string will also vibrate at

ƒ2 = 880, ƒ3 = 1320, ƒ4 = 1760, ƒ5 = 2200, . . . . Hz.

The corresponding pressures of these upper harmonics are as follows.

P2 =0.002

2 sin 1760pt, P3 =

0.0023

sin 2640pt,

P4 =0.002

4 sin 3520pt, and P5 =

0.0025

sin 4400pt

The graph of P = P1 + P2 + P3 + P4 + P5 can be found by entering each Pi as a separate function yi and graphing their sum. The graph, shown in Figure 29, is “saw-toothed.”

−0.005

0.005

0 0.01

Figure 29

−0.005

0.005

0 0.01

Figure 30

(a) Approximate the maximum value of P.

(b) At what values of t = x does this maximum occur over 30, 0.014?SOLUTION

(a) A graphing calculator shows that the maximum value of P is approximately 0.00317. See Figure 30.

(b) The maximum occurs at

t = x ≈ 0.000191, 0.00246, 0.00474, 0.00701, and 0.00928.

Figure 30 shows how the second value is found. The other values are found similarly.

✔ Now Try Exercise 57.

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2796.3 Trigonometric Equations II

6.3 Exercises

CONCEPT PREVIEW Refer to Exercises 1– 6 in the previous section, and use those results to solve each equation over the interval 30, 2p2. 1. cos 2x = 1

2 2. cos 2x = 23

2 3. sin 2x = - 1

4. sin 2x = - 232

5. cos 2x = -1 6. cos 2x = 0

CONCEPT PREVIEW Refer to Exercises 7–12 in the previous section, and use those results to solve each equation over the interval 30°, 360°2. 7. sin

2= 0 8. sin u

2= -1 9. cos u

2= - 1

10. cos u

2= - 22

2 11. sin

2= - 22

2 12. sin

2= - 23

Concept Check Answer each question.

13. Suppose solving a trigonometric equation for solutions over the interval 30, 2p2 leads to 2x = 2p3 , 2p,

8p3 . What are the corresponding values of x?

14. Suppose solving a trigonometric equation for solutions over the interval 30, 2p2 leads to 12 x =

p16 ,

5p12 ,

5p8 . What are the corresponding values of x?

15. Suppose solving a trigonometric equation for solutions over the interval 30°, 360°2 leads to 3u = 180°, 630°, 720°, 930°. What are the corresponding values of u?

16. Suppose solving a trigonometric equation for solutions over the interval 30°, 360°2 leads to 13 u = 45°, 60°, 75°, 90°. What are the corresponding values of u?

Solve each equation in x for exact solutions over the interval 30, 2p2 and each equation in u for exact solutions over the interval 30°, 360°2. See Examples 1– 4.17. 2 cos 2x = 23 18. 2 cos 2x = -1 19. sin 3u = -120. sin 3u = 0 21. 3 tan 3x = 23 22. cot 3x = 2323. 22 cos 2u = -1 24. 223 sin 2u = 23 25. sin x

2= 22 - sin x

26. tan 4x = 0 27. sin x = sin 2x 28. cos 2x - cos x = 0

29. 8 sec2 x2

= 4 30. sin2 x2

- 2 = 0 31. sin u2

= csc u2

32. sec u

2= cos u

233. cos 2x + cos x = 0 34. sin x cos x =

35. 22 sin 3x - 1 = 0 36. -2 cos 2x = 23 37. cos u2

= 1

38. sin u

2= 1 39. 223 sin x

2= 3 40. 223 cos x

2= -3

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280 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

41. 2 sin u = 2 cos 2u 42. cos u - 1 = cos 2u 43. 1 - sin x = cos 2x

44. sin 2x = 2 cos2 x 45. 3 csc2 x2

= 2 sec x 46. cos x = sin2 x2

47. 2 - sin 2u = 4 sin 2u 48. 4 cos 2u = 8 sin u cos u

49. 2 cos2 2u = 1 - cos 2u 50. sin u - sin 2u = 0

Solve each equation for solutions over the interval 30, 2p2. Write solutions as exact values or to four decimal places, as appropriate. See Example 4.

51. sin x2

- cos x2

= 0 52. sin x2

+ cos x2

= 1

53. tan 2x + sec 2x = 3 54. tan 2x - sec 2x = 2

2+ 2x - 2 = - 1

2 x + 2

(Modeling) Solve each problem. See Example 5.

57. Pressure of a Plucked String If a string with a fundamental frequency of 110 Hz is plucked in the middle, it will vibrate at the odd harmonics of 110, 330, 550, . . . Hz but not at the even harmonics of 220, 440, 660, . . . Hz. The resulting pressure P caused by the string is graphed below and can be modeled by the following equation.

P = 0.003 sin 220pt + 0.0033

sin 660pt + 0.0035

sin 1100pt + 0.0037

sin 1540pt

(Source: Benade, A., Fundamentals of Musical Acoustics, Dover Publications. Roederer, J., Introduction to the Physics and Psychophysics of Music, Second Edition, Springer-Verlag.)

(a) Duplicate the graph shown here.

(b) Describe the shape of the sound wave that is produced.

(c) At lower frequencies, the inner ear will hear a tone only when the eardrum is moving outward. This occurs when P is negative. Determine the times over the interval 30, 0.034 when this will occur.

58. Hearing Beats in Music Musicians sometimes tune instruments by playing the same tone on two different instruments and listening for a phenomenon known as beats. Beats occur when two tones vary in frequency by only a few hertz. When the two instruments are in tune, the beats disappear. The ear hears beats because the pressure slowly rises and falls as a result of this slight variation in the frequency. (Source: Pierce, J., The Science of Musical Sound, Scientific American Books.)

(a) Consider the two tones with frequencies of 220 Hz and 223 Hz and pressures P1 = 0.005 sin 440pt and P2 = 0.005 sin 446pt, respectively. A graph of the pressure P = P1 + P2 felt by an eardrum over the 1-sec interval 30.15, 1.154 is shown here. How many beats are there in 1 sec?

(b) Repeat part (a) with frequencies of 220 and 216 Hz.

−0.005

0.005

0 0.03

For x = t,P(t) = 0.003 sin 220pt +

sin 660pt +0.0033

sin 1100pt +0.0035

sin 1540pt0.0037

−0.01

0.01

0.15 1.15

For x = t,P(t) = 0.005 sin 440pt + 0.005 sin 446pt

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2816.3 Trigonometric Equations II

59. Hearing Difference Tones When a musical instrument creates a tone of 110 Hz, it also creates tones at 220, 330, 440, 550, 660, . . . Hz. A small speaker cannot reproduce the 110-Hz vibration but it can reproduce the higher frequencies, which are the upper harmonics. The low tones can still be heard because the speaker produces difference tones of the upper harmonics. The difference between con-secutive frequencies is 110 Hz, and this difference tone will be heard by a listener. (Source: Benade, A., Fundamentals of Musical Acoustics, Dover Publications.)

(a) In the window 30, 0.034 by 3-1, 14, graph the upper harmonics represented by the pressure

P = 12

sin32p12202t4 + 13

sin32p13302t4 + 14

sin32p14402t4.(b) Estimate all t-coordinates where P is maximum.

(d) Graph the pressure produced by a speaker that can vibrate at 110 Hz and above.

60. Daylight Hours in New Orleans The seasonal variation in length of daylight can be modeled by a sine function. For example, the daily number of hours of daylight in New Orleans is given by

h = 353

+ 73

sin 2px365

where x is the number of days after March 21 (disregarding leap year). (Source: Bushaw, D., et al., A Sourcebook of Applications of School Mathematics, Mathemati-cal Association of America.)

(a) On what date will there be about 14 hr of daylight?

(b) What date has the least number of hours of daylight?

61. Average Monthly Temperature in Vancouver The following function approxi-mates average monthly temperature y (in °F) in Vancouver, Canada. Here x represents the month, where x = 1 corresponds to January, x = 2 corresponds to February, and so on. (Source: www.weather.com)

ƒ1x2 = 14 sin cp6

1x - 42 d + 50When is the average monthly temperature (a) 64°F (b) 39°F?

62. Average Monthly Temperature in Phoenix The following function approximates average monthly temperature y (in °F) in Phoenix, Arizona. Here x represents the month, where x = 1 corresponds to January, x = 2 corresponds to February, and so on. (Source: www.weather.com)

ƒ1x2 = 19.5 cos cp6

1x - 72 d + 70.5When is the average monthly temperature (a) 70.5°F (b) 55°F?

(Modeling) Alternating Electric Current The study of alternating electric current requires solving equations of the form

i = Imax sin 2pƒt,

for time t in seconds, where i is instantaneous current in amperes, Imax is maximum cur-rent in amperes, and f is the number of cycles per second. (Source: Hannon, R. H., Basic Technical Mathematics with Calculus, W. B. Saunders Company.) Find the least positive value of t, given the following data.

63. i = 40, Imax = 100, ƒ = 60 64. i = 50, Imax = 100, ƒ = 120

65. i = Imax , ƒ = 60 66. i =12

Imax , ƒ = 60

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http://www.weather.com

282 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Solve each equation for solutions over the interval 30, 2p2. Round approximate answers to four decimal places.

8. tan2 x - 5 tan x + 3 = 0 9. 3 cot 2x - 23 = 0 10. Solve cos

+ 23 = -cos x2

, giving all solutions in radians.

1. Graph y = cos-1 x, and indicate the coordinates of three points on the graph. Give the domain and range.

2. Find the exact value of each real number y. Do not use a calculator.

(a) y = sin-1 ¢ - 222

≤ (b) y = tan-1 23 (c) y = sec-1 ¢ - 2233

≤ 3. Use a calculator to approximate each value in decimal degrees. (a) u = arccos 0.92341853 (b) u = cot-11-1.08867672 4. Evaluate each expression without using a calculator.

(a) cos a tan-1 45b (b) sin acos-1 a - 1

2b + tan-1 A -23 B b

Chapter 6 Quiz (Sections 6.1–6.3)

Solve each equation for exact solutions over the interval 30°, 360°2. 5. 2 sin u - 23 = 0 6. cos u + 1 = 2 sin2 u 7. (Modeling) Electromotive Force In an electric circuit, suppose that

V = cos 2pt

models the electromotive force in volts at t seconds. Find the least value of t where

0 … t … 12 for each value of V.

(a) V = 1 (b) V = 0.30

6.4 Equations Involving Inverse Trigonometric Functions

Solution for x in Terms of y Using Inverse Functions ■ Solution for x in Terms of y Using Inverse Functions

■ Solution of Inverse Trigonometric Equations

EXAMPLE 1 Solving an Equation for a Specified Variable

Solve y = 3 cos 2x for x, where x is restricted to the interval C 0, p2 D .SOLUTION We want to isolate cos 2x on one side of the equation so that we can solve for 2x, and then for x.

y = 3 cos 2x

= cos 2x Divide by 3.

2x = arccos y3

Definition of arccosine

x = 12

arccos y3 Multiply by 12 .

An equivalent form of this answer is x = 12 cos-1 y3 .

Our goal is to isolate x.

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2836.4 Equations Involving Inverse Trigonometric Functions

Because the function y = 3 cos 2x is periodic, with period p, there are infinitely many domain values (x-values) that will result in a given range value ( y-value). For example, the x-values 0 and p both correspond to the y-value 3. See Figure 31. The restriction 0 … x … p2 given in the original problem ensures that this function is one-to-one, and, correspondingly, that

x = 12

arccos y3

has a one-to-one relationship. Thus, each y-value in 3-3, 34 substituted into this equation will lead to a single x-value.

✔ Now Try Exercise 9.

3(0, 3) (P, 3)

–3

pp4

3p4

( , –3)P20 ≤ x ≤

y = 3 cos 2x

Figure 31

Solution of Inverse Trigonometric Equations

EXAMPLE 2 Solving an Equation Involving an Inverse Trigonometric Function

Solve 2 arcsin x = p.

SOLUTION First solve for arcsin x, and then for x.

2 arcsin x = p Original equation

arcsin x = p2

Divide by 2.

x = sin p2

Definition of arcsine

x = 1 arcsin 1 = p2CHECK 2 arcsin x = p Original equation 2 arcsin 1≟ p Let x = 1.

2 ap2b ≟ p Substitute the inverse value.

p = p ✓ TrueThe solution set is 516. ✔ Now Try Exercise 27.

EXAMPLE 3 Solving an Equation Involving Inverse Trigonometric Functions

Solve cos-1 x = sin-1 12

SOLUTION Let sin-1 12 = u. Then sin u =12 , and for u in quadrant I we have

the following.

cos-1 x = sin-1 12 Original equation

cos-1 x = u Substitute. cos u = x Alternative form

Sketch a triangle and label it using the facts that u is in quadrant I and sin u = 12 .

See Figure 32. Because x = cos u, we have x = 232 . The solution set is U 232 V .✔ Now Try Exercise 35.

2 1u

√3

Figure 32

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284 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

EXAMPLE 4 Solving an Inverse Trigonometric Equation Using an Identity

Solve arcsin x - arccos x = p6

SOLUTION Isolate one inverse function on one side of the equation.

arcsin x - arccos x = p6

Original equation

arcsin x = arccos x + p6

Add arccos x. (1)

x = sin aarccos x + p6b Definition of arcsine

Let u = arccos x. The arccosine function yields angles in quadrants I and II, so 0 … u … p by definition.

x = sin au + p6b Substitute.

x = sin u cos p6

+ cos u sin p6

Sine sum identity (2)

Use equation (1) and the definition of the arcsine function.

- p2

… arccos x + p6

… p2

Range of arcsine is C - p2 , p2 D .- 2p

3… arccos x … p

3 Subtract p6 from each part.

Because both 0 … arccos x … p and - 2p3 … arccos x …p3 , the intersection yields

0 … arccos x … p3 . This places u in quadrant I, and we can sketch the triangle in Figure 33. From this triangle we find that sin u = 21 - x2. Now substitute into equation (2) using sin u = 21 - x2, sin p6 = 12 , cos p6 = 232 , and cos u = x.

x = sin u cos p6

+ cos u sin p6

(2)

x = A21 - x2 B 232

+ x # 12

Substitute.

2x = A21 - x2 B 23 + x Multiply by 2. x = A23 B21 - x2 Subtract x; commutative property

x 2 = 311 - x22 Square each side; 1ab22 = a2 b2 x 2 = 3 - 3x2 Distributive property

x 2 = 34

Add 3x2. Divide by 4.

x = B34 Take the square root on each side. x = 23

2 Quotient rule: 3n ab = 2n a2n b

ux0

√1 – x2

Figure 33

Square each factor.

Choose the positive square

root, x 7 0.

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2856.4 Equations Involving Inverse Trigonometric Functions

CHECK A check is necessary because we squared each side when solving the equation.

arcsin x - arccos x = p6 Original equation

arcsin 232

- arccos 232≟ p

6 Let x = 232 .

3- p

6≟ p

6 Substitute inverse values.

6= p

6 ✓ True

The solution set is U 232 V . ✔ Now Try Exercise 37.

CONCEPT PREVIEW Answer each question.

1. Which one of the following equations has solution 0?

A. arctan 1 = x B. arccos 0 = x C. arcsin 0 = x

2. Which one of the following equations has solution p4 ?

A. arcsin 222

= x B. arccos a - 222b = x C. arctan 23

3= x

3. Which one of the following equations has solution 3p4 ?

A. arctan 1 = x B. arcsin 222

= x C. arccos a - 222b = x

4. Which one of the following equations has solution - p6 ?

A. arctan 233

= x B. arccos a - 12b = x C. arcsin a - 1

2b = x

5. Which one of the following equations has solution p?

A. arccos1-12 = x B. arccos 1 = x C. arcsin1-12 = x 6. Which one of the following equations has solution - p2 ?

A. arctan1-12 = x B. arcsin1-12 = x C. arccos1-12 = x

6.4 Exercises

Solve each equation for x, where x is restricted to the given interval. See Example 1.

7. y = 5 cos x, for x in 30, p4 8. y = 14

sin x, for x in c - p2

, p

9. y = 3 tan 2x, for x in a - p4

, p

4b 10. y = 3 sin x

2 , for x in 3-p, p4

11. y = 6 cos x4

, for x in 30, 4p4 12. y = -sin x3

, for x in c - 3p2

, 3p2d

13. y = -2 cos 5x, for x in c 0, p5d 14. y = 3 cot 5x, for x in a0, p

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286 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Use a graphing calculator in each of the following.

45. Provide graphical support for the solution in Example 4 by showing that the graph of

y = sin-1 x - cos-1 x - p6

has a zero of 232≈ 0.8660254.

15. y = sin x - 2, for x in c - p2

, p

2d 16. y = cot x + 1, for x in 10, p2

17. y = -4 + 2 sin x, for x in c - p2

, p

2d 18. y = 4 + 3 cos x, for x in 30, p4

19. y = 12

cot 3x, for x in a0, p3b

20. y = 112

sec x, for x in c 0, p2b ´ ap

2 , p d

24. y = -23 + 2 csc x2

, for x in 3-p, 02 ´ 10, p4

22. y = tan12x - 12, for x in a 12

- p4

, 12

+ p4b

21. y = cos1x + 32, for x in 3-3, p - 34

25. Refer to Exercise 15. A student solving this equation wrote y = sin1x - 22 as the first step, inserting parentheses as shown. Explain why this is incorrect.

23. y = 22 + 3 sec 2x, for x in c 0, p4b ´ ap

4 , p

26. Explain why the equation sin-1 x = cos-1 2 cannot have a solution. (No work is required.)

Solve each equation for exact solutions. See Examples 2 and 3.

27. -4 arcsin x = p 28. 6 arccos x = 5p

29. 43

cos-1 x4

= p 30. 4 tan-1 x = -3p

33. arcsin x = arctan 34

34. arctan x = arccos 513

31. 2 arccos a x3

- p3b = 2p 32. 6 arccos ax - p

3b = p

35. cos-1 x = sin-1 35

36. cot-1 x = tan-1 43

43. cos-1 x + tan-1 x = p2

44. sin-1 x + tan-1 x = 0

Solve each equation for exact solutions. See Example 4.

39. arccos x + 2 arcsin 232

= p 40. arccos x + 2 arcsin 232

= p3

37. sin-1 x - tan-1 1 = - p4

38. sin-1 x + tan-1 23 = 2p3

41. arcsin 2x + arccos x = p6

42. arcsin 2x + arcsin x = p2

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2876.4 Equations Involving Inverse Trigonometric Functions

(Modeling) Solve each problem.

49. Tone Heard by a Listener When two sources located at different positions pro-duce the same pure tone, the human ear will often hear one sound that is equal to the sum of the individual tones. Because the sources are at different locations, they will have different phase angles f. If two speakers located at different positions produce pure tones P1 = A1 sin12pƒt + f12 and P2 = A2 sin12pƒt + f22, where - p4 … f1, f2 …

p4 , then the resulting tone heard by a listener can be written as

P = A sin12pƒt + f2, where A = 21A1 cos f1 + A2 cos f222 + 1A1 sin f1 + A2 sin f222

and f = arctan a A1 sin f1 + A2 sin f2A1 cos f1 + A2 cos f2

b .(Source: Fletcher, N. and T. Rossing, The Physics of Musical Instruments, Second Edition, Springer-Verlag.)

(a) Calculate A and f if A1 = 0.0012, f1 = 0.052, A2 = 0.004, and f2 = 0.61. Also, if ƒ = 220, find an expression for

P = A sin12pƒt + f2.(b) Graph Y1 = P and Y2 = P1 + P2 on the same coordinate axes over the interval 30, 0.014. Are the two graphs the same?

50. Tone Heard by a Listener Repeat Exercise 49. Use A1 = 0.0025 , f1 =p7 , A2 = 0.001 ,

f2 =p6 , and ƒ = 300.

51. Depth of Field When a large-view camera is used to take a picture of an object that is not parallel to the film, the lens board should be tilted so that the planes con-taining the subject, the lens board, and the film intersect in a line. This gives the best “depth of field.” See the figure. (Source: Bushaw, D., et al., A Sourcebook of Applications of School Mathematics, Mathematical Association of America.)

(a) Write two equations, one relating a, x, and z, and the other relating b, x, y, and z.

(b) Eliminate z from the equations in part (a) to get one equation relating a, b, x, and y.

(d) Solve the equation from part (b) for b.

Subject Lens

Film

46. Provide graphical support for the solution in Example 4 by showing that the x-coordinate of the point of intersection of the graphs of

y1 = sin-1 x - cos-1 x and y2 =p

6 is

232≈ 0.8660254.

The following equations cannot be solved by algebraic methods. Use a graphing calcula-tor to find all solutions over the interval 30, 64. Express solutions to four decimal places.47. 1arctan x23 - x + 2 = 0 48. p sin-110.2x2 - 3 = -2x

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288 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

52. Programming Language for Inverse Functions In some programming languages, the only inverse trigonometric function available is arctangent. The other inverse trigonometric functions can be expressed in terms of arctangent.

(a) Let u = arcsin x. Solve the equation for x in terms of u.

(b) Use the result of part (a) to label the three sides of the triangle in the figure in terms of x.

(d) Solve the equation from part (c) for u.

53. Alternating Electric Current In the study of alternating electric current, instanta-neous voltage is modeled by

E = Emax sin 2pƒt,

where ƒ is the number of cycles per second, Emax is the maximum voltage, and t is time in seconds.

(a) Solve the equation for t.

(b) Find the least positive value of t if Emax = 12, E = 5, and ƒ = 100. Use a calcu-lator and round to two significant digits.

54. Viewing Angle of an Observer While visiting a museum, an observer views a painting that is 3 ft high and hangs 6 ft above the ground. See the figure. Assume her eyes are 5 ft above the ground, and let x be the distance from the spot where she is standing to the wall displaying the painting.

x ft

5 ft

6 ft

3 ft

(a) Show that u, the viewing angle subtended by the painting, is given by

u = tan-1 a 4xb - tan-1 a 1

xb .

(b) Find the value of x to the nearest hundredth for each value of u.

(i) u = p6

(ii) u = p8

55. Movement of an Arm In the equation below, t is time (in seconds) and y is the angle formed by a rhythmically moving arm.

y = 13

sin 4pt3

(a) Solve the equation for t.

(b) At what time, to the nearest hundredth of a second, does the arm first form an angle of 0.3 radian?

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289CHAPTER 6 Test Prep

Chapter 6 Test Prep

Key Terms

6.1 one-to-one function inverse function

New Symbols

ƒ−1 inverse of function ƒsin−1 x 1arcsin x 2 inverse sine of xcos−1 x 1arccos x 2 inverse cosine of xtan−1 x 1arctan x 2 inverse tangent of x cot−1 x 1arccot x 2 inverse cotangent of xsec−1 x 1arcsec x 2 inverse secant of xcsc−1 x 1arccsc x 2 inverse cosecant of xQuick ReviewConcepts Examples

Evaluate y = cos-1 0.

Write y = cos-1 0 as cos y = 0. Then

y = p2

because cos p2 = 0 and p2 is in the range of cos

-1 x.

Use a calculator to find y in radians if y = sec-1 1-32.With the calculator in radian mode, enter sec-1 1-32 as cos-1 A 1-3 B to obtain

y ≈ 1.9106332.

Evaluate sin A tan-1 A - 34 B B .Let u = tan-1 A - 34 B . Then tan u = - 34 . Because tan u is negative when u is in quadrant IV, sketch a triangle as shown.

–35

We want sin A tan-1 A - 34 B B = sin u. From the triangle, we have the following.

sin u = - 35

Inverse Circular Functions6.1

y = sin–1 x

–1 0 1

–

(1, )2p

(–1, – )2p x

y = cos–1 x

–1 0 1

(0, )2p

(–1, p)

(1, 0)

y = tan–1 x

–1–2 0 1 2

–

(1, )4p

(–1, – )4p

Range

Inverse Function

Domain

Interval

Quadrants of the Unit Circle

See the section for graphs of the other inverse circular (trigonometric) functions.

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290 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Concepts Examples

Solve tan u + 23 = 223 over the interval 30°, 360°2. tan u + 23 = 223 Original equation

tan u = 23 Subtract 23. u = 60° Definition of inverse tangent

Another solution over 30°, 360°2 isu = 60° + 180° = 240°.

The solution set is 560°, 240°6.Solve 2 cos2 x = 1 for all solutions.

2 cos2 x = 1 Original equation 2 cos2 x - 1 = 0 Subtract 1.

cos 2x = 0 Cosine double-angle identity

2x = p2

+ 2np and 2x = 3p2

+ 2npAdd integer multiples of 2p.

x = p4

+ np and x = 3p4

+ npDivide by 2.

The solution set, where p is the period of cos 2x, isEp4 + np, 3p4 + np, where n is any integer F .

Trigonometric Equations I

Trigonometric Equations II

Solving a Trigonometric Equation

1. Decide whether the equation is linear or quadratic in form in order to determine the solution method.

2. If only one trigonometric function is present, solve the equation for that function.

3. If more than one trigonometric function is present, rewrite the equation so that one side equals 0. Then try to factor and apply the zero-factor property.

4. If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval.

5. Try using identities to change the form of the equation. It may be helpful to square each side of the equation first. In this case, check for extraneous solutions.

6.2

6.3

Equations Involving Inverse Trigonometric Functions6.4

Solve y = 2 sin 3x for x, where x is restricted to the interval C - p6 , p6 D .

y = 2 sin 3x Original equation

2= sin 3x Divide by 2.

3x = arcsin y

2 Definition of arcsine

x = 13

arcsin y

2 Multiply by 13 .

Solve.

4 tan-1 x = p Original equation

tan-1 x = p4

Divide by 4.

x = tan p4

Definition of arctangent

x = 1 Evaluate.

The solution set is 516.

We solve equations of the form y = ƒ1x2, where ƒ1x2 involves a trigonometric function, using inverse trigono-metric functions.

Techniques introduced in this section also show how to solve equations that involve inverse functions.

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291

Chapter 6 Review Exercises 1. Graph the inverse sine, cosine, and tangent functions, indicating the coordinates of

three points on each graph. Give the domain and range for each.

Concept Check Determine whether each statement is true or false. If false, tell why.

2. The ranges of the inverse tangent and inverse cotangent functions are the same.

3. It is true that sin 11p6 = - 12 , and therefore arcsin A - 12 B = 11p6 .

4. For all x, tan1tan-1 x2 = x.Find the exact value of each real number y. Do not use a calculator.

5. y = sin-1 222

6. y = arccos a- 12b 7. y = tan-1 A -23 B

8. y = arcsin1-12 9. y = cos-1 a- 222b 10. y = arctan 23

11. y = sec-11-22 12. y = arccsc 2233

13. y = arccot1-12Give the degree measure of u. Do not use a calculator.

14. u = arccos 12

15. u = arcsin ¢- 232

≤ 16. u = tan-1 0Use a calculator to approximate each value in decimal degrees.

17. u = arctan 1.7804675 18. u = sin-11-0.660453202 19. u = cos-1 0.8039657720. u = cot-1 4.5046388 21. u = arcsec 3.4723155 22. u = csc-1 7.4890096

Evaluate each expression without using a calculator.

23. cos1arccos1-122 24. sin ¢arcsin ¢- 232

≤ ≤ 25. arccos acos 3p4b

26. arcsec1sec p2 27. tan-1 atan p4b 28. cos-11cos 02

29. sin aarccos 34b 30. cos1arctan 32 31. cos1csc-11-222

32. sec a2 sin-1 a- 13b b 33. tan aarcsin 3

5+ arccos 5

Write each trigonometric expression as an algebraic expression in u, for u 7 0.

34. cos ¢arctan u21 - u2 ≤ 35. tan ¢arcsec 2u2 + 1u ≤Solve each equation for exact solutions over the interval 30, 2p2 where appropriate. Give approximate solutions to four decimal places.

36. sin2 x = 1 37. 2 tan x - 1 = 0

38. 3 sin2 x - 5 sin x + 2 = 0 39. tan x = cot x

40. sec2 2x = 2 41. tan2 2x - 1 = 0

CHAPTER 6 Review Exercises

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292 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Solve each equation for all exact solutions, in radians.

42. sec x2

= cos x2

43. cos 2x + cos x = 0 44. 4 sin x cos x = 23Solve each equation for exact solutions over the interval 30° 360°2 where appropriate. Give approximate solutions to the nearest tenth of a degree.

45. sin2 u + 3 sin u + 2 = 0 46. 2 tan2 u = tan u + 1

47. sin 2u = cos 2u + 1 48. 2 sin 2u = 1

49. 3 cos2 u + 2 cos u - 1 = 0 50. 5 cot2 u - cot u - 2 = 0

Solve each equation for all exact solutions, in degrees.

51. 223 cos u2

= -3 52. sin u - cos 2u = 0 53. tan u - sec u = 1

Solve each equation for x.

54. 4p - 4 cot-1 x = p 55. 43

arctan x2

= p

56. arccos x = arcsin 27

57. arccos x + arctan 1 = 11p12

58. y = 3 cos x2

, for x in 30, 2p4 59. y = 12

sin x, for x in c - p2

, p

60. y = 45

sin x - 35

, for x in c - p2

, p

61. y = 12

tan13x + 22, for x in a- 23

- p6

, - 23

+ p6b

62. Solve d = 550 + 450 cos a p50

tb for t, where t is in the interval 30, 504.(Modeling) Solve each problem.

63. Viewing Angle of an Observer A 10-ft-wide chalkboard is situated 5 ft from the left wall of a classroom. See the figure. A student sitting next to the wall x feet from the front of the classroom has a viewing angle of u radians.

(a) Show that the value of u is given by

y1 = tan-1 a15x b - tan-1 a5x b .(b) Graph y1 with a graphing calculator to estimate the

value of x that maximizes the viewing angle.

64. Snell’s Law Snell’s law states that

c1c2

=sin u1sin u2

where c1 is the speed of light in one medium, c2 is the speed of light in a second medium, and u1 and u2 are the angles shown in the figure. Suppose a light is shining up through water into the air as in the figure. As u1 increases, u2 approaches 90°, at which point no light will emerge from the water. Assume the ratio c1c2 in this case is 0.752. For what value of u1, to the nearest tenth, does u2 = 90°? This value of u1 is the critical angle for water.

5 10

Air

Water

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293

65. Snell’s Law Refer to Exercise 64. What happens when u1 is greater than the critical angle?

66. British Nautical Mile The British nautical mile is defined as the length of a minute of arc of a meridian. Because Earth is flat at its poles, the nautical mile, in feet, is given by

L = 6077 - 31 cos 2u,

where u is the latitude in degrees. See the figure. (Source: Bushaw, D., et al., A Sourcebook of Applications of School Mathematics, National Council of Teachers of Mathematics.) Give answers to the nearest tenth if applicable.

(a) Find the latitude between 0° and 90° at which the nautical mile is 6074 ft.(b) At what latitude between 0° and 180° is the nautical mile 6108 ft?(c) In the United States, the nautical mile is defined everywhere as 6080.2 ft. At

what latitude between 0° and 90° does this agree with the British nautical mile?

A nautical mile isthe length on anyof the meridianscut by a centralangle of measure1 minute.

Chapter 6 Test

1. Graph y = sin-1 x, and indicate the coordinates of three points on the graph. Give the domain and range.

2. Find the exact value of each real number y. Do not use a calculator.

(a) y = arccos a- 12b (b) y = sin-1 ¢- 23

2≤

(a) u = arccos 232

(b) u = tan-11-12(c) u = cot-11-12 (d) u = csc-1 ¢- 223

3 ≤

4. Use a calculator to approximate each value in decimal degrees to the nearest hundredth.

(a) sin-1 0.69431882 (b) sec-1 1.0840880

(a) cos aarcsin 23b (b) sin a2 cos-1 1

6. Explain why sin-1 3 is not defined.

7. Explain why arcsin Asin 5p6 B ≠ 5p6 . 8. Write tan1arcsin u2 as an algebraic expression in u, for u 7 0.Solve each equation for exact solutions over the interval 30°, 360°2 where appropriate. Give approximate solutions to the nearest tenth of a degree.

9. -3 sec u + 223 = 0 10. sin2 u = cos2 u + 1 11. csc2 u - 2 cot u = 4

CHAPTER 6 Test

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294 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations

Solve each equation for exact solutions over the interval 30, 2p2 where appropriate. Give approximate solutions to four decimal places.

12. cos x = cos 2x 13. 22 cos 3x - 1 = 0 14. sin x cos x = 13

Solve each equation for all exact solutions in radians (for x) or in degrees (for u). Write answers using the least possible nonnegative angle measures.

15. sin2 u = -cos 2u 16. 223 sin x2

= 3 17. csc x - cot x = 1

Work each problem.

18. Solve each equation for x, where x is restricted to the given interval.

(a) y = cos 3x, for x in c 0, p3d (b) y = 4 + 3 cot x, for x in 10, p2

19. Solve each equation for exact solutions.

(a) arcsin x = arctan 43

(b) arccot x + 2 arcsin 232

= p

20. Upper Harmonics Pressures Suppose that the E key above middle C is played on a piano, and its fundamental frequency is ƒ1 = 330 Hz. Its associated pressure is expressed as

P1 = 0.002 sin 660 pt.

The pressures associated with the next four frequencies are P2 =0.002

2 sin 1320pt,

P3 =0.002

3 sin 1980pt, P4 =0.002

4 sin 2640pt, and P5 =0.002

5 sin 3300pt. Duplicate the graph shown below of

P = P1 + P2 + P3 + P4 + P5.

Approximate the maximum value of P to four significant digits and the least positive value of t for which P reaches this maximum.

−0.005

0.005

0 0.0062

For x = t,y6 = P1 + P2 + P3 + P4 + P5 = P

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295

Surveyors use a method known as triangulation to measure distances when direct measurements cannot be made due to obstructions in the line of sight.

Oblique Triangles and the Law of Sines

The Ambiguous Case of the Law of Sines

The Law of Cosines

Chapter 7 Quiz

Geometrically Defined Vectors and Applications

Algebraically Defined Vectors and the Dot Product

Summary Exercises on Applications of Trigonometry and Vectors

7.1

7.2

7.3

7.4

7.5

Applications of Trigonometry and Vectors7

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296 CHAPTER 7 Applications of Trigonometry and Vectors

7.1 Oblique Triangles and the Law of Sines

Congruency and Oblique Triangles We now turn our attention to solving triangles that are not right triangles. To do this we develop new relationships, or laws, that exist between the sides and angles of any triangle. The congruence axioms assist in this process. Recall that two triangles are congruent if their corresponding sides and angles are equal.

■ Congruency and Oblique Triangles

■ Derivation of the Law of Sines

■ Solutions of SAA and ASA Triangles (Case 1)

■ Area of a Triangle

Congruence Axioms

Side-Angle-Side (SAS)

If two sides and the included angle of one triangle are equal, respectively, to two sides and the included angle of a second triangle, then the triangles are congruent.

Angle-Side-Angle (ASA)

If two angles and the included side of one triangle are equal, respectively, to two angles and the included side of a second triangle, then the triangles are congruent.

Side-Side-Side (SSS)

If three sides of one triangle are equal, respectively, to three sides of a second triangle, then the triangles are congruent.

Examples of congruenttriangles ABC and XYZ

xY Z

If a side and any two angles are given (SAA), the third angle can be determined by the angle sum formula

A + B + C = 180°.

Then the ASA axiom can be applied. Whenever SAS, ASA, or SSS is given, the triangle is unique.

A triangle that is not a right triangle is an oblique triangle. Recall that a triangle can be solved—that is, the measures of the three sides and three angles can be found—if at least one side and any other two measures are known.

NOTE If we know three angles of a triangle, we cannot find unique side lengths because AAA assures us only of similarity, not congruence. For example, there are infinitely many triangles ABC of different sizes with A = 35°, B = 65°, and C = 80°.

Data Required for Solving Oblique Triangles

There are four possible cases.

Case 1 One side and two angles are known (SAA or ASA).

Case 2 Two sides and one angle not included between the two sides are known (SSA). This case may lead to more than one triangle.

Case 3 Two sides and the angle included between the two sides are known (SAS).

Case 4 Three sides are known (SSS).

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2977.1 Oblique Triangles and the Law of Sines

Derivation of the Law of Sines To derive the law of sines, we start with an oblique triangle, such as the acute triangle in Figure 1(a) or the obtuse triangle in Figure 1(b). This discussion applies to both triangles. First, construct the per-pendicular from B to side AC (or its extension). Let h be the length of this perpen-dicular. Then c is the hypotenuse of right triangle ADB, and a is the hypotenuse of right triangle BDC.

In triangle ADB, sin A = hc

, or h = c sin A.

In triangle BDC, sin C = ha

, or h = a sin C.

Because h = c sin A and h = a sin C, we set these two expressions equal.

a sin C = c sin A

sin A=

csin C

Divide each side by sin A sin C.

In a similar way, by constructing perpendicular lines from the other vertices, we can show that these two equations are also true.

asin A

sin B and

bsin B

sin C

This discussion proves the following theorem.

Solving a triangle with given information matching Case 1 or Case 2 requires using the law of sines, while solving a triangle with given information matching Case 3 or Case 4 requires using the law of cosines.

A C

hc a

Acute triangle ABC

(a)

(b)

h c a

A bD

Obtuse triangle ABC

Figure 1

We label oblique triangles as we did right triangles: side a opposite angle A, side b opposite angle B, and side c opposite angle C.

Law of Sines

In any triangle ABC, with sides a, b, and c, the following hold.

asin A

sin B ,

asin A

sin C , and

bsin B

sin C

This can be written in compact form as follows.

asin A

sin B=

csin C

That is, according to the law of sines, the lengths of the sides in a triangle are proportional to the sines of the measures of the angles opposite them.

In practice we can also use an alternative form of the law of sines.

sin Aa

=sin B

sin Cc

Alternative form of the law of sines

NOTE When using the law of sines, a good strategy is to select a form that has the unknown variable in the numerator and where all other variables are known. This makes computation easier.

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298 CHAPTER 7 Applications of Trigonometry and Vectors

Solutions of SAA and ASA Triangles (Case 1)

EXAMPLE 1 Applying the Law of Sines (SAA)

Solve triangle ABC if A = 32.0°, B = 81.8°, and a = 42.9 cm.

SOLUTION Start by drawing a triangle, roughly to scale, and labeling the given parts as in Figure 2. The values of A, B, and a are known, so use the form of the law of sines that involves these variables, and then solve for b.

sin A= b

sin B

Choose a form of the law of sines that has the unknown variable in the numerator.

42.9

sin 32.0°= b

sin 81.8° Substitute the given values.

b = 42.9 sin 81.8°sin 32.0°

Multiply by sin 81.8° and rewrite.

b ≈ 80.1 cm Approximate with a calculator.

To find C, use the fact that the sum of the angles of any triangle is 180°.

A + B + C = 180° Angle sum formula C = 180° - A - B Solve for C. C = 180° - 32.0° - 81.8° Substitute. C = 66.2° Subtract.

Now use the law of sines to find c. (The Pythagorean theorem does not apply because this is not a right triangle.)

sin A= c

sin C Law of sines

42.9

sin 32.0°= c

sin 66.2° Substitute known values.

c = 42.9 sin 66.2°sin 32.0°

Multiply by sin 66.2° and rewrite.

c ≈ 74.1 cm Approximate with a calculator.

■✔ Now Try Exercise 17.

32.0° 81.8°42.9 cm

Figure 2

Be sure to label a sketch carefully to help set up

the correct equation.

EXAMPLE 2 Applying the Law of Sines (ASA)

An engineer wishes to measure the distance across a river. See Figure 3. He determines that C = 112.90°, A = 31.10°, and b = 347.6 ft. Find the distance a.

SOLUTION To use the law of sines, one side and the angle opposite it must be known. Here b is the only side whose length is given, so angle B must be found before the law of sines can be used.

B = 180° - A - C Angle sum formula, solved for B B = 180° - 31.10° - 112.90° Substitute the given values. B = 36.00° Subtract.

b = 347.6 ft

31.10°

a CB

112.90°

Figure 3

CAUTION Whenever possible, use given values in solving triangles, rather than values obtained in intermediate steps, to avoid rounding errors.

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2997.1 Oblique Triangles and the Law of Sines

Solve for a.

Now use the form of the law of sines involving A, B, and b to find side a.

sin A= b

sin B Law of sines

sin 31.10°= 347.6

sin 36.00° Substitute known values.

a = 347.6 sin 31.10°sin 36.00°

Multiply by sin 31.10°.

a ≈ 305.5 ft Use a calculator.

■✔ Now Try Exercise 33.

Recall that bearing is used in navigation to refer to direction of motion or direction of a distant object relative to current course. We consider two methods for expressing bearing.

Method 1 When a single angle is given, such as 220°, this bearing is measured in a clockwise direction from north.

Method 2 Start with a north-south line and use an acute angle to show direction, either east or west, from this line.

Example: 220°

220°

Example: S 40° W

40°

EXAMPLE 3 Applying the Law of Sines (ASA)

Two ranger stations are on an east-west line 110 mi apart. A forest fire is located on a bearing of N 42° E from the western station at A and a bearing of N 15° E from the eastern station at B. To the nearest ten miles, how far is the fire from the western station?

SOLUTION Figure 4 shows the two ranger stations at points A and B and the fire at point C. Angle BAC measures 90° - 42° = 48°, obtuse angle B measures 90° + 15° = 105°, and the third angle, C, measures 180° - 105° - 48° = 27°. We use the law of sines to find side b.

sin B= c

sin C Law of sines

sin 105°= 110

sin 27° Substitute known values.

b = 110 sin 105°sin 27°

Multiply by sin 105°.

b ≈ 230 mi Use a calculator and give two significant digits.A

42° 15°

B110 mi

Figure 4

Solve for b.

■✔ Now Try Exercise 35.

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300 CHAPTER 7 Applications of Trigonometry and Vectors

Area of a Triangle A familiar formula for the area of a triangle is

& =12

bh, where & represents area, b base, and h height.

This formula cannot always be used easily because in practice, h is often unknown. To find another formula, refer to acute triangle ABC in Figure 5(a) or obtuse triangle ABC in Figure 5(b).

A perpendicular has been drawn from B to the base of the triangle (or the extension of the base). Consider right triangle ADB in either figure.

sin A = hc

, or h = c sin A

Substitute into the formula for the area of a triangle.

& = 12

bh = 12

bc sin A

Any other pair of sides and the angle between them could have been used.

Area of a Triangle (SAS)

In any triangle ABC, the area & is given by the following formulas.

& =12

bc sin A, & =12

ab sin C, and & =12

ac sin B

That is, the area is half the product of the lengths of two sides and the sine of the angle included between them.

EXAMPLE 4 Finding the Area of a Triangle (SAS)

Find the area of triangle ABC in Figure 6.

SOLUTION Substitute B = 55° 10′, a = 34.0 ft, and c = 42.0 ft into the area formula.

& = 12

ac sin B = 12

134.02142.02 sin 55° 10′ ≈ 586 ft2■✔ Now Try Exercise 51.

42.0 ft

34.0 ft

55° 10′

Figure 6

b = 27.3 cm

c a

C24° 40! 52° 40!

Figure 7

180° = A + B + C Angle sum formula B = 180° - 24° 40′ - 52° 40′ Substitute and solve for B. B = 102° 40′ Subtract.

EXAMPLE 5 Finding the Area of a Triangle (ASA)

Find the area of triangle ABC in Figure 7.

SOLUTION Before the area formula can be used, we must find side a or c.

First find remaining

angle B.

A C

hc a

Acute triangle ABC

(a)

h c a

A bD

Obtuse triangle ABC

(b)

Figure 5

NOTE If the included angle measures 90°, its sine is 1 and the formula becomes the familiar & = 12 bh.

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3017.1 Oblique Triangles and the Law of Sines

Solve for a.

Now that we know two sides, a and b, and their included angle C, we find the area.

& = 12

ab sin C ≈12

111.72127.32 sin 52° 40′ ≈ 127 cm211.7 is an approximation. In practice,

use the calculator value.

■✔ Now Try Exercise 57.

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. A triangle that is not a right triangle is a(n) triangle.

2. The measures of the three sides and three angles of a triangle can be found if at least one and any other two measures are known.

3. If we know three of a triangle, we cannot find a unique solution for the triangle.

4. In the law of sines, a

sin A= b =

c .

5. An alternative form of the law of sines is sin A =

sin B =

sin C .

6. For any triangle ABC, its area can be found using the formula & = 12 ab .

7.1 Exercises

CONCEPT PREVIEW Consider each case and determine whether there is sufficient information to solve the triangle using the law of sines.

7. Two angles and the side included between them are known.

8. Two angles and a side opposite one of them are known.

9. Two sides and the angle included between them are known.

10. Three sides are known.

Find the length of each side labeled a. Do not use a calculator.

11.

A60° 75°

12.

A45°

105°10

Next use the law of sines to find side a.

sin A= b

sin B Law of sines

sin 24° 40′= 27.3

sin 102° 40′ Substitute known values.

a = 27.3 sin 24° 40′sin 102° 40′

Multiply by sin 24° 40′.

a ≈ 11.7 cm Use a calculator.

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302 CHAPTER 7 Applications of Trigonometry and Vectors

Determine the remaining sides and angles of each triangle ABC. See Example 1.

13.

A37°

48°18 m

14.

B52°

29°43 cm

15.

B115.5°

27.2°76.0 ft

16.

B 18.7° 124.1°94.6 m

17. A = 68.41°, B = 54.23°, 18. C = 74.08°, B = 69.38°, a = 12.75 ft c = 45.38 m

19. A = 87.2°, b = 75.9 yd, 20. B = 38° 40′, a = 19.7 cm, C = 74.3° C = 91° 40′

21. B = 20° 50′, C = 103° 10′, 22. A = 35.3°, B = 52.8°, AC = 132 ft AC = 675 ft

23. A = 39.70°, C = 30.35°, 24. C = 71.83°, B = 42.57°, b = 39.74 m a = 2.614 cm

25. B = 42.88°, C = 102.40°, 26. C = 50.15°, A = 106.1°, b = 3974 ft c = 3726 yd

27. A = 39° 54′, a = 268.7 m, 28. C = 79° 18′, c = 39.81 mm, B = 42° 32′ A = 32° 57′

Concept Check Answer each question.

29. Why can the law of sines not be used to solve a triangle if we are given only the lengths of the three sides of the triangle?

30. In Example 1, we begin (as seen there) by solving for b and C. Why is it a better idea to solve for c by using a and sin A than by using b and sin B?

31. Eli Maor, a perceptive trigonometry student, makes this statement: “If we know any two angles and one side of a triangle, then the triangle is uniquely determined.” Why is this true? Refer to the congruence axioms given in this section.

32. In a triangle, if a is twice as long as b, is A necessarily twice as large as B?

Solve each problem. See Examples 2 and 3.

33. Distance across a River To find the distance AB across a river, a surveyor laid off a dis-tance BC = 354 m on one side of the river. It is found that B = 112° 10′ and C = 15° 20′. Find AB. See the figure.

34. Distance across a Canyon To determine the distance RS across a deep canyon, Rhonda lays off a distance TR = 582 yd. She then finds that T = 32° 50′ and R = 102° 20′. Find RS. See the figure.

B C354 m

112° 10!15° 20!

32° 50!

102° 20!

582 yd

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3037.1 Oblique Triangles and the Law of Sines

35. Distance a Ship Travels A ship is sailing due north. At a certain point the bearing of a lighthouse 12.5 km away is N 38.8° E. Later on, the captain notices that the bearing of the lighthouse has become S 44.2° E. How far did the ship travel between the two observations of the lighthouse?

36. Distance between Radio Direction Finders Radio direction finders are placed at points A and B, which are 3.46 mi apart on an east-west line, with A west of B. From A the bearing of a certain radio transmitter is 47.7°, and from B the bearing is 302.5°. Find the distance of the transmitter from A.

37. Distance between a Ship and a Lighthouse The bearing of a lighthouse from a ship was found to be N 37° E. After the ship sailed 2.5 mi due south, the new bearing was N 25° E. Find the distance between the ship and the lighthouse at each location.

38. Distance across a River Standing on one bank of a river flowing north, Mark notices a tree on the opposite bank at a bearing of 115.45°. Lisa is on the same bank as Mark, but 428.3 m away. She notices that the bearing of the tree is 45.47°. The two banks are parallel. What is the distance across the river?

39. Height of a Balloon A balloonist is directly above a straight road 1.5 mi long that joins two villages. She finds that the town closer to her is at an angle of depression of 35°, and the farther town is at an angle of depression of 31°. How high above the ground is the balloon?

40. Measurement of a Folding Chair A folding chair is to have a seat 12.0 in. deep with angles as shown in the figure. How far down from the seat should the crossing legs be joined? (Find length x in the figure.)

31°35°

1.5 mi

LITTLETON SMALLVILLE

NOT TO SCALE

12.0 in.

54.8°70.4° 17.0 in.

41. Angle Formed by Radii of Gears Three gears are arranged as shown in the figure. Find angle u.

38°

3.6

2.7

1.6

42. Distance between Atoms Three atoms with atomic radii of 2.0, 3.0, and 4.5 are arranged as in the figure. Find the distance between the centers of atoms A and C.

3.0

18°

4.5A

C2.0

43. Distance to the Moon The moon is a relatively close celestial object, so its distance can be measured directly by taking two different photographs at precisely the same time from two different locations. The moon will have a different angle of eleva-tion at each location. On April 29, 1976, at 11:35 a.m., the lunar angles of elevation during a partial solar eclipse at Bochum in upper Germany and at Donaueschingen in lower Germany were measured as 52.6997° and 52.7430°, respectively. The two cities are 398 km apart.

Calculate the distance to the moon, to the nearest thousand kilometers, from Bochum on this day, and compare it with the actual value of 406,000 km. Disregard the curvature of Earth in this calculation. (Source: Scholosser, W., T. Schmidt-Kaler, and E. Milone, Challenges of Astronomy, Springer-Verlag.)

Bochum

Moon

Donaueschingen

NOT TO SCALE

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304 CHAPTER 7 Applications of Trigonometry and Vectors

44. Ground Distances Measured by Aerial Photography The distance covered by an aerial photograph is determined by both the focal length of the camera and the tilt of the camera from the perpendicular to the ground. A camera lens with a 12-in. focal length will have an angular cover-age of 60°. If an aerial photograph is taken with this camera tilted u = 35° at an altitude of 5000 ft, calculate to the nearest foot the ground distance d that will be shown in this photograph. (Source: Brooks, R. and D. Johannes, Phytoarchaeology, Dioscorides Press.)

45. Ground Distances Measured by Aerial Photog-raphy Refer to Exercise 44. A camera lens with a 6-in. focal length has an angular coverage of 86°. Suppose an aerial photograph is taken vertically with no tilt at an altitude of 3500 ft over ground with an increasing slope of 5°, as shown in the figure. Calculate the ground distance CB, to the nearest hundred feet, that will appear in the resulting pho-tograph. (Source: Moffitt, F. and E. Mikhail, Pho-togrammetry, Third Edition, Harper & Row.)

46. Ground Distances Measured by Aerial Photography Repeat Exercise 45 if the camera lens has an 8.25-in. focal length with an angular coverage of 72°.

35°

60°

86°

5° BC

Find the area of each triangle ABC. See Examples 4 and 5.

51. A = 42.5°, b = 13.6 m, c = 10.1 m 52. C = 72.2°, b = 43.8 ft, a = 35.1 ft

53. B = 124.5°, a = 30.4 cm, c = 28.4 cm 54. C = 142.7°, a = 21.9 km, b = 24.6 km

55. A = 56.80°, b = 32.67 in., c = 52.89 in. 56. A = 34.97°, b = 35.29 m, c = 28.67 m

57. A = 30.50°, b = 13.00 cm, C = 112.60° 58. A = 59.80°, b = 15.00 m, C = 53.10°

Find the area of each triangle using the formula & = 12 bh, and then verify that the for-mula & = 12 ab sin C gives the same result.

47.

A60°

48.

A 60° 60°

2 2

49.

C 45°

50. A

C B45°

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3057.1 Oblique Triangles and the Law of Sines

Solve each problem.

59. Area of a Metal Plate A painter is going to apply a special coating to a triangular metal plate on a new building. Two sides measure 16.1 m and 15.2 m. She knows that the angle between these sides is 125°. What is the area of the surface she plans to cover with the coating?

60. Area of a Triangular Lot A real estate agent wants to find the area of a triangular lot. A surveyor takes measurements and finds that two sides are 52.1 m and 21.3 m, and the angle between them is 42.2°. What is the area of the triangular lot?

61. Triangle Inscribed in a Circle For a triangle inscribed in a circle of radius r, the law of sines ratios

asin A

, b

sin B , and

csin C

have value 2r.

The circle in the figure has diameter 1. What are the values of a, b, and c? (Note: This result provides an alternative way to define the sine function for angles between 0° and 180°. It was used nearly 2000 yr ago by the mathematician Ptolemy to construct one of the earliest trigonometric tables.)

62. Theorem of Ptolemy The following theorem is also attributed to Ptolemy:

In a quadrilateral inscribed in a circle, the product of the diagonals is equal to the sum of the products of the opposite sides.

(Source: Eves, H., An Introduction to the History of Math-ematics, Sixth Edition, Saunders College Publishing.) The circle in the figure has diameter 1. Use Ptolemy’s theorem to derive the formula for the sine of the sum of two angles.

63. Law of Sines Several of the exercises on right triangle applications involved a figure similar to the one shown here, in which angles a and b and the length of line segment AB are known, and the length of side CD is to be determined. Use the law of sines to obtain x in terms of a, b, and d.

64. Aerial Photography Aerial photographs can be used to provide coordinates of ordered pairs to determine distances on the ground. Suppose we assign coordinates as shown in the figure. If an object’s photographic coordinates are 1x, y2, then its ground coordinates 1X, Y2 in feet can be com-puted using the following formulas.

X =1a - h2x

ƒ sec u - y sin u , Y =1a - h2y cos u

ƒ sec u - y sin u

Here, ƒ is focal length of the camera in inches, a is altitude in feet of the airplane, and h is elevation in feet of the object. Suppose that a house has photographic coordinates 1xH, yH2 = 10.9, 3.52 with elevation 150 ft, and a nearby forest fire has photographic coordinates 1xF, yF2 = 12.1, -2.42 and is at elevation 690 ft. Also suppose the photograph was taken at 7400 ft by a camera with focal length 6 in. and tilt angle u = 4.1°. (Source: Moffitt, F. and E. Mikhail, Photogrammetry, Third Edition, Harper & Row.)

(a) Use the formulas to find the ground coordinates of the house and the fire to the nearest tenth of a foot.

(b) Use the distance formula d = 21x2 - x122 + 1y2 - y122 to find the distance on the ground between the house and the fire to the nearest tenth of a foot.

ABc

os A

sin A

sin B

cos

sin (A + B)

Aα β

(xF, yF)(xH, yH)

(XF, YF)(XH, YH)

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306 CHAPTER 7 Applications of Trigonometry and Vectors

7.2 The Ambiguous Case of the Law of Sines

Description of the Ambiguous Case We have used the law of sines to solve triangles involving Case 1, given SAA or ASA. If we are given the lengths of two sides and the angle opposite one of them (Case 2, SSA), then zero, one, or two such triangles may exist. (There is no SSA congruence axiom.)

Suppose we know the measure of acute angle A of triangle ABC, the length of side a, and the length of side b, as shown in Figure 8. We must draw the side of length a opposite angle A. The table shows possible outcomes. This situation (SSA) is called the ambiguous case of the law of sines.

■ Description of the Ambiguous Case

■ Solutions of SSA Triangles (Case 2)

■ Analyzing Data for Possible Number of Triangles

B lies along this sideif a triangle exists.

Figure 8

Possible Outcomes for Applying the Law of Sines

Angle A is

Possible Number of Triangles

Sketch

Applying Law of Sines

Leads to

Acute 0

h sin B 7 1, a 6 h 6 b

Acute 1

a = h sin B = 1, a = h and h 6 b

Acute 1

b a

h 0 6 sin B 6 1, a Ú b

Acute 2A

b a

B1B2

a h0 6 sin B1 6 1, h 6 a 6 b,A + B2 6 180°

Obtuse 0

sin B Ú 1, a … b

Obtuse 1a

0 6 sin B 6 1, a 7 b

The following basic facts help determine which situation applies.

Applying the Law of Sines

1. For any angle u of a triangle, 0 6 sin u … 1. If sin u = 1, then u = 90° and the triangle is a right triangle.

2. sin u = sin1180° - u2 (Supplementary angles have the same sine value.)3. The smallest angle is opposite the shortest side, the largest angle is opposite

the longest side, and the middle-valued angle is opposite the intermediate side (assuming the triangle has sides that are all of different lengths).

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3077.2 The Ambiguous Case of the Law of Sines

Solutions of SSA Triangles (Case 2)

EXAMPLE 1 Solving the Ambiguous Case (No Such Triangle)

Solve triangle ABC if B = 55° 40′, b = 8.94 m, and a = 25.1 m.

SOLUTION We are given B, b, and a. We use the law of sines to find angle A.

sin A

a= sin B

b Law of sines (alternative form)

sin A25.1

= sin 55° 40′8.94

Substitute the given values.

sin A = 25.1 sin 55° 40′8.94

Multiply by 25.1.

sin A ≈ 2.3184379 Use a calculator.

Choose a form that has the

unknown variable in the numerator.

Because sin A cannot be greater than 1, there can be no such angle A—and thus no triangle with the given information. An attempt to sketch such a triangle leads to the situation shown in Figure 9.

■✔ Now Try Exercise 17.

55° 40!

b = 8.94

a = 25.1

Figure 9

EXAMPLE 2 Solving the Ambiguous Case (Two Triangles)

Solve triangle ABC if A = 55.3°, a = 22.8 ft, and b = 24.9 ft.

SOLUTION To begin, use the law of sines to find angle B.

sin A

a= sin B

sin 55.3°

22.8= sin B

24.9 Substitute the given values.

sin B = 24.9 sin 55.3°22.8

Multiply by 24.9 and rewrite.

sin B ≈ 0.8978678 Use a calculator.

Solve for sin B.

There are two angles B between 0° and 180° that satisfy this condition. Because sin B ≈ 0.8978678, one value of angle B, to the nearest tenth, is

B1 = 63.9°. Use the inverse sine function.

Supplementary angles have the same sine value, so another possible value of B is

B2 = 180° - 63.9° = 116.1°.

To see whether B2 = 116.1° is a valid possibility, add 116.1° to the measure of A, 55.3°. Because 116.1° + 55.3° = 171.4°, and this sum is less than 180°, it is a valid angle measure for this triangle.

NOTE In the ambiguous case, we are given two sides and an angle opposite one of the sides (SSA). For example, suppose b, c, and angle C are given. This situation represents the ambiguous case because angle C is opposite side c.

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308 CHAPTER 7 Applications of Trigonometry and Vectors

Now separately solve triangles AB1C1 and AB2C2 shown in Figure 10. Begin with AB1C1. Find angle C1 first.

C1 = 180° - A - B1 Angle sum formula, solved for C1 C1 = 180° - 55.3° - 63.9° Substitute. C1 = 60.8° Subtract.

Now, use the law of sines to find side c1.

sin A=

c1sin C1

22.8

sin 55.3°=

c1sin 60.8°

Substitute.

c1 =22.8 sin 60.8°

sin 55.3° Multiply by sin 60.8° and rewrite.

c1 ≈ 24.2 ft Use a calculator.

55.3°B1A c1

b = 24.9 a = 22.8

63.9°

a = 22.8b = 24.9

55.3° 116.1°

Figure 10

Solve for c1.

To solve triangle AB2C2, first find angle C2.

C2 = 180° - A - B2 Angle sum formula, solved for C2 C2 = 180° - 55.3° - 116.1° Substitute. C2 = 8.6° Subtract.

Use the law of sines to find side c2.

sin A=

c2sin C2

22.8

sin 55.3°=

c2sin 8.6°

Substitute.

c2 =22.8 sin 8.6°

sin 55.3° Multiply by sin 8.6° and rewrite.

c2 ≈ 4.15 ft Use a calculator.

Solve for c2 .

■✔ Now Try Exercise 25.

The ambiguous case results in zero, one, or two triangles. The following guide-lines can be used to determine how many triangles there are.

Number of Triangles Satisfying the Ambiguous Case (SSA)

Let sides a and b and angle A be given in triangle ABC. (The law of sines can be used to calculate the value of sin B.)

1. If applying the law of sines results in an equation having sin B 7 1, then no triangle satisfies the given conditions.

2. If sin B = 1, then one triangle satisfies the given conditions and B = 90°.3. If 0 6 sin B 6 1, then either one or two triangles satisfy the given

conditions.

(a) If sin B = k, then let B1 = sin-1 k and use B1 for B in the first triangle. (b) Let B2 = 180° - B1. If A + B2 6 180°, then a second triangle exists.

In this case, use B2 for B in the second triangle.

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3097.2 The Ambiguous Case of the Law of Sines

EXAMPLE 3 Solving the Ambiguous Case (One Triangle)

Solve triangle ABC, given A = 43.5°, a = 10.7 in., and c = 7.2 in.

SOLUTION Find angle C.

sin C

c= sin A

a Law of sines (alternative form)

sin C7.2

= sin 43.5°10.7

Substitute the given values.

sin C = 7.2 sin 43.5°10.7

Multiply by 7.2.

sin C ≈ 0.46319186 Use a calculator.

C ≈ 27.6° Use the inverse sine function.

There is another angle C that has sine value 0.46319186. It is

C = 180° - 27.6° = 152.4°.

However, notice in the given information that c 6 a, meaning that in the tri-angle, angle C must have measure less than angle A. Notice also that when we add this obtuse value to the given angle A = 43.5°, we obtain

152.4° + 43.5° = 195.9°,

which is greater than 180°. Thus either of these approaches shows that there can be only one triangle. See Figure 11. The measure of angle B can be found next.

B = 180° - 27.6° - 43.5° Substitute. B = 108.9° Subtract.

We can find side b with the law of sines.

sin B= a

sin A Law of sines

sin 108.9°= 10.7

sin 43.5° Substitute known values.

b = 10.7 sin 108.9°sin 43.5°

Multiply by sin 108.9°.

b ≈ 14.7 in. Use a calculator.

■✔ Now Try Exercise 21.

7.2 in. 10.7 in.

43.5° 27.6°

Figure 11

Analyzing Data for Possible Number of Triangles

EXAMPLE 4 Analyzing Data Involving an Obtuse Angle

Without using the law of sines, explain why A = 104°, a = 26.8 m, and b = 31.3 m cannot be valid for a triangle ABC.

SOLUTION Because A is an obtuse angle, it is the largest angle, and so the longest side of the triangle must be a. However, we are given b 7 a.

Thus, B 7 A, which is impossible if A is obtuse.

Therefore, no such triangle ABC exists. ■✔ Now Try Exercise 33.

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310 CHAPTER 7 Applications of Trigonometry and Vectors

1. CONCEPT PREVIEW Which one of the following sets of data does not determine a unique triangle?

A. A = 50°, b = 21, a = 19 B. A = 45°, b = 10, a = 12C. A = 130°, b = 4, a = 7 D. A = 30°, b = 8, a = 4

2. CONCEPT PREVIEW Which one of the following sets of data determines a unique triangle?

A. A = 50°, B = 50°, C = 80° B. a = 3, b = 5, c = 20C. A = 40°, B = 20°, C = 30° D. a = 7, b = 24, c = 25

7.2 Exercises

CONCEPT PREVIEW In each figure, a line segment of length L is to be drawn from the given point to the positive x-axis in order to form a triangle. For what value(s) of L can we draw the following?

(a) two triangles (b) exactly one triangle (c) no triangle

(3, 4)

y 4.

(–3, 4)

CONCEPT PREVIEW Determine the number of triangles ABC possible with the given parts.

5. a = 50, b = 26, A = 95° 6. a = 35, b = 30, A = 40°

7. a = 31, b = 26, B = 48° 8. B = 54°, c = 28, b = 23

9. c = 50, b = 61, C = 58° 10. c = 60, a = 82, C = 100°

Find each angle B. Do not use a calculator.

11. C

A B60°

2 6

12. C

A B45°

3 23

Find the unknown angles in triangle ABC for each triangle that exists. See Examples 1–3.

13. A = 29.7°, b = 41.5 ft, a = 27.2 ft

14. B = 48.2°, a = 890 cm, b = 697 cm

15. C = 41° 20′, b = 25.9 m, c = 38.4 m

16. B = 48° 50′, a = 3850 in., b = 4730 in.

17. B = 74.3°, a = 859 m, b = 783 m

18. C = 82.2°, a = 10.9 km, c = 7.62 km

19. A = 142.13°, b = 5.432 ft, a = 7.297 ft

20. B = 113.72°, a = 189.6 yd, b = 243.8 yd

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3117.2 The Ambiguous Case of the Law of Sines

Solve each triangle ABC that exists. See Examples 1–3.

21. A = 42.5°, a = 15.6 ft, b = 8.14 ft 22. C = 52.3°, a = 32.5 yd, c = 59.8 yd

23. B = 72.2°, b = 78.3 m, c = 145 m 24. C = 68.5°, c = 258 cm, b = 386 cm

25. A = 38° 40′, a = 9.72 m, b = 11.8 m

26. C = 29° 50′, a = 8.61 m, c = 5.21 m

27. A = 96.80°, b = 3.589 ft, a = 5.818 ft

28. C = 88.70°, b = 56.87 m, c = 112.4 m

29. B = 39.68°, a = 29.81 m, b = 23.76 m

30. A = 51.20°, c = 7986 cm, a = 7208 cm

Concept Check Answer each question.

31. Apply the law of sines to the following: a = 25, c = 225, A = 30°. What is the value of sin C ? What is the measure of C ? Based on its angle measures, what kind of triangle is triangle ABC ?

32. What condition must exist to determine that there is no triangle satisfying the given values of a, b, and B, once the value of sin A is found by applying the law of sines?

33. Without using the law of sines, why can no triangle ABC exist that satisfies A = 103° 20′, a = 14.6 ft, b = 20.4 ft?

34. If the law of sines is applied to the data given in Example 4, what happens when we try to find the measure of angle B using a calculator?

Use the law of sines to solve each problem.

35. Distance between Inaccessible Points To find the distance between a point X and an inacces-sible point Z, a line segment XY is constructed. It is found that XY = 960 m, angle XYZ = 43° 30′, and angle YZX = 95° 30′. Find the distance be-tween X and Z to the nearest meter.

36. Height of an Antenna Tower The angle of elevation from the top of a building 45.0 ft high to the top of a nearby antenna tower is 15° 20′. From the base of the building, the angle of elevation of the tower is 29° 30′. Find the height of the tower.

37. Height of a Building A flagpole 95.0 ft tall is on the top of a building. From a point on level ground, the angle of elevation of the top of the flagpole is 35.0°, and the angle of elevation of the bottom of the flagpole is 26.0°. Find the height of the building.

38. Flight Path of a Plane A pilot flies her plane on a bearing of 35° 00′ from point X to point Y, which is 400 mi from X. Then she turns and flies on a bearing of 145° 00′ to point Z, which is 400 mi from her starting point X. What is the bearing of Z from X, and what is the distance YZ?

Use the law of sines to prove that each statement is true for any triangle ABC, with cor-responding sides a, b, and c.

39. a + b

b= sin A + sin B

sin B 40.

a - ba + b =

sin A - sin Bsin A + sin B

95° 30′

43° 30′

960 m

45.0 ft

29° 30′

15° 20′

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312 CHAPTER 7 Applications of Trigonometry and Vectors

Relating Concepts

For individual or collaborative investigation. (Exercises 41—44)

Colors of the U.S. Flag The flag of the United States includes the colors red, white, and blue.

Which color is predominant?

Clearly the answer is either red or white. (It can be shown that only 18.73% of the total area is blue.) (Source: Banks, R., Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics, Princeton University Press.)

To answer this question, work Exercises 41– 44 in order.

41. Let R denote the radius of the circumscribing circle of a five-pointed star ap-pearing on the American flag. The star can be decomposed into ten congruent triangles. In the figure, r is the radius of the circumscribing circle of the penta-gon in the interior of the star. Show that the area of a star is

! = c 5 sin A sin Bsin1A + B2 dR2 . (Hint: sin C = sin3180° - 1A + B24 = sin1A + B2.)

42. Angles A and B have values 18° and 36°, respectively. Express the area ! of a star in terms of its radius, R.

43. To determine whether red or white is predominant, we must know the measure-ments of the flag. Consider a flag of width 10 in., length 19 in., length of each upper stripe 11.4 in., and radius R of the circumscribing circle of each star 0.308 in. The thirteen stripes consist of six matching pairs of red and white stripes and one additional red, upper stripe. Therefore, we must compare the area of a red, upper stripe with the total area of the 50 white stars.

(a) Compute the area of the red, upper stripe.

(b) Compute the total area of the 50 white stars.

44. Which color occupies the greatest area on the flag?

7.3 The Law of Cosines

If we are given two sides and the included angle (Case 3) or three sides (Case 4) of a triangle, then a unique triangle is determined. These are the SAS and SSS cases, respectively. Both require using the law of cosines to solve the triangle.

The following property is important when applying the law of cosines.

■ Derivation of the Law of Cosines

■ Solutions of SAS and SSS Triangles (Cases 3 and 4)

■ Heron’s Formula for the Area of a Triangle

■ Derivation of Heron’s Formula

Triangle Side Length Restriction

In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.

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(Video) Mathematics Trigonometry Grade 11 Revision: How to approach your final exam.

3137.3 The Law of Cosines

As an example of this property, it would be impossible to construct a tri-angle with sides of lengths 3, 4, and 10. See Figure 12.

a = 3

c = 10

b = 4

No triangle is formed.

Figure 12

Law of Cosines

In any triangle ABC, with sides a, b, and c, the following hold.

a2 = b2 + c2 − 2bc cos A

b2 = a2 + c2 − 2ac cos B

c2 = a2 + b2 − 2ab cos C

That is, according to the law of cosines, the square of a side of a triangle is equal to the sum of the squares of the other two sides, minus twice the product of those two sides and the cosine of the angle included between them.

Derivation of the Law of Cosines To derive the law of cosines, let ABC be any oblique triangle. Choose a coordinate system so that vertex B is at the origin and side BC is along the positive x-axis. See Figure 13.

Let 1x, y2 be the coordinates of vertex A of the triangle. Then the following are true for angle B, whether obtuse or acute.

sin B =yc and cos B = x

c Definition of sine and cosine

y = c sin B and x = c cos B Here x is negative when B is obtuse.

Thus, the coordinates of point A become 1c cos B, c sin B2.Point C in Figure 13 has coordinates 1a, 02, AC has length b, and point A

has coordinates 1c cos B, c sin B2. We can use the distance formula to write an equation.

b = 21c cos B - a22 + 1c sin B - 022 d = 21x2 - x122 + 1y2 - y122 b2 = 1c cos B - a22 + 1c sin B22 Square each side. b2 = 1c2 cos2 B - 2ac cos B + a22 + c2 sin2 B Multiply; 1x - y22 = x2 - 2xy + y2 b2 = a2 + c21cos2 B + sin2 B2 - 2ac cos B Properties of real numbers b2 = a2 + c2112 - 2ac cos B Fundamental identity b2 = a2 + c2 - 2ac cos B Law of cosines

This result is one of three possible forms of the law of cosines. In our work, we could just as easily have placed vertex A or C at the origin. This would have given the same result, but with the variables rearranged.

c90° (0, 0)

BxD a

(a, 0)

(c cos B, c sin B)

Figure 13

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314 CHAPTER 7 Applications of Trigonometry and Vectors

Solutions of SAS and SSS Triangles (Cases 3 and 4)

NOTE If we let C = 90° in the third form of the law of cosines, then cos C = cos 90° = 0, and the formula becomes

c2 = a2 + b2. Pythagorean theorem

The Pythagorean theorem is a special case of the law of cosines.

EXAMPLE 1 Applying the Law of Cosines (SAS)

A surveyor wishes to find the distance between two inaccessible points A and B on opposite sides of a lake. While standing at point C, she finds that b = 259 m, a = 423 m, and angle ACB measures 132° 40′. Find the distance c. See Figure 14.

SOLUTION We can use the law of cosines here because we know the lengths of two sides of the triangle and the measure of the included angle.

c2 = a2 + b2 - 2ab cos C Law of cosines c2 = 4232 + 2592 - 21423212592 cos 132° 40′ Substitute. c2 ≈ 394,510.6 Use a calculator.

c ≈ 628 Take the square root of each side. Choose the positive root.The distance between the points is approximately 628 m. ■✔ Now Try Exercise 39.

b = 259 m

a = 423 m

132° 40!

Figure 14

EXAMPLE 2 Applying the Law of Cosines (SAS)

Solve triangle ABC if A = 42.3°, b = 12.9 m, and c = 15.4 m.

SOLUTION See Figure 15. We start by finding side a with the law of cosines.

a2 = b2 + c2 - 2bc cos A Law of cosines a2 = 12.92 + 15.42 - 2112.92115.42 cos 42.3° Substitute. a2 ≈ 109.7 Use a calculator.

a ≈ 10.47 m Take square roots and choose the positive root.Of the two remaining angles B and C, B must be the smaller because it is oppo-site the shorter of the two sides b and c. Therefore, B cannot be obtuse.

sin A

a= sin B

b Law of sines (alternative form)

sin 42.3°

10.47= sin B

12.9 Substitute.

sin B = 12.9 sin 42.3°10.47

Multiply by 12.9 and rewrite.

B ≈ 56.0° Use the inverse sine function.

The easiest way to find C is to subtract the measures of A and B from 180°.

C = 180° - A - B Angle sum formula, solved for C C ≈ 180° - 42.3° - 56.0° Substitute. C ≈ 81.7° Subtract. ■✔ Now Try Exercise 19.

b = 12.9 m a

c = 15.4 mA B

42.3°

Figure 15

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3157.3 The Law of Cosines

CAUTION Had we used the law of sines to find C rather than B in Example 2, we would not have known whether C was equal to 81.7° or to its supplement, 98.3°.

EXAMPLE 3 Applying the Law of Cosines (SSS)

Solve triangle ABC if a = 9.47 ft, b = 15.9 ft, and c = 21.1 ft.

SOLUTION We can use the law of cosines to solve for any angle of the triangle. We solve for C, the largest angle. We will know that C is obtuse if cos C 6 0.

c2 = a2 + b2 - 2ab cos C Law of cosines

cos C = a2 + b2 - c2

2ab Solve for cos C.

cos C = 9.472 + 15.92 - 21.12219.472115.92 Substitute.

cos C ≈ -0.34109402 Use a calculator. C ≈ 109.9° Use the inverse cosine function.

Now use the law of sines to find angle B.

sin B

b= sin C

c Law of sines (alternative form)

sin B15.9

= sin 109.9°21.1

Substitute.

sin B = 15.9 sin 109.9°21.1

Multiply by 15.9.

B ≈ 45.1° Use the inverse sine function.

Since A = 180° - B - C, we have A ≈ 180° - 45.1° - 109.9° ≈ 25.0°.■✔ Now Try Exercise 23.

Trusses are frequently used to support roofs on buildings, as illustrated in Figure 16. The simplest type of roof truss is a triangle, as shown in Figure 17. (Source: Riley, W., L. Sturges, and D. Morris, Statics and Mechanics of Materials, John Wiley and Sons.)Figure 16

9 ft 6 ft

11 ft

B C

Figure 17

EXAMPLE 4 Designing a Roof Truss (SSS)

Find angle B to the nearest degree for the truss shown in Figure 17.

SOLUTION

b2 = a2 + c2 - 2ac cos B Law of cosines

cos B = a2 + c2 - b2

2ac Solve for cos B.

cos B = 112 + 92 - 6221112192 Let a = 11, b = 6, and c = 9.

cos B ≈ 0.83838384 Use a calculator.

B ≈ 33° Use the inverse cosine function.

■✔ Now Try Exercise 49.

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316 CHAPTER 7 Applications of Trigonometry and Vectors

Four possible cases can occur when we solve an oblique triangle. They are summarized in the following table. In all four cases, it is assumed that the given information actually produces a triangle.

Four Cases for Solving Oblique Triangles

Oblique Triangle Suggested Procedure for Solving

Case 1: One side and two angles are known. (SAA or ASA)

Step 1 Find the remaining angle using the angle sum formula 1A + B + C = 180°2.

Step 2 Find the remaining sides using the law of sines.

Case 2: Two sides and one angle (not included between the two sides) are known. (SSA)

This is the ambiguous case. There may be no triangle, one triangle, or two triangles.

Step 1 Find an angle using the law of sines.

Step 2 Find the remaining angle using the angle sum formula.

Step 3 Find the remaining side using the law of sines.

If two triangles exist, repeat Steps 2 and 3.

Case 3: Two sides and the included angle are known. (SAS)

Step 1 Find the third side using the law of cosines.

Step 2 Find the smaller of the two remaining angles using the law of sines.

Step 3 Find the remaining angle using the angle sum formula.

Case 4: Three sides are known. (SSS)

Step 1 Find the largest angle using the law of cosines.

Step 2 Find either remaining angle using the law of sines.

Step 3 Find the remaining angle using the angle sum formula.

Heron of Alexandria (c. 62 CE)

Heron (also called Hero), a Greek geometer and inventor, produced writings that contain knowledge of the mathematics and engineering of Babylonia, ancient Egypt, and the Greco-Roman world.

Heron’s Area Formula (SSS)

If a triangle has sides of lengths a, b, and c, with semiperimeter

s =12

1a + b + c 2 ,then the area & of the triangle is given by the following formula.

& = !s 1s − a 2 1s − b 2 1s − c 2 That is, according to Heron’s formula, the area of a triangle is the square root of the product of four factors: (1) the semiperimeter, (2) the semiperimeter minus the first side, (3) the semiperimeter minus the second side, and (4) the semiperimeter minus the third side.

Heron’s Formula for the Area of a Triangle A formula for finding the area of a triangle given the lengths of the three sides, known as Heron’s for-mula, is named after the Greek mathematician Heron of Alexandria. It is found in his work Metrica. Heron’s formula can be used for the case SSS.

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3177.3 The Law of Cosines

EXAMPLE 5 Using Heron’s Formula to Find an Area (SSS)

The distance “as the crow flies” from Los Angeles to New York is 2451 mi, from New York to Montreal is 331 mi, and from Montreal to Los Angeles is 2427 mi. What is the area of the triangular region having these three cities as vertices? (Ignore the curvature of Earth.)

SOLUTION In Figure 18, we let a = 2451, b = 331, and c = 2427.

Montreal

Los AngelesNewYork

NOT TO SCALE

c = 2427 mi

a = 2451 mi

b = 331 mi

Figure 18

First, find the semiperimeter s.

s = 12

1a + b + c2 Semiperimeter s = 1

2 12451 + 331 + 24272 Substitute the given values.

s = 2604.5 Add, and then multiply.

Now use Heron’s formula to find the area &.

& = 2s1s - a21s - b21s - c2 & = 22604.512604.5 - 2451212604.5 - 331212604.5 - 24272 & ≈ 401,700 mi2 Use a calculator.

Don’t forget the factor s.

■✔ Now Try Exercise 73.

Derivation of Heron’s Formula A trigonometric derivation of Heron’s formula illustrates some ingenious manipulation.

Let triangle ABC have sides of lengths a, b, and c. Apply the law of cosines.

a2 = b2 + c2 - 2bc cos A Law of cosines

cos A = b2 + c2 - a2

2bc Solve for cos A. (1)

The perimeter of the triangle is a + b + c, so half of the perimeter (the semi-perimeter) is given by the formula in equation (2) below.

s = 12

1a + b + c2 (2) 2s = a + b + c Multiply by 2. (3)

b + c - a = 2s - 2a Subtract 2a from each side and rewrite. b + c - a = 21s - a2 Factor. (4)

Subtract 2b and 2c in a similar way in equation (3) to obtain the following.

a - b + c = 21s - b2 (5) a + b - c = 21s - c2 (6)

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318 CHAPTER 7 Applications of Trigonometry and Vectors

Now we obtain an expression for 1 - cos A.

1 - cos A = 1 - b2 + c2 - a2

2bc(1111)1111* cos A, from (1)

= 2bc + a2 - b2 - c22bc

Find a common denominator, and distribute the - sign.

=a2 - 1b2 - 2bc + c22

2bc Regroup.

=a2 - 1b - c22

2bc

Factor the perfect square trinomial.

=3a - 1b - c243a + 1b - c24

2bc

Factor the difference of squares.

=1a - b + c21a + b - c2

2bc Distributive property

=21s - b2 # 21s - c2

2bc Use equations (5) and (6).

1 - cos A =21s - b21s - c2

bc Lowest terms (7)

Pay attention to signs.

Similarly, it can be shown that

1 + cos A =2s1s - a2

bc . (8)

Recall the double-angle identities for cos 2u.

cos 2u = 2 cos2 u - 1

cos A = 2 cos2 aA2b - 1 Let u = A2 .

1 + cos A = 2 cos2 aA2b Add 1.

2s1s - a2

bc= 2 cos2 aA

2b Substitute.

(1+)11*From (8)

s1s - a2

bc= cos2 aA

2b Divide by 2.

cos aA2b = B s1s - a2bc (9)

cos 2u = 1 - 2 sin2 u

cos A = 1 - 2 sin2 aA2b Let u = A2 .

1 - cos A = 2 sin2 aA2b Subtract 1. Multiply by -1.

21s - b21s - c2

bc= 2 sin2 aA

2b Substitute.

(111111)111111*From (7)

1s - b21s - c2

bc= sin2 aA

2b Divide by 2.

sin aA2b = B1s - b21s - c2bc (10)

The area of triangle ABC can be expressed as follows.

& = 12

bc sin A Area formula

2& = bc sin A Multiply by 2.

2&bc

= sin A Divide by bc. (11)

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3197.3 The Law of Cosines

Recall the double-angle identity for sin 2u.

sin 2u = 2 sin u cos u

sin A = 2 sin aA2b cos aA

2b Let u = A2 .

2&bc

= 2 sin aA2b cos aA

2b Use equation (11).

2&bc

= 2 B1s - b21s - c2bc # B s1s - a2bc Use equations (9) and (10). 2&bc

= 2 B s1s - a21s - b21s - c2b2c 2 Multiply. 2&bc

=22s1s - a21s - b21s - c2

bc Simplify the denominator.

& = !s 1s − a 2 1s − b 2 1s − c 2 Multiply by bc. Divide by 2.Heron’s formula results.CONCEPT PREVIEW Assume a triangle ABC has standard labeling.

(a) Determine whether SAA, ASA, SSA, SAS, or SSS is given.

(b) Decide whether the law of sines or the law of cosines should be used to begin solving the triangle.

1. a, b, and C 2. A, C, and c 3. a, b, and A 4. a, B, and C

5. A, B, and c 6. a, c, and A 7. a, b, and c 8. b, c, and A

7.3 Exercises

Find the length of the remaining side of each triangle. Do not use a calculator.

45°

4√2

10.

60°8

Find the measure of u in each triangle. Do not use a calculator.

11.

12. 1

Solve each triangle. Approximate values to the nearest tenth.

13. C

A B3

121°

14.

A 6

B61°

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320 CHAPTER 7 Applications of Trigonometry and Vectors

15.

A 10

10 12

C 16.

A 8

104

17.

55°A B

100

18.

A 9

7 5

Solve each triangle. See Examples 2 and 3.

19. A = 41.4°, b = 2.78 yd, c = 3.92 yd

20. C = 28.3°, b = 5.71 in., a = 4.21 in.

21. C = 45.6°, b = 8.94 m, a = 7.23 m

22. A = 67.3°, b = 37.9 km, c = 40.8 km

23. a = 9.3 cm, b = 5.7 cm, c = 8.2 cm

24. a = 28 ft, b = 47 ft, c = 58 ft

25. a = 42.9 m, b = 37.6 m, c = 62.7 m

26. a = 189 yd, b = 214 yd, c = 325 yd

27. a = 965 ft, b = 876 ft, c = 1240 ft

28. a = 324 m, b = 421 m, c = 298 m

29. A = 80° 40′, b = 143 cm, c = 89.6 cm

30. C = 72° 40′, a = 327 ft, b = 251 ft

31. B = 74.8°, a = 8.92 in., c = 6.43 in.

32. C = 59.7°, a = 3.73 mi, b = 4.70 mi

33. A = 112.8°, b = 6.28 m, c = 12.2 m

34. B = 168.2°, a = 15.1 cm, c = 19.2 cm

35. a = 3.0 ft, b = 5.0 ft, c = 6.0 ft

36. a = 4.0 ft, b = 5.0 ft, c = 8.0 ft

Concept Check Answer each question.

37. Refer to Figure 12. If we attempt to find any angle of a triangle with the values a = 3, b = 4, and c = 10 using the law of cosines, what happens?

38. “The shortest distance between two points is a straight line.” How is this statement related to the geometric property that states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side?

Solve each problem. See Examples 1–4.

39. Distance across a River Points A and B are on opposite sides of False River. From a third point, C, the angle between the lines of sight to A and B is 46.3°. If AC is 350 m long and BC is 286 m long, find AB.

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3217.3 The Law of Cosines

40. Distance across a Ravine Points X and Y are on opposite sides of a ravine. From a third point Z, the angle between the lines of sight to X and Y is 37.7°. If XZ is 153 m long and YZ is 103 m long, find XY.

41. Angle in a Parallelogram A parallelogram has sides of length 25.9 cm and 32.5 cm. The longer diagonal has length 57.8 cm. Find the measure of the angle opposite the longer diagonal.

42. Diagonals of a Parallelogram The sides of a parallelogram are 4.0 cm and 6.0 cm. One angle is 58°, while another is 122°. Find the lengths of the diagonals of the parallelogram.

43. Flight Distance Airports A and B are 450 km apart, on an east-west line. Tom flies in a northeast direction from airport A to airport C. From C he flies 359 km on a bearing of 128° 40′ to B. How far is C from A?

44. Distance Traveled by a Plane An airplane flies 180 mi from point X at a bearing of 125°, and then turns and flies at a bearing of 230° for 100 mi. How far is the plane from point X?

45. Distance between Ends of the Vietnam Memorial The Vietnam Veterans Memorial in Washington, D.C., is V-shaped with equal sides of length 246.75 ft. The angle between these sides measures 125° 12′. Find the distance between the ends of the two sides. (Source: Pamphlet obtained at Vietnam Veterans Memorial.)

246.75 ft 246.75 ft

125° 12′

46. Distance between Two Ships Two ships leave a harbor together, traveling on courses that have an angle of 135° 40′ between them. If each travels 402 mi, how far apart are they?

135° 40!

Harbor

402 mi402 mi

47. Distance between a Ship and a Rock A ship is sailing east. At one point, the bear-ing of a submerged rock is 45° 20′. After the ship has sailed 15.2 mi, the bearing of the rock has become 308° 40′. Find the distance of the ship from the rock at the latter point.

E15.2 mi

Rock

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322 CHAPTER 7 Applications of Trigonometry and Vectors

48. Distance between a Ship and a Submarine From an airplane flying over the ocean, the angle of depression to a submarine lying under the surface is 24° 10′. At the same moment, the angle of depression from the airplane to a battleship is 17° 30′. See the figure. The distance from the airplane to the battleship is 5120 ft. Find the distance between the battleship and the submarine. (Assume the airplane, subma-rine, and battleship are in a vertical plane.)

Submarine

24° 10!5120 ft

17° 30!

Battleship

49. Truss Construction A triangular truss is shown in the figure. Find angle u.

50. Truss Construction Find angle b in the truss shown in the figure.

51. Distance between a Beam and Cables A weight is supported by cables attached to both ends of a bal-ance beam, as shown in the figure. What angles are formed between the beam and the cables?

52. Distance between Points on a Crane A crane with a counterweight is shown in the figure. Find the horizontal distance between points A and B to the nearest foot.

u b

20 ft

16 ft 13 ft

90 ft

? ?

45 ft 60 ft

10 ft 10 ft

A B

128°

53. Distance on a Baseball Diamond A baseball diamond is a square, 90.0 ft on a side, with home plate and the three bases as vertices. The pitcher’s position is 60.5 ft from home plate. Find the distance from the pitcher’s position to each of the bases.

3rd base

Homeplate

1st base

90.0

2nd base

60.5 ft

54. Distance on a Softball Diamond A softball diamond is a square, 60.0 ft on a side, with home plate and the three bases as vertices. The pitcher’s position is 46.0 ft from home plate. Find the distance from the pitcher’s position to each of the bases.

3rd base

Homeplate

1st base

60.0

2nd base

46.0 ft

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3237.3 The Law of Cosines

55. Distance between a Ship and a Point Starting at point A, a ship sails 18.5 km on a bearing of 189°, then turns and sails 47.8 km on a bearing of 317°. Find the distance of the ship from point A.

56. Distance between Two Factories Two factories blow their whistles at exactly 5:00. A man hears the two blasts at 3 sec and 6 sec after 5:00, respectively. The angle between his lines of sight to the two factories is 42.2°. If sound travels 344 m per sec, how far apart are the factories?

57. Measurement Using Triangulation Surveyors are often confronted with obstacles, such as trees, when measuring the boundary of a lot. One technique used to obtain an accurate measurement is the triangulation method. In this technique, a triangle is constructed around the obstacle and one angle and two sides of the triangle are measured. Use this technique to find the length of the property line (the straight line between the two markers) in the figure. (Source: Kavanagh, B., Surveying Principles and Applications, Sixth Edition, Prentice-Hall.)

14.0 ft 13.0 ft70.0

NOT TO SCALE

Marker Marker

18.0 ft 14.0 ft

58. Path of a Ship A ship sailing due east in the North Atlantic has been warned to change course to avoid icebergs. The captain turns and sails on a bearing of 62°, then changes course again to a bearing of 115° until the ship reaches its original course. See the figure. How much farther did the ship have to travel to avoid the icebergs?

Icebergs

50 mi

59. Length of a Tunnel To measure the distance through a mountain for a proposed tunnel, a point C is chosen that can be reached from each end of the tunnel. See the figure. If AC = 3800 m, BC = 2900 m, and angle C = 110°, find the length of the tunnel.

A B

110°2900 m3800 m

Tunnel

60. Distance between an Airplane and a Mountain A person in a plane flying straight north observes a mountain at a bearing of 24.1°. At that time, the plane is 7.92 km from the mountain. A short time later, the bearing to the mountain becomes 32.7°. How far is the airplane from the mountain when the second bearing is taken?

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324 CHAPTER 7 Applications of Trigonometry and Vectors

Solve each problem. See Example 5.

71. Perfect Triangles A perfect triangle is a triangle whose sides have whole number lengths and whose area is numerically equal to its perimeter. Show that the triangle with sides of length 9, 10, and 17 is perfect.

72. Heron Triangles A Heron triangle is a triangle having integer sides and area. Show that each of the following is a Heron triangle.

(a) a = 11, b = 13, c = 20 (b) a = 13, b = 14, c = 15(c) a = 7, b = 15, c = 20 (d) a = 9, b = 10, c = 17

73. Area of the Bermuda Triangle Find the area of the Bermuda Triangle if the sides of the triangle have approximate lengths 850 mi, 925 mi, and 1300 mi.

74. Required Amount of Paint A painter needs to cover a triangular region 75 m by 68 m by 85 m. A can of paint covers 75 m2 of area. How many cans (to the next higher number of cans) will be needed?

75. Consider triangle ABC shown here.

(a) Use the law of sines to find candidates for the value of angle C. Round angle measures to the nearest tenth of a degree.

(b) Rework part (a) using the law of cosines.

76. Show that the measure of angle A is twice the measure of angle B. (Hint: Use the law of cosines to find cos A and cos B, and then show that cos A = 2 cos2 B - 1.)

Find the measure of each angle u to two decimal places.

61. (6, 8)

(4, 3)

y 62.

(8, 6)(12, 5)

Find the exact area of each triangle using the formula & = 12 bh, and then verify that Heron’s formula gives the same result.

63.

6 14

64.

614

Find the area of each triangle ABC. See Example 5.

65. a = 12 m, b = 16 m, c = 25 m 66. a = 22 in., b = 45 in., c = 31 in.

67. a = 154 cm, b = 179 cm, c = 183 cm 68. a = 25.4 yd, b = 38.2 yd, c = 19.8 yd

69. a = 76.3 ft, b = 109 ft, c = 98.8 ft 70. a = 15.8 m, b = 21.7 m, c = 10.9 m

A C

1315

60°

A B

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325CHAPTER 7 Quiz

Relating Concepts

For individual or collaborative investigation (Exercises 77—80)

We have introduced two new formulas for the area of a triangle in this chapter. We can now find the area & of a triangle using one of three formulas.

(a) & = 12 bh

(b) & = 12 ab sin C 1or & = 12 ac sin B or & = 12 bc sin A2(c) & = 2s1s - a21s - b21s - c2 (Heron’s formula)If the coordinates of the vertices of a triangle are given, then the following area formula is also valid.

(d) & = 122 1x1y2 - y1x2 + x2 y3 - y2x3 + x3y1 - y3 x12 2 The vertices are the ordered pairs 1x1, y12, 1x2, y22, and 1x3, y32.

Work Exercises 77–80 in order, showing that the various formulas all lead to the same area.

77. Draw a triangle with vertices A12, 52, B1-1, 32, and C14, 02, and use the dis-tance formula to find the lengths of the sides a, b, and c.

78. Find the area of triangle ABC using formula (b). (First use the law of cosines to find the measure of an angle.)

79. Find the area of triangle ABC using formula (c) —that is, Heron’s formula.

80. Find the area of triangle ABC using new formula (d).

Find the indicated part of each triangle ABC.

1. Find A if B = 30.6°, b = 7.42 in., and c = 4.54 in.

2. Find a if A = 144°, c = 135 m, and b = 75.0 m.

3. Find C if a = 28.4 ft, b = 16.9 ft, and c = 21.2 ft.

Solve each problem.

4. Find the area of the triangle shown here.

7 150°

5. Find the area of triangle ABC if a = 19.5 km, b = 21.0 km, and c = 22.5 km.

6. For triangle ABC with c = 345, a = 534, and C = 25.4°, there are two possible val-ues for angle A. What are they?

7. Solve triangle ABC if c = 326, A = 111°, and B = 41.0°.

8. Height of a Balloon The angles of elevation of a hot air balloon from two observation points X and Y on level ground are 42° 10′ and 23° 30′, respectively. As shown in the figure, points X, Y, and Z are in the same vertical plane and points X and Y are 12.2 mi apart. Approximate the height of the balloon to the nearest tenth of a mile.

Chapter 7 Quiz (Sections 7.1—7.3)

12.2 mi23° 30!42° 10!

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326 CHAPTER 7 Applications of Trigonometry and Vectors

9. Volcano Movement To help predict eruptions from the volcano Mauna Loa on the island of Hawaii, scientists keep track of the volcano’s movement by using a “super triangle” with vertices on the three volcanoes shown on the map at the right. Find BC given that AB = 22.47928 mi, AC = 28.14276 mi, and A = 58.56989°.

10. Distance between Two Towns To find the distance between two small towns, an electronic distance mea-suring (EDM) instrument is placed on a hill from which both towns are visible. The distance to each town from the EDM and the angle between the two lines of sight are measured. See the figure. Find the distance between the towns.

Mauna Loa

Mauna Kea

Hualalai

Town A

Town B

43.33° 3428

5631 mHill

7.4 Geometrically Defined Vectors and Applications

Basic Terminology Quantities that involve magnitudes, such as 45 lb or 60 mph, can be represented by real numbers called scalars. Other quantities, called vector quantities, involve both magnitude and direction. Typical vector quantities are velocity, acceleration, and force. For example, traveling 50 mph east represents a vector quantity.

A vector quantity can be represented with a directed line segment (a seg-ment that uses an arrowhead to indicate direction) called a vector. The length of the vector represents the magnitude of the vector quantity. The direction of the vector, indicated by the arrowhead, represents the direction of the quantity. See Figure 19.

■ Basic Terminology■ The Equilibrant■ Incline Applications■ Navigation

Applications

10 lb

Horizontal

This vector represents a forceof 10 lb applied at an angle30° above the horizontal.

30°

Figure 19

OVector OP,or vector u

OVector PO,or vector A

Vectors may be named with two uppercase letters or with one lowercase or uppercase letter.

Figure 20

When we indicate vectors in print, it is customary to use boldface type or an arrow over the letter or letters. Thus, OP and OP

> both represent the vector OP.

When two letters name a vector, the first indicates the initial point and the second indicates the terminal point of the vector. Knowing these points gives the direc-tion of the vector. For example, vectors OP and PO in Figure 20 are not the same vector. They have the same magnitude but opposite directions. The magnitude of vector OP is written ∣OP ∣ .

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3277.4 Geometrically Defined Vectors and Applications

Two vectors are equal if and only if they have the same direction and the same magnitude. In Figure 21, vectors A and B are equal, as are vectors C and D. As Figure 21 shows, equal vectors need not coincide, but they must be paral-lel and in the same direction. Vectors A and E are unequal because they do not have the same direction, while A ≠ F because they have different magnitudes.

A B C D E F

Figure 21

The sum of two vectors is also a vector. There are two ways to find the sum of two vectors A and B geometrically.

1. Place the initial point of vector B at the terminal point of vector A, as shown in Figure 22(a). The vector with the same initial point as A and the same terminal point as B is the sum A + B.

2. Use the parallelogram rule. Place vectors A and B so that their initial points coincide, as in Figure 22(b). Then, complete a parallelogram that has A and B as two sides. The diagonal of the parallelogram with the same initial point as A and B is the sum A + B.

Parallelograms can be used to show that vector B + A is the same as vec-tor A + B, or that A + B = B + A, so vector addition is commutative. The vector sum A + B is the resultant of vectors A and B.

For every vector v there is a vector -v that has the same magnitude as v but opposite direction. Vector -v is the opposite of v. See Figure 23. The sum of v and -v has magnitude 0 and is the zero vector. As with real numbers, to subtract vector B from vector A, find the vector sum A + 1-B2. See Figure 24.

A + B

(a)

A + B

(b)

Figure 22

–vv

Vectors v and –vare opposites.

Figure 23

–B

A + (–B)

Figure 24

2uu –2uuu

Figure 25

The product of a real number (or scalar) k and a vector u is the vector k # u, which has magnitude + k + times the magnitude of u. The vector k # u has the same direction as u if k 7 0 and the opposite direction if k 6 0. See Figure 25.

The following properties are helpful when solving vector applications.

Geometric Properties of Parallelograms

1. A parallelogram is a quadrilateral whose opposite sides are parallel.

2. The opposite sides and opposite angles of a parallelogram are equal, and adjacent angles of a parallelogram are supplementary.

3. The diagonals of a parallelogram bisect each other, but they do not nec-essarily bisect the angles of the parallelogram.

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328 CHAPTER 7 Applications of Trigonometry and Vectors

EXAMPLE 1 Finding the Magnitude of a Resultant

Two forces of 15 and 22 newtons act on a point in the plane. (A newton is a unit of force that equals 0.225 lb.) If the angle between the forces is 100°, find the magnitude of the resultant force.

SOLUTION As shown in Figure 26, a parallelogram that has the forces as adjacent sides can be formed. The angles of the parallelogram adjacent to angle P measure 80° because adjacent angles of a parallelogram are supple-mentary. Opposite sides of the parallelogram are equal in length. The resultant force divides the parallelogram into two triangles. Use the law of cosines with either triangle.

+v +2 = 152 + 222 - 211521222 cos 80° Law of cosines+v +2 ≈ 225 + 484 - 115 Evaluate powers

and cos 80°. Multiply.+v +2 ≈ 594 Add and subtract.

+v + ≈ 24 Take square roots and choose the positive square root.To the nearest unit, the magnitude of the resultant force is 24 newtons.

■✔ Now Try Exercise 27.

80°

100°

Figure 26

The Equilibrant The previous example showed a method for finding the resultant of two vectors. Sometimes it is necessary to find a vector that will counterbalance the resultant. This opposite vector is the equilibrant. That is, the equilibrant of vector u is the vector -u.

EXAMPLE 2 Finding the Magnitude and Direction of an Equilibrant

Find the magnitude of the equilibrant of forces of 48 newtons and 60 newtons acting on a point A, if the angle between the forces is 50°. Then find the angle between the equilibrant and the 48-newton force.

SOLUTION As shown in Figure 27, the equilibrant is -v.

v–v

48 60

4860

130°

50°

Figure 27

The magnitude of v, and hence of -v, is found using triangle ABC and the law of cosines.

+v +2 = 482 + 602 - 214821602 cos 130° Law of cosines +v +2≈ 9606.5 Use a calculator.

+v + ≈ 98 Square root property; Give two significant digits.To the nearest unit, the magnitude is 98 newtons.

The required angle, labeled a in Figure 27, can be found by subtracting angle CAB from 180°. Use the law of sines to find angle CAB.

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3297.4 Geometrically Defined Vectors and Applications

sin CAB

60= sin 130°

98 Law of sines (alternative form)

sin CAB ≈ 0.46900680 Multiply by 60 and use a calculator.

CAB ≈ 28° Use the inverse sine function.

Finally, a ≈ 180° - 28° = 152°. ■✔ Now Try Exercise 31.

Incline Applications We can use vectors to solve incline problems.

EXAMPLE 3 Finding a Required Force

Find the force required to keep a 50-lb wagon from sliding down a ramp inclined at 20° to the horizontal. (Assume there is no friction.)

SOLUTION In Figure 28, the vertical 50-lb force BA represents the force of gravity. It is the sum of vectors BC and -AC. The vector BC represents the force with which the weight pushes against the ramp. The vector BF represents the force that would pull the weight up the ramp. Because vectors BF and AC are equal, +AC + gives the magnitude of the required force.

Vectors BF and AC are parallel, so angle EBD equals angle A by alternate interior angles. Because angle BDE and angle C are right angles, triangles CBA and DEB have two corresponding angles equal and, thus, are similar triangles. Therefore, angle ABC equals angle E, which is 20°. From right triangle ABC, we have the following.

sin 20° =+AC +

50 sin B = side opposite Bhypotenuse

+AC + = 50 sin 20° Multiply by 50 and rewrite. +AC + ≈17 Use a calculator.

A force of approximately 17 lb will keep the wagon from sliding down the ramp.

■✔ Now Try Exercise 39.

Ramp

20°E D

20°

Figure 28

EXAMPLE 4 Finding an Incline Angle

A force of 16.0 lb is required to hold a 40.0-lb lawn mower on an incline. What angle does the incline make with the horizontal?

SOLUTION This situation is illustrated in Fig-ure 29. Consider right triangle ABC. Angle B equals angle u, the magnitude of vector BA represents the weight of the mower, and vector AC equals vector BE, which represents the force required to hold the mower on the incline.

sin B = 16.040.0

sin B = side opposite Bhypotenuse

sin B = 0.4 Simplify. B ≈ 23.6° Use the inverse sine function.

The hill makes an angle of about 23.6° with the horizontal.

■✔ Now Try Exercise 41.

16.0

40.0

Figure 29

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330 CHAPTER 7 Applications of Trigonometry and Vectors

Navigation Applications Problems that involve bearing can also be solved using vectors.

EXAMPLE 5 Applying Vectors to a Navigation Problem

A ship leaves port on a bearing of 28.0° and travels 8.20 mi. The ship then turns due east and travels 4.30 mi. How far is the ship from port? What is its bearing from port?

SOLUTION In Figure 30, vectors PA and AE represent the ship’s path. The magnitude and bear-ing of the resultant PE can be found as follows. Triangle PNA is a right triangle, so

angle NAP = 90° - 28.0° = 62.0°,and angle PAE = 180° - 62.0° = 118.0°.

Use the law of cosines to find +PE + , the magnitude of vector PE.

+PE +2 = 8.202 + 4.302 - 218.20214.302 cos 118.0° Law of cosines +PE +2 ≈ 118.84 Use a calculator.

+PE + ≈ 10.9 Square root property

The ship is about 10.9 mi from port.To find the bearing of the ship from port, find angle APE.

sin APE

4.30= sin 118.0°

10.9 Law of sines

sin APE = 4.30 sin 118.0°10.9

Multiply by 4.30.

APE ≈ 20.4° Use the inverse sine function.

Finally, 28.0° + 20.4° = 48.4°, so the bearing is 48.4°.■✔ Now Try Exercise 45.

E4.30

8.20

28.0°

118.0°

Figure 30

In air navigation, the airspeed of a plane is its speed relative to the air, and the ground speed is its speed relative to the ground. Because of wind, these two speeds are usually different. The ground speed of the plane is represented by the vector sum of the airspeed and windspeed vectors. See Figure 31.

Course and ground speed(actual direction of plane)

Wind directionand speed

Bearingand airspeed

Drift angle

Figure 31

EXAMPLE 6 Applying Vectors to a Navigation Problem

An airplane that is following a bearing of 239° at an airspeed of 425 mph encounters a wind blowing at 36.0 mph from a direction of 115°. Find the result-ing bearing and ground speed of the plane.

SOLUTION An accurate sketch is essential to the solution of this problem. We have included two sets of geographical axes, which enable us to determine mea-sures of necessary angles. Analyze Figure 32 on the next page carefully.

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3317.4 Geometrically Defined Vectors and Applications

425124°

36.0

E1W1

239°

59°

25°

115°

NOT TO SCALE

Figure 32

Vector c represents the airspeed and bearing of the plane, vector a repre-sents the speed and direction of the wind, and vector b represents the resulting bearing and ground speed of the plane. Angle ABC has as its measure the sum of angle ABN1 and angle N1BC.

Angle SAB measures 239° - 180° = 59°. Because angle ABN1 is an alternate interior angle to it, ABN1 = 59°.Angle E1BF measures 115° - 90° = 25°. Thus, angle CBW1 also measures 25° because it is a vertical angle. Angle N1BC is the complement of 25°, which is 90° - 25° = 65°.

By these results,

angle ABC = 59° + 65° = 124°.

To find +b + , we use the law of cosines.

+b +2 = +a +2 + + c +2 - 2 +a + + c + cos ABC Law of cosines

+b +2 = 36.02 + 4252 - 2136.0214252 cos 124° Substitute. +b +2 ≈ 199,032 Use a calculator.

+b + ≈ 446 Square root property

The ground speed is approximately 446 mph.To find the resulting bearing of b, we must find the measure of angle a in

Figure 32 and then add it to 239°. To find a, we use the law of sines.

sin a36.0

= sin 124°446

sin a = 36.0 sin 124°446

Multiply by 36.0.

a = sin-1 a 36.0 sin 124°446

b Use the inverse sine function. a ≈ 4° Use a calculator.

To maintain accuracy, use all the significant digits that

a calculator allows.

Add 4° to 239° to find the resulting bearing of 243°. ■✔ Now Try Exercise 51.

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332 CHAPTER 7 Applications of Trigonometry and Vectors

CONCEPT PREVIEW Refer to the vectors m through t below.

1. Name all pairs of vectors that appear to be equal.

2. Name all pairs of vectors that are opposites.

3. Name all pairs of vectors where the first is a scalar multiple of the other, with the scalar positive.

4. Name all pairs of vectors where the first is a scalar multiple of the other, with the scalar negative.

7.4 Exercises

nm o

r s tqp

For each pair of vectors u and v with angle u between them, sketch the resultant.

19. +u + = 12, +v + = 20, u = 27° 20. +u + = 8, +v + = 12, u = 20°

21. +u + = 20, +v + = 30, u = 30° 22. +u + = 50, +v + = 70, u = 40°

Use the parallelogram rule to find the magnitude of the resultant force for the two forces shown in each figure. Round answers to the nearest tenth.

23. 40 lb

60 lb

40°

24.

85 lb

102 lb

65°

25.

15 lb

25 lb

110°

26.

1500 lb

2000 lb

140°

b d

g h

a c

a +

CONCEPT PREVIEW Refer to vectors a through h below. Make a copy or a sketch of each vector, and then draw a sketch to represent each of the following. For example, find a + e by placing a and e so that their initial points coincide. Then use the parallelogram rule to find the resultant, as shown in the figure on the right.

5. -b 6. -g 7. 2c 8. 2h

9. a + b 10. h + g 11. a - c 12. d - e

13. a + 1b + c2 14. 1a + b2 + c 15. c + d 16. d + c 17. From the results of Exercises 13 and 14, does it appear that vector addition is

associative?

18. From the results of Exercises 15 and 16, does it appear that vector addition is commutative?

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3337.4 Geometrically Defined Vectors and Applications

Two forces act at a point in the plane. The angle between the two forces is given. Find the magnitude of the resultant force. See Example 1.

27. forces of 250 and 450 newtons, forming an angle of 85°

28. forces of 19 and 32 newtons, forming an angle of 118°

29. forces of 116 and 139 lb, forming an angle of 140° 50′

30. forces of 37.8 and 53.7 lb, forming an angle of 68.5°

Solve each problem. See Examples 1–4.

31. Direction and Magnitude of an Equilibrant Two tugboats are pulling a disabled speedboat into port with forces of 1240 lb and 1480 lb. The angle between these forces is 28.2°. Find the direction and magnitude of the equilibrant.

32. Direction and Magnitude of an Equilibrant Two rescue vessels are pulling a broken-down motorboat toward a boathouse with forces of 840 lb and 960 lb. The angle between these forces is 24.5°. Find the direction and magnitude of the equilibrant.

33. Angle between Forces Two forces of 692 newtons and 423 newtons act at a point. The resultant force is 786 newtons. Find the angle between the forces.

34. Angle between Forces Two forces of 128 lb and 253 lb act at a point. The resultant force is 320 lb. Find the angle between the forces.

35. Magnitudes of Forces A force of 176 lb makes an angle of 78° 50′ with a second force. The resultant of the two forces makes an angle of 41° 10′ with the first force. Find the magnitudes of the second force and of the resultant.

36. Magnitudes of Forces A force of 28.7 lb makes an angle of 42° 10′ with a second force. The resultant of the two forces makes an angle of 32° 40′ with the first force. Find the magni-tudes of the second force and of the resultant.

37. Angle of a Hill Slope A force of 25 lb is required to hold an 80-lb crate on a hill. What angle does the hill make with the horizontal?

38. Force Needed to Keep a Car Parked Find the force required to keep a 3000-lb car parked on a hill that makes an angle of 15° with the horizontal.

39. Force Needed for a Monolith To build the pyramids in Egypt, it is believed that giant causeways were constructed to transport the building materials to the site. One such causeway is said to have been 3000 ft long, with a slope of about 2.3°. How much force would be required to hold a 60-ton monolith on this causeway?

60 tons2.3°

3000 ft

176 lbFirst force

41° 10!

78° 50!

Resul

tant

Secondforce

28.7 lbFirst force

32° 40′42° 10′

Resul

tant

Secondforce

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334 CHAPTER 7 Applications of Trigonometry and Vectors

40. Force Needed for a Monolith If the causeway in Exercise 39 were 500 ft longer and the monolith weighed 10 tons more, how much force would be required?

41. Incline Angle A force of 18.0 lb is required to hold a 60.0-lb stump grinder on an incline. What angle does the incline make with the horizontal?

42. Incline Angle A force of 30.0 lb is required to hold an 80.0-lb pressure washer on an incline. What angle does the incline make with the horizontal?

43. Weight of a Box Two people are carrying a box. One person exerts a force of 150 lb at an angle of 62.4° with the horizontal. The other person exerts a force of 114 lb at an angle of 54.9°. Find the weight of the box.

44. Weight of a Crate and Tension of a Rope A crate is supported by two ropes. One rope makes an angle of 46° 20′ with the horizontal and has a tension of 89.6 lb on it. The other rope is horizontal. Find the weight of the crate and the tension in the hori-zontal rope.

Solve each problem. See Examples 5 and 6.

45. Distance and Bearing of a Ship A ship leaves port on a bearing of 34.0° and travels 10.4 mi. The ship then turns due east and travels 4.6 mi. How far is the ship from port, and what is its bearing from port?

46. Distance and Bearing of a Luxury Liner A luxury liner leaves port on a bearing of 110.0° and travels 8.8 mi. It then turns due west and travels 2.4 mi. How far is the liner from port, and what is its bearing from port?

47. Distance of a Ship from Its Starting Point Starting at point A, a ship sails 18.5 km on a bearing of 189°, then turns and sails 47.8 km on a bearing of 317°. Find the distance of the ship from point A.

48. Distance of a Ship from Its Starting Point Starting at point X, a ship sails 15.5 km on a bearing of 200°, then turns and sails 2.4 km on a bearing of 320°. Find the dis-tance of the ship from point X.

49. Distance and Direction of a Motor-boat A motorboat sets out in the direction N 80° 00′ E. The speed of the boat in still water is 20.0 mph. If the current is flowing directly south, and the actual direction of the motorboat is due east, find the speed of the current and the actual speed of the motorboat.

50. Movement of a Motorboat Suppose we would like to cross a 132-ft-wide river in a motorboat. Assume that the motor-boat can travel at 7.0 mph relative to the water and that the current is flowing west at the rate of 3.0 mph. The bearing u is chosen so that the motorboat will land at a point exactly across from the starting point.

(a) At what speed will the motorboat be traveling relative to the banks?

(b) How long will it take for the motorboat to make the crossing?

150 lb

114 lb

62.4° 54.9°

Box

3.0

7.0

Ending point

Starting point

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3357.4 Geometrically Defined Vectors and Applications

51. Bearing and Ground Speed of a Plane An airline route from San Francisco to Honolulu is on a bearing of 233.0°. A jet flying at 450 mph on that bearing encounters a wind blowing at 39.0 mph from a direction of 114.0°. Find the resulting bearing and ground speed of the plane.

52. Path Traveled by a Plane The aircraft carrier Tallahassee is traveling at sea on a steady course with a bearing of 30° at 32 mph. Patrol planes on the carrier have enough fuel for 2.6 hr of flight when traveling at a speed of 520 mph. One of the pilots takes off on a bearing of 338° and then turns and heads in a straight line, so as to be able to catch the carrier and land on the deck at the exact instant that his fuel runs out. If the pilot left at 2 p.m., at what time did he turn to head for the carrier?

53. Airspeed and Ground Speed A pilot wants to fly on a bearing of 74.9°. By fly-ing due east, he finds that a 42.0-mph wind, blowing from the south, puts him on course. Find the airspeed and the ground speed.

54. Bearing of a Plane A plane flies 650 mph on a bearing of 175.3°. A 25-mph wind, from a direction of 266.6°, blows against the plane. Find the resulting bearing of the plane.

55. Bearing and Ground Speed of a Plane A pilot is flying at 190.0 mph. He wants his flight path to be on a bearing of 64° 30′. A wind is blowing from the south at 35.0 mph. Find the bearing he should fly, and find the plane’s ground speed.

56. Bearing and Ground Speed of a Plane A pilot is flying at 168 mph. She wants her flight path to be on a bearing of 57° 40′. A wind is blowing from the south at 27.1 mph. Find the bearing she should fly, and find the plane’s ground speed.

57. Bearing and Airspeed of a Plane What bearing and airspeed are required for a plane to fly 400 mi due north in 2.5 hr if the wind is blowing from a direction of 328° at 11 mph?

58. Ground Speed and Bearing of a Plane A plane is headed due south with an air-speed of 192 mph. A wind from a direction of 78.0° is blowing at 23.0 mph. Find the ground speed and resulting bearing of the plane.

59. Ground Speed and Bearing of a Plane An airplane is headed on a bearing of 174° at an airspeed of 240 km per hr. A 30-km-per-hr wind is blowing from a direction of 245°. Find the ground speed and resulting bearing of the plane.

60. Velocity of a Star The space velocity v of a star rela-tive to the sun can be expressed as the resultant vector of two perpendicular vectors—the radial velocity vr and the tangential velocity vt , where v = vr + vt . If a star is located near the sun and its space velocity is large, then its motion across the sky will also be large. Barnard’s Star is a relatively close star with a distance of 35 trillion mi from the sun. It moves across the sky through an angle of 10.34″ per year, which is the largest motion of any known star. Its radial velocity vr is 67 mi per sec toward the sun. (Sources: Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Second Edition, Saunders College Publishing; Acker, A. and C. Jaschek, Astronomical Methods and Calculations, John Wiley and Sons.)

(a) Approximate the tangential velocity vt of Barnard’s Star. (Hint: Use the arc length formula s = r u.)

(b) Compute the magnitude of v.

30°Plane

Carrier

338°

Barnard's Star

Sun

vrvt

NOT TO SCALE

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336 CHAPTER 7 Applications of Trigonometry and Vectors

LOOKING AHEAD TO CALCULUSIn addition to two-dimensional vectors

in a plane, calculus courses introduce

three-dimensional vectors in space.

The magnitude of the two-dimensional

vector 8a, b9 is given by2a2 + b2.If we extend this to the three-

dimensional vector 8a, b, c9 , the expression becomes2a2 + b2 + c2.Similar extensions are made for other

concepts.

7.5 Algebraically Defined Vectors and the Dot Product

Algebraic Interpretation of Vectors A vector with initial point at the origin in a rectangular coordi-nate system is a position vector . A position vector u with endpoint at the point 1a, b2 is written 8a, b9 , so

u = 8a, b9 .This means that every vector in the real plane cor-responds to an ordered pair of real numbers. Thus, geometrically a vector is a directed line segment while algebraically it is an ordered pair. The numbers a and b are the horizontal component and the vertical component, respectively, of vector u.

Figure 33 shows the vector u = 8a, b9 . The positive angle between the x-axis and a position vector is the direction angle for the vector. In Figure 33, u is the direction angle for vector u. The magnitude and direction angle of a vector are related to its horizontal and vertical components.

■ Algebraic Interpretation of Vectors

■ Operations with Vectors■ The Dot Product and

the Angle between Vectors

u = ⟨a, b⟩

(a, b)

Figure 33

Magnitude and Direction Angle of a Vector 8 a, b 9The magnitude (length) of vector u = 8a, b9 is given by the following.

∣u ∣ = !a2 + b2The direction angle u satisfies tan u = ba , where a ≠ 0.

EXAMPLE 1 Finding Magnitude and Direction Angle

Find the magnitude and direction angle for u = 83, -29 .ALGEBRAIC SOLUTION

The magnitude is +u + = 232 + 1-222 = 213. To find the direction angle u, start with tan u = ba =

-23 =

- 23 . Vector u has a positive horizontal component and a negative vertical component, which places the position vector in quadrant IV. A calculator then gives

tan-1 1- 232 ≈ -33.7°. Adding 360° yields the direction angle u ≈ 326.3°. See Figure 34.

GRAPHING CALCULATOR SOLUTION

The TI-84 Plus calculator can find the magnitude and direction angle using rectangular to polar conversion (which is covered in detail in the next chapter). An approximation for 213 is given, and the TI-84 Plus gives the direction angle with the least possible absolute value. We must add 360° to the given value -33.7° to obtain the positive direction angle u ≈ 326.3°.

–2

u = !3, –2"

(3, –2)

Figure 34Figure 35

■✔ Now Try Exercise 9.

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3377.5 Algebraically Defined Vectors and the Dot Product

Horizontal and Vertical Components

The horizontal and vertical components, respectively, of a vector u having magnitude ∣u ∣ and direction angle u are the following.

a = ∣u ∣ cos U and b = ∣u ∣ sin U

That is, u = 8a, b9 = 8 +u + cos u, +u + sin u9 .yx

25.0 b

41.7°

Figure 36

EXAMPLE 3 Writing Vectors in the Form 8 a, b 9Write each vector in Figure 38 in the form 8a, b9 .SOLUTION

u = 85 cos 60°, 5 sin 60°9 = h 5 # 12

, 5 # 232i = h 5

2 ,

5232i

v = 82 cos 180°, 2 sin 180°9 = 821-12, 21029 = 8 -2, 09 w = 86 cos 280°, 6 sin 280°9 ≈ 81.0419, -5.90889 Use a calculator.

■✔ Now Try Exercises 19 and 21.

6280°

60°

180°

Figure 38

Operations with Vectors As shown in Figure 39,

m = 8a, b9 , n = 8c, d9 , and p = 8a + c, b + d9 .Using geometry, we can show that the endpoints of the three vectors and the origin form a parallel-ogram. A diagonal of this parallelogram gives the resultant of m and n, so we have p = m + n or8a + c, b + d9 = 8a, b9 + 8c, d9 .Similarly, we can verify the following operations.

!a, b"

n !c, d "

!a + c, b + d "

Figure 39

EXAMPLE 2 Finding Horizontal and Vertical Components

Vector w in Figure 36 has magnitude 25.0 and direction angle 41.7°. Find the horizontal and vertical components.

ALGEBRAIC SOLUTION

Use the formulas below, with +w + = 25.0 and u = 41.7°.

a = +w + cos u b = +w + sin u a = 25.0 cos 41.7° b = 25.0 sin 41.7° a ≈ 18.7 b ≈ 16.6

Therefore, w = 818.7, 16.69 . The horizontal compo-nent is 18.7, and the vertical component is 16.6 (rounded to the nearest tenth).

GRAPHING CALCULATOR SOLUTION

See Figure 37. The results support the algebraic solution.

Figure 37

■✔ Now Try Exercise 13.

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338 CHAPTER 7 Applications of Trigonometry and Vectors

Vector Operations

Let a, b, c, d, and k represent real numbers.

8 a, b 9 + 8 c, d 9 = 8 a + c, b + d 9 k # 8 a, b 9 = 8 ka, kb 9

If u = 8 a1 , a2 9 , then −u = 8−a1 , −a2 9 .8 a, b9 − 8 c, d 9 = 8 a, b9 + 1− 8 c, d 9 2 = 8 a − c, b − d 9yx

!–2, 1"

!4, 3"

Figure 40

EXAMPLE 4 Performing Vector Operations

Let u = 8 -2, 19 and v = 84, 39 . See Figure 40. Find and illustrate each of the following.

(a) u + v (b) -2u (c) 3u - 2v

SOLUTION See Figure 41.

(a) u + v = 8 -2, 19 + 84, 39 = 8 -2 + 4, 1 + 39 = 82, 49

(b) -2u = -2 # 8 -2, 19 = 8 -21-22, -21129 = 84, -29

u + v

!–2, 1"

!4, 3"!2, 4"

–2u 0

!–2, 1"

!4, –2"

–2v

3u – 2v 0

!–6, 3"

!8, 6"

!–8, –6"

!–14, –3"

(a) (b) (c)

Figure 41 ■✔ Now Try Exercises 35, 37, and 39.

A unit vector is a vector that has magnitude 1. Two very important unit vectors are defined as follows and shown in Figure 42(a).

i = 81, 09 j = 80, 19

0 ij

(3, 4)

u = 3i + 4ju 4j

(a) (b)

Figure 42

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3397.5 Algebraically Defined Vectors and the Dot Product

With the unit vectors i and j, we can express any other vector 8a, b9 in the form ai + bj, as shown in Figure 42(b), where 83, 49 = 3i + 4j. The vector operations previously given can be restated, using ai + bj notation.

i, j Form for Vectors

If v = 8a, b9 , thenv = ai + bj, where i = 81, 09 and j = 80, 19 .

The Dot Product and the Angle between Vectors The dot product of two vectors is a real number, not a vector. It is also known as the inner product. Dot products are used to determine the angle between two vectors, to derive geometric theorems, and to solve physics problems.

Dot Product

The dot product of the two vectors u = 8a, b9 and v = 8c, d9 is denoted u # v, read “u dot v,” and given by the following.

u # v = ac + bd

That is, the dot product of two vectors is the sum of the product of their first components and the product of their second components.

EXAMPLE 5 Finding Dot Products

Find each dot product.

(a) 82, 39 # 84, -19 (b) 86, 49 # 8 -2, 39SOLUTION

(a) 82, 39 # 84, -19 = 2142 + 31-12 = 5

(b) 86, 49 # 8 -2, 39 = 61-22 + 4132 = 0

■✔ Now Try Exercises 47 and 49.

The following properties of dot products can be verified using the defini-tions presented so far.

Properties of the Dot Product

For all vectors u, v, and w and real numbers k, the following hold.

(a) u # v = v # u (b) u # 1v + w 2 = u # v + u # w(c) 1u + v 2 # w = u # w + v # w (d) 1ku 2 # v = k 1u # v 2 = u # 1kv 2(e) 0 # u = 0 (f) u # u = ∣u ∣2

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340 CHAPTER 7 Applications of Trigonometry and Vectors

For example, to prove the first part of property (d),1k u2 # v = k1u # v2,we let u = 8a, b9 and v = 8c, d9 .

1k u2 # v = Ak8a, b9 B # 8c, d9 Substitute. = 8ka, kb9 # 8c, d9 Multiply by scalar k. = kac + kbd Dot product

= k1ac + bd2 Distributive property = kA 8a, b9 # 8c, d9 B Dot product = k1u # v2 Substitute.

The proofs of the remaining properties are similar.The dot product of two vectors can be positive,

0, or negative. A geometric interpretation of the dot product explains when each of these cases occurs. This interpretation involves the angle between the two vectors.

Consider the two vectors u = 8a1, a29 and v = 8b1, b29 , as shown in Figure 43. The angle U between u and v is defined to be the angle having the two vectors as its sides for which 0° … u … 180°.

We can use the law of cosines to develop a formula to find angle u in Figure 43.

+u - v +2 = +u +2 + +v +2 - 2 +u + +v + cos u Law of cosines applied to

Figure 43

A21a1 - b122 + 1a2 - b222 B2 = A2a1 2 + a2 2 B2 + A2b1 2 + b2 2 B2 Magnitude of a vector

- 2 +u + +v + cos u

a1 2 - 2a1b1 + b1 2 + a2 2 - 2a2 b2 + b2 2 Square.

= a1 2 + a2 2 + b1 2 + b2 2 - 2 +u + +v + cos u

-2a1b1 - 2a2 b2 = -2 +u + +v + cos u Subtract like terms from each side.

a1b1 + a2 b2 = +u + +v + cos u Divide by -2.

u # v = +u + +v + cos u Definition of dot product

cos u = u# v

+u + +v + Divide by +u + +v + and

rewrite.

vu – v

!b1, b2"

!a1, a2"

Figure 43

Geometric Interpretation of Dot Product

If u is the angle between the two nonzero vectors u and v, where 0° … u … 180°, then the following holds.

cos U =u # v∣u ∣ ∣v ∣

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3417.5 Algebraically Defined Vectors and the Dot Product

EXAMPLE 6 Finding the Angle between Two Vectors

Find the angle u between the two vectors.

(a) u = 83, 49 and v = 82, 19 (b) u = 82, -69 and v = 86, 29SOLUTION

(a) cos u = u# v

+u + +v + Geometric interpretation

of the dot product

cos u =83, 49 # 82, 19+83, 49 + +82, 19 + Substitute values.

cos u =3122 + 411229 + 16 # 24 + 1 Use the definitions.

cos u = 10525 Simplify.

cos u ≈ 0.894427191 Use a calculator.

u ≈ 26.57° Use the inverse cosine function.

(b) cos u = u# v

+u + +v + Geometric interpretation

of the dot product

cos u =82, -69 # 86, 29+82, -69 + +86, 29 + Substitute values.

cos u =2162 + 1-6212224 + 36 # 236 + 4 Use the definitions.

cos u = 0 Evaluate. The numerator is equal to 0. u = 90° cos-1 0 = 90°

■✔ Now Try Exercises 53 and 55.

For angles U between 0° and 180°, cos U is positive, 0, or negative when U is less than, equal to, or greater than 90°, respectively. Therefore, the dot prod-uct of nonzero vectors is positive, 0, or negative according to this table.

Dot Product Angle between Vectors

Positive Acute

0 Right

Negative Obtuse

Thus, in Example 6, the vectors in part (a) form an acute angle, and those in part (b) form a right angle. If u # v = 0 for two nonzero vectors u and v, then cos u = 0 and u = 90°. Thus, u and v are perpen-dicular vectors, also called orthogonal vectors. See Figure 44.

u · v = 0

k2, –6l

k6, 2l

Orthogonal vectors

Figure 44

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342 CHAPTER 7 Applications of Trigonometry and Vectors

7.5 Exercises

u = ⟨Ë3, 1⟩

Ë3

CONCEPT PREVIEW Fill in the blank to correctly complete each sentence.

45°

1. The magnitude of vector u is .

2. The direction angle of vector u is .

3. The horizontal component, a, of vector v is .

4. The vertical component, b, of vector v is .

5. The sum of the vectors u = 8 -3, 59 and v = 87, 49 is u + v = . 6. The vector u = 84, -29 is written in i, j form as . 7. The formula for the dot product of the two vectors u = 8a, b9 and v = 8c, d9 is

u # v = . 8. If the dot product of two vectors is a positive number, then the angle between them

is . (acute/obtuse)

Find the magnitude and direction angle for each vector. See Example 1.

9. 815, -89 10. 8 -7, 249 11. 8 -4, 4239 12. 8822, -8229Vector v has the given direction and magnitude. Find the horizontal and vertical compo-nents of v, if u is the direction angle of v from the horizontal. See Example 2.

13. u = 20°, +v + = 50 14. u = 50°, +v + = 26

15. u = 35° 50′, +v + = 47.8 16. u = 27° 30′, +v + = 15.4

17. u = 128.5°, + v + = 198 18. u = 146.3°, + v + = 238

Write each vector in the form 8a, b9 . Round to four decimal places as applicable. See Example 3.

19. y

030°

20. y

60°0

21.

4 140°x

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3437.5 Algebraically Defined Vectors and the Dot Product

22.

3130°

23.

–35°x

24.

–140°

Use the figure to find each vector: (a) u + v (b) u - v (c) -u. Use vector notation as in Example 4.

25.

–4 4

–8

4u v x

y 26. 27.

28. 29. 30.

Given vectors u and v, find: (a) 2u (b) 2u + 3v (c) v - 3u. See Example 4.

31. u = 2i, v = i + j 32. u = - i + 2j, v = i - j

33. u = 8 -1, 29 , v = 83, 09 34. u = 8 -2, -19 , v = 8 -3, 29Given u = 8 -2, 59 and v = 84, 39 , find each of the following. See Example 4.35. u - v 36. v - u 37. -4u 38. -5v

39. 3u - 6v 40. -2u + 4v 41. u + v - 3u 42. 2u + v - 6v

Write each vector in the form ai + bj.

43. 8 -5, 89 44. 86, -39 45. 82, 09 46. 80, -49Find the dot product for each pair of vectors. See Example 5.

47. 86, -19 , 82, 59 48. 8 -3, 89 , 87, -59 49. 85, 29 , 8 -4, 10950. 87, -29 , 84, 149 51. 4i, 5i - 9j 52. 2i + 4j, - jFind the angle between each pair of vectors. See Example 6.

53. 82, 19 , 8 -3, 19 54. 81, 79 , 81, 19 55. 81, 29 , 8 -6, 3956. 84, 09 , 82, 29 57. 3i + 4j, j 58. -5i + 12j, 3i + 2jLet u = 8 -2, 19 , v = 83, 49 , and w = 8 -5, 129 . Evaluate each expression.59. 13u2 # v 60. u # 13v2 61. u # v - u # w 62. u # 1v - w2

8–4

v u x

–4 4

–8

–8 4u

–8

–4 4u v x

8–4

–8

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344 CHAPTER 7 Applications of Trigonometry and Vectors

Determine whether each pair of vectors is orthogonal. See Example 6(b).

63. 81, 29 , 8 -6, 39 64. 81, 19 , 81, -1965. 81, 09 , 822, 09 66. 83, 49 , 86, 8967. 25 i - 2j, -5i + 225 j 68. -4i + 3j, 8i - 6j69. (Modeling) Measuring Rainfall Suppose that vector R models the

amount of rainfall in inches and the direction it falls, and vector A models the area in square inches and the orientation of the opening of a rain gauge, as illustrated in the figure. The total volume V of water collected in the rain gauge is given by

V = +R # A + . This formula calculates the volume of water collected even if the

wind is blowing the rain in a slanted direction or the rain gauge is not exactly vertical. Let R = i - 2j and A = 0.5i + j.(a) Find +R + and +A + to the nearest tenth. Interpret the results.(b) Calculate V to the nearest tenth, and interpret this result.

70. Concept Check In Exercise 69, for the rain gauge to collect the maximum amount of water, what should be true about vectors R and A?

7654321

Relating Concepts

For individual or collaborative investigation (Exercises 71—76)

Consider the two vectors u and v shown. Assume all values are exact. Work Exercises 71–76 in order.

71. Use trigonometry alone (without using vector notation) to find the magnitude and direction angle of u + v. Use the law of cosines and the law of sines in your work.

72. Find the horizontal and vertical components of u, using a calculator.

73. Find the horizontal and vertical components of v, using a calculator.

74. Find the horizontal and vertical components of u + v by adding the results obtained in Exercises 72 and 73.

75. Use a calculator to find the magnitude and direction angle of the vector u + v.

76. Compare the answers in Exercises 71 and 75. What do you notice? Which method of solution do you prefer?

12 u1 = 1108u2 = 2608

NOT TO SCALE

These summary exercises provide practice with applications that involve solving trian-gles and using vectors.

1. Wires Supporting a Flagpole A flagpole stands vertically on a hillside that makes an angle of 20° with the horizontal. Two supporting wires are attached as shown in the figure. What are the lengths of the supporting wires?

Summary Exercises on Applications of Trigonometry and Vectors

30 ft

15 ft15 ft

20°

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345Summary Exercises on Applications of Trigonometry and Vectors

2. Distance between a Pin and a Rod A slider crank mechanism is shown in the figure. Find the distance between the wrist pin W and the connecting rod center C.

3. Distance between Two Lighthouses Two lighthouses are located on a north-south line. From lighthouse A, the bearing of a ship 3742 m away is 129° 43′. From light-house B, the bearing of the ship is 39° 43′. Find the distance between the lighthouses.

4. Hot-Air Balloon A hot-air balloon is rising straight up at the speed of 15.0 ft per sec. Then a wind starts blowing horizontally at 5.00 ft per sec. What will the new speed of the balloon be and what angle with the horizontal will the balloon’s path make?

5. Playing on a Swing Mary is playing with her daughter Brittany on a swing. Starting from rest, Mary pulls the swing through an angle of 40° and holds it briefly before releasing the swing. If Brittany weighs 50 lb, what horizontal force, to the nearest pound, must Mary apply while hold-ing the swing?

6. Height of an Airplane Two observation points A and B are 950 ft apart. From these points the angles of elevation of an airplane are 52° and 57°. See the figure. Find the height of the airplane.

7. Wind and Vectors A wind can be described by v = 6i + 8j, where vector j points north and represents a south wind of 1 mph.

(a) What is the speed of the wind?

(b) Find 3v and interpret the result.

8. Ground Speed and Bearing A plane with an airspeed of 355 mph is on a bearing of 62°. A wind is blowing from west to east at 28.5 mph. Find the ground speed and the actual bearing of the plane.

9. Property Survey A surveyor reported the following data about a piece of property: “The property is triangular in shape, with dimensions as shown in the figure.” Use the law of sines to see whether such a piece of property could exist.

21.9 yd 78.3 yd

Can such a triangle exist?

38° 50!

10. Property Survey A triangular piece of property has the dimensions shown. It turns out that the surveyor did not consider every possible case. Use the law of sines to show why.

C W

Fixed pin Slider25.5°11

.2 cm 28.6 cm

Track

40°

52° 57°

950 ft

21.2 yd 26.5 yd

28° 10!

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346 CHAPTER 7 Applications of Trigonometry and Vectors

Chapter 7 Test Prep

Key Terms

7.1 Side-Angle-Side (SAS)

Angle-Side-Angle (ASA)

Side-Side-Side (SSS) oblique triangle Side-Angle-Angle

(SAA) 7.2 ambiguous case

7.3 semiperimeter 7.4 scalar

vector quantity vector magnitude initial point terminal point parallelogram rule resultant

opposite (of a vector)

zero vector equilibrant

airspeed ground speed

7.5 position vector horizontal

component

vertical component

direction angle unit vector dot product

(inner product)angle between two

vectors orthogonal vectors

New Symbols

OP or OP> vector OP

∣OP ∣ magnitude of vector OP8 a, b 9 position vectori, j unit vectors

Quick ReviewConcepts Examples

Oblique Triangles and the Law of Sines7.1

In triangle ABC, find c, to the nearest hundredth, if A = 44°, C = 62°, and a = 12.00 units. Then find its area.

sin A= c

sin C Law of sines

12.00sin 44°

= csin 62°

Substitute.

c = 12.00 sin 62°sin 44°

Multiply by sin 62° and rewrite.

c ≈ 15.25 units Use a calculator.

For triangle ABC above, apply the appropriate area formula.

& = 12

ac sin B Area formula

& = 12

112.002115.252 sin 74° B = 180° - 44° - 62° B = 74°

& ≈ 87.96 sq units Use a calculator.

Law of SinesIn any triangle ABC, with sides a, b, and c, the following holds.

sin A=

bsin B

sin C

sin A

sin Bb

=sin C

c Alternative

form

Area of a TriangleIn any triangle ABC, the area & is half the product of the lengths of two sides and the sine of the angle between them.

& =12

bc sin A, & =12

ab sin C, & =12

ac sin B

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347CHAPTER 7 Test Prep

Concepts Examples

Solve triangle ABC, given A = 44.5°, a = 11.0 in., and c = 7.0 in.

Find angle C.

sin C7.0

= sin 44.5°11.0

Law of sines

sin C ≈ 0.4460 Solve for sin C.

C ≈ 26.5° Use the inverse sine function.

Another angle with this sine value is

180° - 26.5° ≈ 153.5°.

However, 153.5° + 44.5° 7 180°, so there is only one triangle.

B = 180° - 44.5° - 26.5° Angle sum formula

B = 109° Subtract.

Use the law of sines again to solve for b.

b ≈ 14.8 in.

Ambiguous CaseIf we are given the lengths of two sides and the angle oppo-site one of them (for example, A, a, and b in triangle ABC), then it is possible that zero, one, or two such triangles exist. If A is acute, h is the altitude from C, and

a 6 h 6 b, then there is no triangle.a = h and h 6 b, then there is one triangle (a right triangle).

a Ú b, then there is one triangle.h 6 a 6 b, then there are two triangles.

If A is obtuse and

a … b, then there is no triangle.a 7 b, then there is one triangle.

See the guidelines in this section that illustrate the possible outcomes.

The Ambiguous Case of the Law of Sines7.2

The Law of Cosines

Law of CosinesIn any triangle ABC, with sides a, b, and c, the following hold.

a2 = b2 + c2 − 2bc cos A

b2 = a2 + c2 − 2 ac cos B

c2 = a2 + b2 − 2 ab cos C

Heron’s Area FormulaIf a triangle has sides of lengths a, b, and c, with semi- perimeter

s =12

1a + b + c 2 ,then the area & of the triangle is given by the following.

& = !s 1s − a 2 1s − b 2 1s − c 2

7.3

In triangle ABC, find C if a = 11 units, b = 13 units, and c = 20 units. Then find its area.

c2 = a2 + b2 - 2ab cos C Law of cosines

202 = 112 + 132 - 211121132 cos C Substitute.

400 = 121 + 169 - 286 cos C Square and multiply.

cos C = 400 - 121 - 169-286 Solve for cos C.

cos C ≈ -0.38461538 Use a calculator.

C ≈ 113° Use the inverse cosine function.

The semiperimeter s of the above triangle is

s = 12

111 + 13 + 202 = 22,so the area is

& = 222122 - 112122 - 132122 - 202 & = 66 sq units.

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348 CHAPTER 7 Applications of Trigonometry and Vectors

Concepts Examples

Two forces of 25 newtons and 32 newtons act on a point in a plane. If the angle between the forces is 62°, find the magnitude of the resultant force.

118°

62°

The resultant force divides a parallelogram into two tri-angles. The measure of angle Q in the figure is 118°. We use the law of cosines to find the desired magnitude.

+v +2 = 252 + 322 - 212521322 cos 118° +v +2 ≈ 2400

+v + ≈ 49

The magnitude of the resultant force is 49 newtons.

Vector SumThe sum of two vectors is also a vector. There are two ways to find the sum of two vectors A and B geometrically.

1. The vector with the same initial point as A and the same terminal point as B is the sum A + B.

A + B

2. The diagonal of the parallelogram with the same initial point as A and B is the sum A + B. This is the paral-lelogram rule.

A + B

Geometrically Defined Vectors and Applications7.4

Algebraically Defined Vectors and the Dot Product7.5

Find the magnitude and direction angle of vector u in the figure.

+u + = 3A223 B2 + 22 = 216= 4 Magnitude

tan u = 2223 = 123 # 2323 = 233 , so u = 30°.

For u defined above,

u = 84 cos 30°, 4 sin 30°9 = 8223, 29 . cos 30° = 232 ; sin 30° = 12

Find each of the following.

84, 69 + 8 -8, 39 = 8 -4, 99 58 -2, 19 = 8 -10, 59

- 8 -9, 69 = 89, -69 84, 69 - 8 -8, 39 = 812, 39

If u = 8223, 29 as above, thenu = 223 i + 2j.

Magnitude and Direction Angle of a VectorThe magnitude (length) of vector u = 8a, b9 is given by the following.

∣u ∣ = !a2 + b2The direction angle u satisfies tan u = ba , where a≠ 0.

If u = 8a, b9 has direction angle u, thenu = 8 +u + cos u, +u + sin u9 .

Vector OperationsLet a, b, c, d, and k represent real numbers.

8 a, b 9 + 8 c, d 9 = 8 a + c, b + d 9 k # 8 a, b 9 = 8 ka, kb 9

If u = 8 a1 , a2 9 , then −u = 8−a1 , −a2 9 .8 a, b 9 − 8 c, d 9 = 8 a, b 9 + 1− 8 c, d 9 2 = 8 a − c, b − d 9i, j Form for VectorsIf v = 8a, b9, then

v = ai + bj, where i = 81, 09 and j = 80, 19.

2 30u

u = !2Ë3, 2"

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349CHAPTER 7 Review Exercises

Concepts Examples

Find the dot product.82, 19 # 85, -29 = 2 # 5 + 11-22 = 8

Find the angle u between u = 83, 19 and v = 82, -39 . cos u = u

# v!u ! !v !

Geometric interpretation of the dot product

cos u =3122 + 11-32232 + 12 # 222 + 1-322 Use the definitions.

cos u = 32130 Simplify. cos u ≈ 0.26311741 Use a calculator.

u ≈ 74.7° Use the inverse cosine function.

Dot ProductThe dot product of the two vectors u = 8a, b9 and v = 8c, d9, denoted u # v, is given by the following.

u # v = ac + bd

Geometric Interpretation of the Dot ProductIf u is the angle between the two nonzero vectors u and v, where 0° … u … 180°, then the following holds.

cos U =u # v∣u ∣ ∣ v ∣

Chapter 7 Review ExercisesUse the law of sines to find the indicated part of each triangle ABC.

1. Find b if C = 74.2°, c = 96.3 m, B = 39.5°.

2. Find B if A = 129.7°, a = 127 ft, b = 69.8 ft.

3. Find B if C = 51.3°, c = 68.3 m, b = 58.2 m.

4. Find b if a = 165 m, A = 100.2°, B = 25.0°.

5. Find A if B = 39° 50′, b = 268 m, a = 340 m.

6. Find A if C = 79° 20′, c = 97.4 mm, a = 75.3 mm.

Use the law of cosines to find the indicated part of each triangle ABC.

11. Find A if a = 86.14 in., b = 253.2 in., c = 241.9 in.

12. Find b if B = 120.7°, a = 127 ft, c = 69.8 ft.

13. Find a if A = 51° 20′, c = 68.3 m, b = 58.2 m.

Answer each question.

7. If we are given a, A, and C in a triangle ABC, does the possibility of the ambiguous case exist? If not, explain why.

8. Can triangle ABC exist if a = 4.7, b = 2.3, and c = 7.0? If not, explain why. Answer this question without using trigonometry.

9. Given a = 10 and B = 30° in triangle ABC, for what values of b does A have(a) exactly one value (b) two possible values (c) no value?

10. Why can there be no triangle ABC satisfying A = 140°, a = 5, and b = 7?

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350 CHAPTER 7 Applications of Trigonometry and Vectors

14. Find B if a = 14.8 m, b = 19.7 m, c = 31.8 m.

15. Find a if A = 60°, b = 5.0 cm, c = 21 cm.

16. Find A if a = 13 ft, b = 17 ft, c = 8 ft.

Solve each triangle ABC.

17. A = 25.2°, a = 6.92 yd, b = 4.82 yd 18. A = 61.7°, a = 78.9 m, b = 86.4 m

19. a = 27.6 cm, b = 19.8 cm, C = 42° 30′ 20. a = 94.6 yd, b = 123 yd, c = 109 yd

Find the area of each triangle ABC.

21. b = 840.6 m, c = 715.9 m, A = 149.3° 22. a = 6.90 ft, b = 10.2 ft, C = 35° 10′

23. a = 0.913 km, b = 0.816 km, c = 0.582 km

24. a = 43 m, b = 32 m, c = 51 m

Solve each problem.

25. Distance across a Canyon To measure the distance AB across a canyon for a power line, a surveyor mea-sures angles B and C and the distance BC, as shown in the figure. What is the distance from A to B?

26. Length of a Brace A banner on an 8.0-ft pole is to be mounted on a building at an angle of 115°, as shown in the figure. Find the length of the brace.

27. Height of a Tree A tree leans at an angle of 8.0° from the vertical. From a point 7.0 m from the bottom of the tree, the angle of elevation to the top of the tree is 68°. Find the slanted height x in the figure.

28. Hanging Sculpture A hanging sculp-ture is to be hung in an art gallery with two wires of lengths 15.0 ft and 12.2 ft so that the angle between them is 70.3°. How far apart should the ends of the wire be placed on the ceiling?

Canyon

58.4° 27.9°125 ft

B C

8.0 ft

22°

Brace

115°

68°

8.0°

7.0 m

70.3°15.0 ft12.2 ft

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351CHAPTER 7 Review Exercises

29. Height of a Tree A hill makes an angle of 14.3° with the horizontal. From the base of the hill, the angle of elevation to the top of a tree on top of the hill is 27.2°. The distance along the hill from the base to the tree is 212 ft. Find the height of the tree.

30. Pipeline Position A pipeline is to run between points A and B, which are separated by a protected wetlands area. To avoid the wetlands, the pipe will run from point A to C and then to B. The distances involved are AB = 150 km, AC = 102 km, and BC = 135 km. What angle should be used at point C?

31. Distance between Two Boats Two boats leave a dock together. Each travels in a straight line. The angle between their courses measures 54° 10′. One boat travels 36.2 km per hr, and the other travels 45.6 km per hr. How far apart will they be after 3 hr?

54° 10!

32. Distance from a Ship to a Lighthouse A ship sailing parallel to shore sights a light-house at an angle of 30° from its direction of travel. After the ship travels 2.0 mi farther, the angle has increased to 55°. At that time, how far is the ship from the lighthouse?

30°55°

2.0 mi

33. Area of a Triangle Find the area of the triangle shown in the figure using Heron’s area formula.

y(–8, 6)

(3, 4)

(0, 0)

34. Show that the triangle in Exercise 33 is a right triangle. Then use the formula & = 12 ac sin B, with B = 90°, to find the area.

Use the given vectors to sketch each of the following.

35. a - b

36. a + 3c

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352 CHAPTER 7 Applications of Trigonometry and Vectors

Vector v has the given magnitude and direction angle. Find the horizontal and vertical components of v.

39. !v ! = 964, u = 154° 20′ 40. ! v ! = 50, u = 45° (Give exact values.)

Find the magnitude and direction angle for u rounded to the nearest tenth.

41. u = 8 -9, 129 42. u = 821, -20943. Let v = 2i - j and u = -3i + 2j. Express each in terms of i and j.

(a) 2v + u (b) 2v (c) v - 3u

Find the angle between the vectors. Round to the nearest tenth of a degree. If the vectors are orthogonal, say so.

44. 83, -29 , 8 -1, 39 45. 85, -39 , 83, 59 46. 80, 49 , 8 -4, 49Solve each problem.

47. Weight of a Sled and Passenger Paula and Steve are pulling their daughter Jessie on a sled. Steve pulls with a force of 18 lb at an angle of 10°. Paula pulls with a force of 12 lb at an angle of 15°. Find the magnitude of the resultant force on Jessie and the sled.

15°10°

12 lb

18 lb

Sled

48. Force Placed on a Barge One boat pulls a barge with a force of 100 newtons. Another boat pulls the barge at an angle of 45° to the first force, with a force of 200 newtons. Find the resultant force acting on the barge, to the nearest unit, and the angle between the resultant and the first boat, to the nearest tenth.

49. Direction and Speed of a Plane A plane has an airspeed of 520 mph. The pilot wishes to fly on a bearing of 310°. A wind of 37 mph is blowing from a bearing of 212°. In what direction should the pilot fly, and what will be her ground speed?

50. Angle of a Hill A 186-lb force is required to hold a 2800-lb car on a hill. What angle does the hill make with the horizontal?

51. Incline Force Find the force required to keep a 75-lb sled from sliding down an incline that makes an angle of 27° with the horizontal. (Assume there is no friction.)

52. Speed and Direction of a Boat A boat travels 15 km per hr in still water. The boat is traveling across a large river, on a bearing of 130°. The current in the river, com-ing from the west, has a speed of 7 km per hr. Find the resulting speed of the boat and its resulting direction of travel.

Given two forces and the angle between them, find the magnitude of the resultant force.

37.

52°130 lb

100 lb

38. forces of 142 and 215 newtons, forming an angle of 112°

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353CHAPTER 7 Test

Other Formulas from Trigonometry The following identities involve all six parts of a triangle ABC and are useful for checking answers.

a + b

cos 12 1A - B2sin 12 C

Newton’s formula

a - b

sin 12 1A - B2cos 12 C

Mollweide’s formula

53. Apply Newton’s formula to the given triangle to verify the accuracy of the information.

54. Apply Mollweide’s formula to the given triangle to verify the accuracy of the information.

55. Law of Tangents In addition to the law of sines and the law of cosines, there is a law of tangents. In any triangle ABC,

tan 12 1A - B2tan 12 1A + B2 = a - ba + b .

Verify this law for the triangle ABC with a = 2, b = 223, A = 30°, and B = 60°.

C B

30°

60°a = 7

c = 14b = 7√3

Find the indicated part of each triangle ABC.

1. Find C if A = 25.2°, a = 6.92 yd, and b = 4.82 yd.

2. Find c if C = 118°, a = 75.0 km, and b = 131 km.

3. Find B if a = 17.3 ft, b = 22.6 ft, c = 29.8 ft.

Solve each problem.

4. Find the area of triangle ABC if a = 14, b = 30, and c = 40.

5. Find the area of triangle XYZ shown here.

6. Given a = 10 and B = 150° in triangle ABC, determine the values of b for which A has(a) exactly one value (b) two possible values (c) no value.

Solve each triangle ABC.

7. A = 60°, b = 30 m, c = 45 m 8. b = 1075 in., c = 785 in., C = 38° 30′

Work each problem.

9. Find the magnitude and the direction angle, to the nearest tenth, for the vector shown in the figure.

Chapter 7 Test

YX630°

–6

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354 CHAPTER 7 Applications of Trigonometry and Vectors

10. Use the given vectors to sketch a + b.

11. For the vectors u = 8 -1, 39 and v = 82, -69 , find each of the following.(a) u + v (b) -3v (c) u # v (d) +u +

12. Find the measure of the angle u between u = 84, 39 and v = 81, 59 to the nearest tenth.

13. Show that the vectors u = 8 -4, 79 and v = 8 -14, -89 are orthogonal vectors.Solve each problem.

14. Height of a Balloon The angles of elevation of a balloon from two points A and B on level ground are 24° 50′ and 47° 20′, respectively. As shown in the figure, points A, B, and C are in the same vertical plane and points A and B are 8.4 mi apart. Approximate the height of the balloon above the ground to the nearest tenth of a mile.

15. Horizontal and Vertical Components Find the horizontal and vertical components of the vector with magnitude 569 and direction angle 127.5° from the horizontal. Give your answer in the form 8a, b9 to the nearest unit.

16. Radio Direction Finders Radio direction finders are placed at points A and B, which are 3.46 mi apart on an east-west line, with A west of B. From A, the bearing of a certain illegal pirate radio transmitter is 48°, and from B the bearing is 302°. Find the distance between the transmitter and A to the nearest hundredth of a mile.

17. Height of a Tree A tree leans at an angle of 8.0° from the vertical, as shown in the figure. From a point 8.0 m from the bottom of the tree, the angle of elevation to the top of the tree is 66°. Find the slanted height x in the figure.

18. Walking Dogs on Leashes While Michael is walking his two dogs, Gus and Dotty, they reach a corner and must wait for a WALK sign. Michael is holding the two leashes in the same hand, and the dogs are pulling on their leashes at the angles and forces shown in the figure. Find the magni-tude of the equilibrant force (to the nearest tenth of a pound) that Michael must apply to restrain the dogs.

19. Bearing and Airspeed Find the bearing and airspeed required for a plane to fly 630 mi due north in 3.0 hr if the wind is blowing from a direction of 318° at 15 mph. Approximate the bearing to the nearest degree and the airspeed to the nearest 10 mph.

20. Incline Angle A force of 16.0 lb is required to hold a 50.0-lb wheelbarrow on an incline. What angle does the incline make with the horizontal?

AB47°209

8.4 mi24°509

8.0°

66°

8.0 m

25°35°

15 lb

20 lb

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355

High-resolution computer graphics and complex numbers make it possible to produce the endless self-similarity property of a fractal image.

Complex Numbers

Trigonometric (Polar) Form of Complex Numbers

The Product and Quotient Theorems

De Moivre’s Theorem; Powers and Roots of Complex Numbers

Chapter 8 Quiz

Polar Equations and Graphs

Parametric Equations, Graphs, and Applications

8.1

8.2

8.3

8.4

8.5

8.6

Complex Numbers, Polar Equations, and Parametric Equations8

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356 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Basic Concepts of Complex Numbers There is no real number solution of the equation

x2 = −1

because no real number, when squared, gives -1. To extend the real number system to include solutions of equations of this type, the number i is defined.

8.1 Complex Numbers■ Basic Concepts of

Complex Numbers■ Complex Solutions of

Quadratic Equations (Part 1)

■ Operations on Complex Numbers

■ Complex Solutions of Quadratic Equations (Part 2)

■ Powers of i

Two complex numbers a + bi and c + di are equal provided that their real parts are equal and their imaginary parts are equal.

The following important concepts apply to a complex number a + bi.

1. If b = 0, then a + bi = a, which is a real number. (This means that the set of real numbers is a subset of the set of complex numbers. See Figure 2 on the next page.)

2. If b ≠ 0, then a + bi is a nonreal complex number. Examples: 7 + 2i, -1 - i3. If a = 0 and b ≠ 0, then the nonreal complex number is a pure imaginary

number. Examples: 3i, -16i

The form a + bi (or a + ib) is standard form. (The form a + ib is used to write expressions such as i 25 because 25i could be mistaken for 25i .)

Imaginary Unit i

i = !−1, and therefore i2 = −1.(Note that - i is also a square root of -1.)

Complex Number

If a and b are real numbers, then any number of the form a + bi is a complex number. In the complex number a + bi, a is the real part and b is the imaginary part.*

*In some texts, the term bi is defined to be the imaginary part.

LOOKING AHEAD TO CALCULUSThe letters j and k are also used to

represent 2-1 in calculus and some applications (electronics, for example).

Square roots of negative numbers were not incorporated into an integrated number system until the 16th century. They were then used as solutions of equa-tions. Today, complex numbers are used extensively in science and engineering.

The calculator is in complex number mode. This screen supports the definition of i and shows how the calculator returns the real and imaginary parts of the complex number 7 + 2i.

Figure 1

Equality of Complex Numbers

a + bi = c + di if and only if a = c and b = d.

Some graphing calculators, such as the TI-84 Plus, are capable of working with complex numbers, as seen in Figure 1. ■

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3578.1 Complex Numbers

Figure 2 shows the relationships among subsets of the complex numbers.

For a positive real number a, the expression 2-a is defined as follows.

Complex Numbers a + bi, for a and b Real

Real numbersa + bi, b = 0

Irrationalnumbers

Integers–11, –6, –3, –2, –1

Nonreal complex numbersa + bi, b ≠ 0

Whole numbers

Natural numbers

1, 2, 3, 4, 5, 37, 40

!15

!!8

7 + 2i, 5 – i!3, – i1232+

Pure imaginary numbersa + bi, a = 0 and b ≠ 0

3i, – i, – i, i!523

Rational numbers

, – ,49

117

Figure 2

Meaning of !−aIf a 7 0, then !−a = i !a.

EXAMPLE 1 Writing !−a as i !aWrite as the product of a real number and i, using the definition of 2-a.(a) 2-16 (b) 2-70 (c) 2-48SOLUTION

(a) 2-16 = i216 = 4i (b) 2-70 = i270(c) 2-48 = i248 = i216 # 3 = 4i23 Product rule for radicals: 2n ab = 2n a # 2n b

■✔ Now Try Exercises 21, 23, and 25.

Complex Solutions of Quadratic Equations (Part 1) Such solutions are expressed in the form a + bi or a + ib.

EXAMPLE 2 Solving Quadratic Equations (Complex Solutions)

Solve each equation over the set of complex numbers.

(a) x2 = -9 (b) x2 + 24 = 0SOLUTION

(a) x2 = -9

x = {2-9 Square root property x = { i29 2-a = i2a x = {3i 29 = 3

The solution set is 5{3i6.Take both square

roots, indicated by the { symbol.

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358 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

(b) x2 + 24 = 0

x2 = -24 Subtract 24.

x = {2-24 Square root property x = { i224 2-a = i2a x = { i24 # 26 Product rule for radicals x = {2i26 24 = 2

The solution set is E{2i26 F . ■✔ Now Try Exercises 85 and 87.

CAUTION When working with negative radicands, use the definition !−a = i !a before using any of the other rules for radicals. In par-ticular, the rule 2c # 2d = 2cd is valid only when c and d are not both negative. For example, consider the following.

2-4 # 2-9 = 2i # 3i = 6i2 = -6 Correct 2-4 # 2-9 = 21-421-92 = 236 = 6 Incorrect

Operations on Complex Numbers Products or quotients with negative radicands are simplified by first rewriting2-a as i2a, for a positive number a. Then the properties of real numbers and the fact that i2 = -1 are applied.

EXAMPLE 3 Finding Products and Quotients Involving !−aFind each product or quotient. Simplify the answers.

(a) 2-7 # 2-7 (b) 2-6 # 2-10 (c) 2-202-2 (d) 2-48224SOLUTION

(a) 2-7 # 2-7 = i27 # i27 = i2 # A27 B2 = -1 # 7 i2 = -1; A2a B2 = a = -7 Multiply.

(b) 2-6 # 2-10 = i26 # i210 = i2 # 260 = -124 # 15 = -1 # 2215 = -2215

(d) 2-48224 = i248224 = i B4824 = i22 Quotient rule for radicals

■✔ Now Try Exercises 29, 31, 33, and 35.

First write all square roots in terms of i.

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3598.1 Complex Numbers

With the definitions i2 = -1 and 2-a = i2a for a 7 0, all properties of real numbers are extended to complex numbers.

EXAMPLE 4 Simplifying a Quotient Involving !−aWrite

-8 + 2-1284

in standard form a + bi.

SOLUTION -8 + 2-128

= -8 + 2-64 # 24

Product rule for radicals

= -8 + 8i224

2-64 = 8i =

4 A -2 + 2i22 B4

Factor.

= -2 + 2i22 Lowest terms; standard form■✔ Now Try Exercise 41.

Be sure to factor before dividing.

That is, to add or subtract complex numbers, add or subtract the real parts, and add or subtract the imaginary parts.

Addition and Subtraction of Complex Numbers

For complex numbers a + bi and c + di, the following hold.

1a + bi 2 + 1c + di 2 = 1a + c 2 + 1b + d 2 i 1a + bi 2 − 1c + di 2 = 1a − c 2 + 1b − d 2 i

EXAMPLE 5 Adding and Subtracting Complex Numbers

Find each sum or difference. Write answers in standard form.

(a) 13 - 4i2 + 1-2 + 6i2 (b) 1-4 + 3i2 - 16 - 7i2SOLUTION

(a) 13 - 4i2 + 1-2 + 6i2 Add real Add imaginary parts. parts.

= 33 + 1-224 + 3-4 + 64i = 1 + 2i Standard form

(b) 1-4 + 3i2 - 16 - 7i2 = 1-4 - 62 + 33 - 1-724i Subtract real parts. Subtract imaginary parts. = -10 + 10i Standard form

■✔ Now Try Exercises 47 and 49.

(++)++* (+)+* Commutative, associative and distributive properties

This screen supports the results in Example 5.

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360 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

The product of two complex numbers is found by multiplying as though the numbers were binomials and using the fact that i2 = -1, as follows.1a + bi21c + di2

= ac + adi + bic + bidi FOIL method (Multiply First, Outer, Inner, Last terms.)

= ac + adi + bci + bdi2 Commutative property; Multiply. = ac + 1ad + bc2i + bd1-12 Distributive property; i2 = -1 = 1ac - bd2 + 1ad + bc2i Group like terms.

Multiplication of Complex Numbers

For complex numbers a + bi and c + di, the following holds.1a + bi 2 1c + di 2 = 1ac − bd 2 + 1ad + bc 2 iTo find a given product in routine calculations, it is often easier to multiply

as with binomials and use the fact that i2 = -1.

EXAMPLE 6 Multiplying Complex Numbers

Find each product. Write answers in standard form.

(a) 12 - 3i213 + 4i2 (b) 14 + 3i22 (c) 16 + 5i216 - 5i2SOLUTION

(a) 12 - 3i213 + 4i2 = 2132 + 214i2 - 3i132 - 3i14i2 FOIL method = 6 + 8i - 9i - 12i2 Multiply. = 6 - i - 121-12 Combine like terms; i2 = -1. = 18 - i Standard form

(b) 14 + 3i22 = 42 + 214213i2 + 13i22 Square of a binomial: 1x + y22 = x2 + 2xy + y2 = 16 + 24i + 9i2 Multiply; 13i22 = 32i2. = 16 + 24i + 91-12 i2 = -1 = 7 + 24i Standard form

(c) 16 + 5i216 - 5i2 = 62 - 15i22 Product of the sum and difference of two terms: 1x + y21x - y2 = x2 - y2 = 36 - 251-12 Square; 15i22 = 52i2 = 251-12. = 36 + 25 Multiply. = 61, or 61 + 0i Standard form

■✔ Now Try Exercises 55, 59, and 63.

Remember to add twice the product of the two terms.

This screen supports the results found in Example 6.

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3618.1 Complex Numbers

Example 6(c) showed that 16 + 5i216 - 5i2 = 61. The numbers 6 + 5i and 6 - 5i differ only in the sign of their imaginary parts and are called complex con-jugates. The product of a complex number and its conjugate is always a real number. This product is the sum of the squares of the real and imaginary parts.

The complex conjugate of 6 + 5i is 6 - 5i. Property of Complex Conjugates

For real numbers a and b,1a + bi 2 1a − bi 2 = a2 + b2.EXAMPLE 7 Dividing Complex Numbers

Find each quotient. Write answers in standard form.

(a) 3 + 2i5 - i (b)

SOLUTION

(a) 3 + 2i5 - i

=13 + 2i215 + i215 - i215 + i2 Multiply by the complex conjugate of the denominator in both the numerator and the denominator.

= 15 + 3i + 10i + 2i2

25 - i2 Multiply.

= 13 + 13i26

Combine like terms; i2 = -1.

= 1326

+ 13i26

a + bic =ac +

bic

= 12

+ 12

i Lowest terms; standard form

CHECK a 12

+ 12

ib15 - i2 = 3 + 2i ✓ Quotient * Divisor = Dividend(b)

=31- i2i1- i2 - i is the conjugate of i.

= -3i- i2 Multiply.

= -3i1

- i2 = -1-12 = 1 = -3i, or 0 - 3i Standard form ■✔ Now Try Exercises 73 and 79.

To find the quotient of two complex numbers in standard form, we multi-ply both the numerator and the denominator by the complex conjugate of the denominator.

This screen supports the results in Example 7.

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362 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Complex Solutions of Quadratic Equations (Part 2)

EXAMPLE 9 Simplifying Powers of i

Simplify each power of i.

(a) i15 (b) i -3

SOLUTION

(a) Because i4 = 1, write the given power as a product involving i4.i15 = i12 # i3 = 1i423 # i3 = 131- i2 = - i

(b) Multiply i -3 by 1 in the form of i4 to create the least positive exponent for i.

i -3 = i -3 # 1 = i -3 # i4 = i i4 = 1■✔ Now Try Exercises 97 and 105.

Powers of i Powers of i can be simplified using the facts

i2 = -1 and i4 = 1i222 = 1-122 = 1.Consider the following powers of i.

i1 = i i5 = i4 # i = 1 # i = i i2 = -1 i6 = i4 # i2 = 11-12 = -1 i3 = i2 # i = 1-12 # i = - i i7 = i4 # i3 = 1 # 1- i2 = - ii4 = i2 # i2 = 1-121-12 = 1 i8 = i4 # i4 = 1 # 1 = 1 and so on.

Powers of i cycle through the same four outcomes (i, −1, − i, and 1) because i4 has the same multiplicative property as 1. It follows that a power of i with an exponent that is a multiple of 4 has value 1.

Powers of i can be found on a TI-84 Plus calculator.

EXAMPLE 8 Solving a Quadratic Equation (Complex Solutions)

Solve 9x2 + 5 = 6x over the set of complex numbers.

SOLUTION 9x2 - 6x + 5 = 0 Standard form ax2 + bx + c = 0

x = -b { 2b2 - 4ac2a

Use the quadratic formula.

=-1-62 { 21-622 - 4192152

2192 Substitute a = 9, b = -6, c = 5. = 6 { 2-144

18 Simplify.

= 6 { 12i18

2-144 = 12i =

611 { 2i26 # 3 Factor.

x = 1 { 2i3

Write in lowest terms.

The solution set is E 13 { 23 i F . ■✔ Now Try Exercise 89.

The fraction bar extends

under -b.

Factor first, then divide out the

common factor.

a { bic =

ac { bc i

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3638.1 Complex Numbers

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

1. By definition, i = , and therefore i2 = .

2. In -4 - 8i, the real part is and the imaginary part is .

3. In terms of i, 2-100 = . 4. The complex conjugate of 6 - 2i is .

8.1 Exercises

CONCEPT PREVIEW Determine whether each statement is true or false. If it is false, tell why.

5. Every real number is a complex number.

6. No real number is a pure imaginary number.

7. Every pure imaginary number is a complex number.

8. A number can be both real and complex.

9. There is no real number that is a complex number.

10. A complex number might not be a pure imaginary number.

Concept Check Identify each number as real, complex, pure imaginary, or nonreal com-plex. (More than one of these descriptions will apply.)

11. -4 12. 0 13. 13i 14. -7i 15. 5 + i

16. -6 - 2i 17. p 18. 224 19. 2-25 20. 2-36Write each number as the product of a real number and i. See Example 1.

21. 2-25 22. 2-36 23. 2-10 24. 2-1525. 2-288 26. 2-500 27. -2-18 28. -2-80Find each product or quotient. Simplify the answers. See Example 3.

29. 2-13 # 2-13 30. 2-17 # 2-17 31. 2-3 # 2-832. 2-5 # 2-15 33. 2-302-10 34. 2-702-735.

2-2428 36. 2-54227 37. 2-102-4038.

2-82-72 39. 2-6 # 2-223 40. 2-12 # 2-628Write each number in standard form a + bi. See Example 4.

41. -6 - 2-24

242.

-9 - 2-183

43. 10 + 2-200

44. 20 + 2-8

245.

-3 + 2-1824

46. -5 + 2-50

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364 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Find each sum or difference. Write answers in standard form. See Example 5.

47. 13 + 2i2 + 19 - 3i2 48. 14 - i2 + 18 + 5i2 49. 1-2 + 4i2 - 1-4 + 4i2 50. 1-3 + 2i2 - 1-4 + 2i2 51. 12 - 5i2 - 13 + 4i2 - 1-1 - 9i2 52. 1-4 - i2 - 12 + 3i2 + 16 + 4i2 53. - i22 - 2 - A6 - 4i22 B - A5 - i22 B 54. 327 - A427 - i B - 4i + A -227 + 5i B

Find each quotient. Write answers in standard form. See Example 7.

73. 6 + 2i1 + 2i 74.

14 + 5i3 + 2i

75. 2 - i2 + i

76. 4 - 3i4 + 3i 77.

1 - 3i1 + i 78.

-3 + 4i2 - i

79. -5i

80. -6i

81. 8- i

82. 12- i 83.

23i

84. 59i

Simplify each power of i. See Example 9.

97. i25 98. i29 99. i22 100. i26

101. i23 102. i27 103. i32 104. i40

105. i -13 106. i -14 107. 1

i -11108.

1i -12

Solve each equation over the set of complex numbers. See Examples 2 and 8.

85. x2 = -16 86. x2 = -36

87. x2 + 12 = 0 88. x2 + 48 = 0

89. 3x2 + 2 = -4x 90. 2x2 + 3x = -2

91. x2 - 6x + 14 = 0 92. x2 + 4x + 11 = 0

93. 41x2 - x2 = -7 94. 313x2 - 2x2 = -7 95. x2 + 1 = -x 96. x2 + 2 = 2x

Find each product. Write answers in standard form. See Example 6.

55. 12 + i213 - 2i2 56. 1-2 + 3i214 - 2i2 57. 12 + 4i21-1 + 3i2 58. 11 + 3i212 - 5i2 59. 13 - 2i22 60. 12 + i22 61. 13 + i213 - i2 62. 15 + i215 - i2 63. 1-2 - 3i21-2 + 3i2 64. 16 - 4i216 + 4i2 65. A26 + i B A26 - i B 66. A22 - 4i B A22 + 4i B 67. i13 - 4i213 + 4i2 68. i12 + 7i212 - 7i2 69. 3i12 - i22 70. -5i14 - 3i22 71. 12 + i212 - i214 + 3i2 72. 13 - i213 + i212 - 6i2

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3658.1 Complex Numbers

Concept Check Work each problem.

109. Suppose a friend says that she has discovered a method of simplifying a positive power of i:

“Just divide the exponent by 4. The answer is i raised to the remainder.”

Explain why her method works.

110. Why does the following method of simplifying i -42 work?

i -42 = i -42 # i44 = i2 = -1(Modeling) Impedance Impedance is a measure of the opposition to the flow of alter-nating electrical current found in common electrical outlets. It consists of two parts, resistance and reactance. Resistance occurs when a light bulb is turned on, while reac-tance is produced when electricity passes through a coil of wire like that found in electric motors. Impedance Z in ohms 1Ω2 can be expressed as a complex number, where the real part represents resistance and the imaginary part represents reactance.

For example, if the resistive part is 3 ohms and the reactive part is 4 ohms, then the impedance could be described by the complex number Z = 3 + 4i. In the series circuit shown in the figure, the total impedance will be the sum of the individual impedances. (Source: Wilcox, G. and C. Hesselberth, Electricity for Engineering Technology, Allyn & Bacon.)

111. The circuit contains two light bulbs and two electric motors. Assuming that the light bulbs are pure resistive and the motors are pure reactive, find the total imped-ance in this circuit and express it in the form Z = a + bi.

112. The phase angle u measures the phase difference between the voltage and the cur-rent in an electrical circuit. Angle u (in degrees) can be determined by the equation

tan u = ba . Find u, to the nearest hundredth, for this circuit.

50 Ω 60 Ω

15 Ω 17 Ω

(Modeling) Ohm’s Law Complex numbers are used to describe current I, voltage E, and impedance Z (the opposition to current). These three quantities are related by the equation

E = IZ, which is known as Ohm’s law.

Thus, if any two of these quantities are known, the third can be found. In each exercise, solve the equation E = IZ for the missing variable.

113. I = 8 + 6i, Z = 6 + 3i 114. I = 10 + 6i, Z = 8 + 5i

115. I = 7 + 5i, E = 28 + 54i 116. E = 35 + 55i, Z = 6 + 4i

Work each problem.

117. Show that 222 + 222 i is a square root of i.118. Show that 232 + 12 i is a cube root of i.119. Show that -2 + i is a solution of the equation x2 + 4x + 5 = 0.

120. Show that -3 + 4i is a solution of the equation x2 + 6x + 25 = 0.

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366 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Recall that the sum of the two complex numbers 4 + i and 1 + 3i is14 + i2 + 11 + 3i2 = 5 + 4i.Graphically, the sum of two complex numbers is represented by the vector that is the resultant of the vectors corresponding to the two numbers. See Figure 4.

8.2 Trigonometric (Polar) Form of Complex Numbers

The Complex Plane and Vector Representation Unlike real numbers, complex numbers cannot be ordered. One way to organize and illustrate them is by using a graph in a rectangular coordinate system.

To graph a complex number such as 2 - 3i, we modify the coordinate system by calling the horizontal axis the real axis and the vertical axis the imaginary axis. Then complex numbers can be graphed in this complex plane, as shown in Figure 3. Each complex number a + bi determines a unique position vector with initial point 10, 0 2 and terminal point 1a, b 2 .

■ The Complex Plane and Vector Representation

■ Trigonometric (Polar) Form

■ Converting between Rectangular and Trigonometric (Polar) Forms

■ An Application of Complex Numbers to Fractals

P2 – 3i

–3

2 Real axis

Imaginary axis

Figure 3

NOTE This geometric representation is the reason that a + bi is called the rectangular form of a complex number. (Rectangular form is also known as standard form.)

1 + 3i

1 2 3 4 5

5 + 4i

4 + i

Figure 4

EXAMPLE 1 Expressing the Sum of Complex Numbers Graphically

Find the sum of 6 - 2i and -4 - 3i. Graph both complex numbers and their resultant.

SOLUTION The sum is found by adding the two numbers.16 - 2i2 + 1-4 - 3i2 = 2 - 5i Add real parts, and add imaginary parts.The graphs are shown in Figure 5. ■✔ Now Try Exercise 17.

x02 4 6–2–4

–2

–6

–4 – 3i

2 – 5i

6 – 2i

–4

Figure 5

x + yi

Figure 6

Relationships among x, y, r, and U�

x = r cos U y = r sin U

r = !x2 + y2 tan U = yx

, if x ≠ 0

Substituting x = r cos u and y = r sin u into x + yi gives the following.

x + yi = r cos u + 1r sin u2i Substitute. = r 1cos u + i sin u2 Factor out r.

Trigonometric (Polar) Form Figure 6 shows the complex number x + yi that corresponds to a vector OP with direction angle u and magnitude r. The fol-lowing relationships among x, y, r, and u can be verified from Figure 6.

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3678.2 Trigonometric (Polar) Form of Complex Numbers

Trigonometric (Polar) Form of a Complex Number

The trigonometric form (or polar form) of the complex number x + yi isr 1cos U + i sin U 2 .

The expression cos u + i sin u is sometimes abbreviated cis U. Using this notation, r 1cos U + i sin U 2 is written r cis U.

The number r is the absolute value (or modulus) of x + yi, and u is the argument of x + yi. In this section, we choose the value of u in the interval 30°, 360°2. Any angle coterminal with u also could serve as the argument.

Converting from Rectangular to Trigonometric Form

Step 1 Sketch a graph of the number x + yi in the complex plane.

Step 2 Find r by using the equation r = 2x2 + y2.Step 3 Find u by using the equation tan u = yx , where x ≠ 0, choosing the

quadrant indicated in Step 1.

CAUTION Errors often occur in Step 3. Be sure to choose the correct quadrant for U by referring to the graph sketched in Step 1.

Converting between Rectangular and Trigonometric (Polar) Forms To convert from rectangular form to trigonometric form, we use the following procedure.

ALGEBRAIC SOLUTION

21cos 300° + i sin 300°2= 2¢ 1

2- i 23

2 ≤ cos 300° = 12 ; sin 300° = - 232

= 1 - i23 Distributive propertyNote that the real part is positive and the imaginary part is negative. This is consistent with 300° being a quad-rant IV angle. For a 300° angle, the reference angle is 60°. Thus the function values cos 300° and sin 300° cor-respond in absolute value to those of cos 60° and sin 60°, with the first of these equal to 12 and the second equal

to - 232 .

EXAMPLE 2 Converting from Trigonometric Form to Rectangular Form

Write 21cos 300° + i sin 300°2 in rectangular form.GRAPHING CALCULATOR SOLUTION

In Figure 7, the first result confirms the algebraic solu-tion, where an approximation for -23 is used for the imaginary part (from the second result). The TI-84 Plus also converts from polar to rectangular form, as seen in the third and fourth results.

Figure 7

■✔ Now Try Exercise 33.

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368 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

(b) See Figure 9. Because -3i = 0 - 3i, we have x = 0 and y = -3.

r = 202 + 1-322 = 20 + 9 = 29 = 3 Substitute.We cannot find u by using tan u = yx because x = 0. However, the graph shows that the least positive value for u is 270°.

-3i = 31cos 270° + i sin 270°2, or 3 cis 270° Trigonometric form■✔ Now Try Exercises 45 and 51.

See Example 3(a). The TI-84 Plus converts from rectangular form to polar form. The value of u in the second result is an approximation for 5p6 , as shown in the third result.

EXAMPLE 3 Converting from Rectangular to Trigonometric Form

Write each complex number in trigonometric form.

(a) -23 + i (Use radian measure.) (b) -3i (Use degree measure.)SOLUTION

(a) We start by sketching the graph of -23 + i in the complex plane, as shown in Figure 8. Next, we use x = -23 and y = 1 to find r and u.

r = 2x2 + y2 = 4A -23 B2 + 12 = 23 + 1 = 2 tan u =

= 1-23 = - 123 # 2323 = - 233Rationalize the denominator.

Because tan u = - 233 , the reference angle for u in radians is p6 . From the graph, we see that u is in quadrant II, so u = p - p6 =

5p6 .

-23 + i = 2 acos 5p6

+ i sin 5p6b , or 2 cis 5p

x0–2

y = 1r = 2

u =

–√3 + i

x = – √3

5p6

Figure 8

r = 30 – 3i

u = 270°

Figure 9

Compare to the result in Example 3(b). The angle -90° is coterminal with 270°. The calculator returns u values between -180° and 180°.

EXAMPLE 4 Converting between Trigonometric and Rectangular Forms Using Calculator Approximations

Write each complex number in its alternative form, using calculator approxima-tions as necessary.

(a) 61cos 125° + i sin 125°2 (b) 5 - 4iSOLUTION

(a) Because 125° does not have a special angle as a reference angle, we cannot find exact values for cos 125° and sin 125°.

61cos 125° + i sin 125°2 ≈ 61-0.5735764364 + 0.8191520443i2 Use a calculator set to degree mode. ≈ -3.4415 + 4.9149i Four decimal places

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3698.2 Trigonometric (Polar) Form of Complex Numbers

(b) A sketch of 5 - 4i shows that u must be in quadrant IV. See Figure 10.

r = 252 + 1-422 = 241 and tan u = - 45

Use a calculator to find that one measure of u is -38.66°. In order to express u in the interval 30, 360°2, we find u = 360° - 38.66° = 321.34°.

5 - 4i = 241 cis 321.34°■✔ Now Try Exercises 57 and 61.

–4

5 – 4i

u = 321.34° 50

r = √41

Figure 10

EXAMPLE 5 Deciding Whether a Complex Number Is in the Julia Set

The fractal called the Julia set is shown in Figure 11. To determine whether a complex number z = a + bi is in this Julia set, perform the following sequence of calculations.

z2 - 1, 1z2 - 122 - 1, 31z2 - 122 - 142 - 1, . . .If the absolute values of any of the resulting complex numbers exceed 2, then the complex number z is not in the Julia set. Otherwise z is part of this set and the point 1a, b2 should be shaded in the graph.

Figure 11

Determine whether each number belongs to the Julia set.

(a) z = 0 + 0i (b) z = 1 + 1iSOLUTION

(a) Here z = 0 + 0i = 0,

z2 - 1 = 0 2 - 1 = -1, 1z2 - 122 - 1 = 1-122 - 1 = 0, 31z2 - 122 - 142 - 1 = 0 2 - 1 = -1, and so on.

We see that the calculations repeat as 0, -1, 0, -1, and so on. The absolute values are either 0 or 1, which do not exceed 2, so 0 + 0i is in the Julia set and the point 10, 02 is part of the graph.

An Application of Complex Numbers to Fractals At its basic level, a fractal is a unique, enchanting geometric figure with an endless self-similarity property. A fractal image repeats itself infinitely with ever-decreasing dimensions. If we look at smaller and smaller portions, we will continue to see the whole—it is much like looking into two parallel mirrors that are facing each other.

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370 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

(b) For z = 1 + 1i, we have the following.z2 - 1

= 11 + i22 - 1 Substitute for z; 1 + 1i = 1 + i. = 11 + 2i + i22 - 1 Square the binomial; 1x + y22 = x2 + 2xy + y2. = -1 + 2i i2 = -1

The absolute value is 21-122 + 22 = 25.Because 25 is greater than 2, the number 1 + 1i is not in the Julia set and 11, 12 is not part of the graph.

■✔ Now Try Exercise 67.

CONCEPT PREVIEW For each complex number shown, give (a) its rectangular form and (b) its trigonometric (polar) form with r 7 0, 0° … u 6 360°.

1. Imaginary

–2

–1

–1–2 1 2Real0

2. Imaginary

–2

–1

–1–2 1 2Real0

4. Imaginary

–2

–1

–1–2 1 2Real0

5. Imaginary

–2

–1

–1–2 1 2Real0

Imaginary

–2

–1

–1–2 1 2Real0

Imaginary

–2

–1

–2 1 2Real0

–√3

8.2 Exercises

CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.

7. The absolute value (or modulus) of a complex number represents the of the vector representing it in the complex plane.

8. The geometric interpretation of the argument of a complex number is the formed by the vector and the positive -axis.

Graph each complex number. See Example 1.

9. -3 + 2i 10. 6 - 5i 11. 22 + 22i 12. 2 - 2i2313. -4i 14. 3i 15. -8 16. 2

Find the sum of each pair of complex numbers. In Exercises 17–20, graph both complex numbers and their resultant. See Example 1.

17. 4 - 3i, -1 + 2i 18. 2 + 3i, -4 - i 19. 5 - 6i, -5 + 3i

20. 7 - 3i, -4 + 3i 21. -3, 3i 22. 6, -2i

23. -5 - 8i, -1 24. 4 - 2i, 5 25. 7 + 6i, 3i

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3718.2 Trigonometric (Polar) Form of Complex Numbers

26. 2 + 6i, -2i 27. 12

+ 23

i, 23

+ 12

i 28. - 15

+ 27

i, 37

- 34

Write each complex number in trigonometric form r1cos u + i sin u2, with u in the inter-val 30°, 360°2. See Example 3.43. -3 - 3i23 44. 1 + i23 45. 23 - i 46. 423 + 4i47. -5 - 5i 48. -2 + 2i 49. 2 + 2i 50. 4 + 4i51. 5i 52. -2i 53. -4 54. 7

Write each complex number in its alternative form, using a calculator to approximate answers to four decimal places as necessary. See Example 4.

Rectangular Form Trigonometric Form

55. 2 + 3i

56. cos 35° + i sin 35°

57. 31cos 250° + i sin 250°258. -4 + i

59. 12i

60. 3 cis 180°

61. 3 + 5i

62. cis 110.5°

Concept Check The complex number z, where z = x + yi, can be graphed in the plane as 1x, y2. Describe the graphs of all complex numbers z satisfying the given conditions.63. The absolute value of z is 1. 64. The real and imaginary parts of z are equal.

65. The real part of z is 1. 66. The imaginary part of z is 1.

Julia Set Refer to Example 5.

67. Is z = -0.2i in the Julia set?

68. The graph of the Julia set in Figure 11 appears to be symmetric with respect to both the x-axis and the y-axis. Complete the following to show that this is true.

(a) Show that complex conjugates have the same absolute value.

(b) Compute z1 2 - 1 and z2 2 - 1, where z1 = a + bi and z2 = a - bi.(c) Discuss why if 1a, b2 is in the Julia set, then so is 1a, -b2.(d) Conclude that the graph of the Julia set must be symmetric with respect to the

x-axis.

(e) Using a similar argument, show that the Julia set must also be symmetric with respect to the y-axis.

Write each complex number in rectangular form. See Example 2.

29. 21cos 45° + i sin 45°2 30. 41cos 60° + i sin 60°233. 41cos 240° + i sin 240°2 34. 21cos 330° + i sin 330°235. 3 cis 150° 36. 2 cis 30° 37. 5 cis 300°

38. 6 cis 135° 39. 22 cis 225° 40. 23 cis 315°41. 41cos1-30°2 + i sin1-30°22 42.221cos1-60°2 + i sin1-60°2231. 101cos 90° + i sin 90°2 32. 81cos 270° + i sin 270°2

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372 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Concept Check Identify the geometric condition (A, B, or C) that implies the situation.

A. The corresponding vectors have opposite directions.

B. The terminal points of the vectors corresponding to a + bi and c + di lie on a horizontal line.

C. The corresponding vectors have the same direction.

69. The difference between two nonreal complex numbers a + bi and c + di is a real number.

70. The absolute value of the sum of two complex numbers a + bi and c + di is equal to the sum of their absolute values.

71. The absolute value of the difference of two complex numbers a + bi and c + di is equal to the sum of their absolute values.

72. Concept Check Show that z and iz have the same absolute value. How are the graphs of these two numbers related?

8.3 The Product and Quotient Theorems

Products of Complex Numbers in Trigonometric Form Using the FOIL method to multiply complex numbers in rectangular form, we find the prod-

uct of 1 + i23 and -223 + 2i as follows.A1 + i23 B A -223 + 2i B = -223 + 2i - 2i132 + 2i223 FOIL method = -223 + 2i - 6i - 223 i2 = -1 = -423 - 4i Combine like terms.

We can also find this same product by first converting the complex numbers 1 + i23 and -223 + 2i to trigonometric form. 1 + i23 = 21cos 60° + i sin 60°2 -223 + 2i = 41cos 150° + i sin 150°2

If we multiply the trigonometric forms and use identities for the cosine and the sine of the sum of two angles, then the result is as follows.321cos 60° + i sin 60°24341cos 150° + i sin 150°24

= 2 # 41cos 60° # cos 150° + i sin 60° # cos 150° + i cos 60° # sin 150° + i2 sin 60° # sin 150°2

= 831cos 60° # cos 150° - sin 60° # sin 150°2 i2 = -1; Factor out i.

+ i1sin 60° # cos 150° + cos 60° # sin 150°24 = 83cos160° + 150°2 + i sin160° + 150°24

cos1A + B2 = cos A # cos B - sin A # sin B; sin1A + B2 = sin A # cos B + cos A # sin B

= 81cos 210° + i sin 210°2 Add.The absolute value of the product, 8, is equal to the product of the absolute val-ues of the factors, 2 # 4. The argument of the product, 210°, is equal to the sum of the arguments of the factors, 60° + 150°.

■ Products of Complex Numbers in Trigonometric Form

■ Quotients of Complex Numbers in Trigonometric Form

With the TI-84 Plus calculator in complex and degree modes, the MATH menu can be used to find the angle and the magnitude (absolute value) of a complex number.

Multiply the absolute values and the binomials.

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3738.3 The Product and Quotient Theorems

The product obtained when multiplying by the first method is the rectangu-lar form of the product obtained when multiplying by the second method.

81cos 210° + i sin 210°2= 8a - 23

2- 1

2 ib cos 210° = - 232 ; sin 210° = - 12

= -423 - 4i Rectangular formProduct Theorem

If r11cos u1 + i sin u12 and r21cos u2 + i sin u22 are any two complex num-bers, then the following holds.3r1 1cos U1 + i sin U1 2 4 # 3r2 1cos U2 + i sin U2 2 4

= r1 r2 3cos 1U1 + U2 2 + i sin 1U1 + U2 2 4In compact form, this is written1r1 cis U1 2 1r2 cis U2 2 = r1 r2 cis 1U1 + U2 2 .

That is, to multiply complex numbers in trigonometric form, multiply their absolute values and add their arguments.

EXAMPLE 1 Using the Product Theorem

Find the product of 31cos 45° + i sin 45°2 and 21cos 135° + i sin 135°2. Write the answer in rectangular form.

SOLUTION331cos 45° + i sin 45°24321cos 135° + i sin 135°24 Write as a product.= 3 # 2 3cos145° + 135°2 + i sin145° + 135°24 Product theorem= 61cos 180° + i sin 180°2 Multiply and add.= 61-1 + i # 02 cos 180° = -1; sin 180° = 0= -6 Rectangular form

■✔ Now Try Exercise 11.

Quotients of Complex Numbers in Trigonometric Form The rectan-

gular form of the quotient of 1 + i23 and -223 + 2i is found as follows.1 + i23

-223 + 2i=A1 + i23 B A -223 - 2i BA -223 + 2i B A -223 - 2i B Multiply both numerator and denominator by the conjugate of the denominator.

= -223 - 2i - 6i - 2i22312 - 4i2 FOIL method; 1x + y21x - y2 = x2 - y2

= -8i16

Simplify.

= - 12

i Lowest terms

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374 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Writing 1 + i23, -223 + 2i, and - 12 i in trigonometric form gives 1 + i23 = 21cos 60° + i sin 60°2, -223 + 2i = 41cos 150° + i sin 150°2, Use r =2x2 + y2 and tan u = yx .and - 1

2 i = 1

2 3cos1-90°2 + i sin1-90°24. -90° can be replaced by 270°.

Quotient Theorem

If r11cos u1 + i sin u12 and r21cos u2 + i sin u22 are any two complex num-bers, where r21cos u2 + i sin u22 ≠ 0, then the following holds.

r1 1cos U1 + i sin U1 2r2 1cos U2 + i sin U2 2 = r1r2 3cos 1U1 − U2 2 + i sin 1U1 − U2 2 4

In compact form, this is written

r1 cis U1r2 cis U2

=r1r2

cis 1U1 − U2 2 .That is, to divide complex numbers in trigonometric form, divide their abso-lute values and subtract their arguments.

EXAMPLE 2 Using the Quotient Theorem

Find the quotient 10 cis1-60°2

5 cis 150° . Write the answer in rectangular form.

SOLUTION

10 cis1-60°25 cis 150°

= 105

cis1-60° - 150°2 Quotient theorem = 2 cis1-210°2 Divide and subtract. = 2 3cos1-210°2 + i sin1-210°24 Rewrite. = 2 c - 23

2+ i a 1

2b d cos1-210°2 = - 232 ;

sin1-210°2 = 12 = -23 + i Distributive property

■✔ Now Try Exercise 21.

Here, the absolute value of the quotient, 12 , is the quotient of the two absolute values, 24 =

12 . The argument of the quotient, -90°, is the difference of the two

arguments,

60° - 150° = -90°.

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3758.3 The Product and Quotient Theorems

EXAMPLE 3 Using the Product and Quotient Theorems with a Calculator

Use a calculator to find the following. Write the answers in rectangular form.

(a) 19.3 cis 125.2°212.7 cis 49.8°2 (b) 10.42 Acos 3p4 + i sin 3p4 B5.21 Acos p5 + i sin p5 B

SOLUTION

(a) 19.3 cis 125.2°212.7 cis 49.8°2 = 9.312.72 cis1125.2° + 49.8°2 Product theorem = 25.11 cis 175° Multiply. Add. = 25.111cos 175° + i sin 175°2 Equivalent form ≈ 25.113-0.99619470 + i10.0871557424 Use a calculator. ≈ -25.0144 + 2.1885i Rectangular form

Multiply the absolute values and add the arguments.

(b) 10.42Acos 3p4 + i sin 3p4 B

5.21Acos p5 + i sin p5 B = 10.42

5.21 c cos a 3p

4- p

5b + i sin a 3p

4- p

5b d Quotient theorem

= 2 acos 11p20

+ i sin 11p20b 3p4 = 15p20 ; p5 = 4p20

≈ -0.3129 + 1.9754i Rectangular form■✔ Now Try Exercises 31 and 33.

Divide the absolute values and subtract

the arguments.

CONCEPT PREVIEW Fill in the blanks to correctly complete each problem.

1. When multiplying two complex numbers in trigonometric form, we their absolute values and their arguments.

2. When dividing two complex numbers in trigonometric form, we their absolute values and their arguments.

3. 351cos 150° + i sin 150°24321cos 30° + i sin 30°24 = 1cos + i sin 2 = + i

4. 61cos 120° + i sin 120°221cos 30° + i sin 30°2

= 1cos + i sin 2 = + i

8.3 Exercises

5. cis1-1000°2 # cis 1000° = cis = + i

6. 5 cis 50,000°cis 50,000°

= 5 cis = + i

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376 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Find each product. Write answers in rectangular form. See Example 1.

7. 331cos 60° + i sin 60°24321cos 90° + i sin 90°24 8. 341cos 30° + i sin 30°24351cos 120° + i sin 120°24 9. 341cos 60° + i sin 60°24361cos 330° + i sin 330°24 10. 381cos 300° + i sin 300°24351cos 120° + i sin 120°24 11. 321cos 135° + i sin 135°24321cos 225° + i sin 225°24 12. 381cos 210° + i sin 210°24321cos 330° + i sin 330°24 13. A23 cis 45° B A23 cis 225° B 14. A26 cis 120° B C26 cis1-30°2 D 15. 15 cis 90°213 cis 45°2 16. 13 cis 300°217 cis 270°2Find each quotient. Write answers in rectangular form. In Exercises 23–28, first convert the numerator and the denominator to trigonometric form. See Example 2.

17. 41cos 150° + i sin 150°221cos 120° + i sin 120°2 18. 241cos 150° + i sin 150°221cos 30° + i sin 30°2

19. 101cos 50° + i sin 50°251cos 230° + i sin 230°2 20. 121cos 23° + i sin 23°261cos 293° + i sin 293°2

21. 3 cis 305°9 cis 65°

22. 16 cis 310°8 cis 70°

23. 823 + i 24. 2i-1 - i23 25. - i1 + i

26. 1

2 - 2i27.

226 - 2i2222 - i26 28. -322 + 3i2626 + i22Use a calculator to perform the indicated operations. Write answers in rectangular form, expressing real and imaginary parts to four decimal places. See Example 3.

29. 32.51cos 35° + i sin 35°2433.01cos 50° + i sin 50°2430. 34.61cos 12° + i sin 12°2432.01cos 13° + i sin 13°2431. 112 cis 18.5°213 cis 12.5°2 32. 14 cis 19.25°217 cis 41.75°233.

45 Acos 2p3 + i sin 2p3 B22.5 Acos 3p5 + i sin 3p5 B 34. 30 Acos 2p5 + i sin 2p5 B10 Acos p7 + i sin p7 B

35. c 2 cis 5p9d 2 36. c 24.3 cis 7p

12d 2

Work each problem.

37. Note that 1r cis u22 = 1r cis u21r cis u2 = r2 cis1u + u2 = r2 cis 2u. Explain how we can square a complex number in trigonometric form. (In the next section, we will develop this idea more fully.)

38. Without actually performing the operations, state why the following products are the same. 321cos 45° + i sin 45°24 # 351cos 90° + i sin 90°24and C 23cos1-315°2 + i sin1-315°24 D # C 53cos1-270°2 + i sin1-270°24 D

39. Show that 1z =1r 1cos u - i sin u2, where z = r 1cos u + i sin u2.

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3778.3 The Product and Quotient Theorems

40. The complex conjugate of r 1cos u + i sin u2 is r 1cos u - i sin u2. Use these trigo-nometric forms to show that the product of complex conjugates is always a real number.

(Modeling) Electrical Current Solve each problem.

41. The alternating current in an electric inductor is I = EZ amperes, where E is voltage and Z = R + XLi is impedance. If E = 81cos 20° + i sin 20°2, R = 6, and XL = 3, find the current. Give the answer in rectangular form, with real and imaginary parts to the nearest hundredth.

42. The current I in a circuit with voltage E, resistance R, capacitive reactance Xc, and inductive reactance XL is

I = ER + 1XL - Xc2i .

Find I if E = 121cos 25° + i sin 25°2, R = 3, XL = 4, and Xc = 6. Give the answer in rectangular form, with real and imaginary parts to the nearest tenth.

(Modeling) Impedance In the parallel electrical circuit shown in the figure, the impedance Z can be calculated using the equation

Z = 11Z1

+ 1Z2

where Z1 and Z2 are the impedances for the branches of the circuit.

43. If Z1 = 50 + 25i and Z2 = 60 + 20i, approximate Z to the nearest hundredth.

44. Determine the angle u, to the nearest hundredth, for the value of Z found in Exercise 43.

50 Ω25 Ω

20 Ω60 Ω

Relating Concepts

For individual or collaborative investigation (Exercises 45–52)

Consider the following complex numbers, and work Exercises 45–52 in order.

w = -1 + i and z = -1 - i

45. Multiply w and z using their rectangular forms and the FOIL method. Leave the product in rectangular form.

46. Find the trigonometric forms of w and z.

47. Multiply w and z using their trigonometric forms and the method described in this section.

48. Use the result of Exercise 47 to find the rectangular form of wz. How does this compare to the result in Exercise 45?

49. Find the quotient wz using their rectangular forms and multiplying both the numerator and the denominator by the conjugate of the denominator. Leave the quotient in rectangular form.

50. Use the trigonometric forms of w and z, found in Exercise 46, to divide w by z using the method described in this section.

51. Use the result in Exercise 50 to find the rectangular form of wz .

52. How does the result in Exercise 51 compare to the result in Exercise 49?

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378 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

8.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers

Powers of Complex Numbers (De Moivre’s Theorem) Because raising a number to a positive integer power is a repeated application of the product rule, it would seem likely that a theorem for finding powers of complex numbers exists. Consider the following.3r 1cos u + i sin u242

= 3r 1cos u + i sin u243r 1cos u + i sin u24 a2 = a # a = r # r3cos1u + u2 + i sin1u + u24 Product theorem = r21cos 2u + i sin 2u2 Multiply and add.

In the same way,3r 1cos u + i sin u243 is equivalent to r31cos 3u + i sin 3u2.These results suggest the following theorem for positive integer values of n. Although the theorem is stated and can be proved for all n, we use it only for positive integer values of n and their reciprocals.

■ Powers of Complex Numbers (De Moivre’s Theorem)

■ Roots of Complex Numbers

Abraham De Moivre (1667–1754)

Named after this French expatriate friend of Isaac Newton, De Moivre’s theorem relates complex numbers and trigonometry.

De Moivre’s Theorem

If r 1cos u + i sin u2 is a complex number, and if n is any real number, then the following holds.3r 1cos U + i sin U 2 4n = rn 1cos nU + i sin nU 2In compact form, this is written3r cis U 4n = rn 1cis nU 2 .

EXAMPLE 1 Finding a Power of a Complex Number

Find A1 + i23 B8 and write the answer in rectangular form.SOLUTION Using earlier methods, write 1 + i23 in trigonometric form.

21cos 60° + i sin 60°2 Trigonometric form of 1 + i 23Now, apply De Moivre’s theorem.A1 + i23 B8

= 321cos 60° + i sin 60°248 Trigonometric form = 283cos18 # 60°2 + i sin18 # 60°24 De Moivre’s theorem = 2561cos 480° + i sin 480°2 Apply the exponent and multiply. = 2561cos 120° + i sin 120°2 480° and 120° are coterminal. = 256 ¢ - 1

2+ i 23

2 ≤ cos 120° = - 12 ; sin 120° = 232

= -128 + 128i23 Rectangular form■✔ Now Try Exercise 13.

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3798.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers

Roots of Complex Numbers Every nonzero complex number has exactly n distinct complex nth roots. De Moivre’s theorem can be extended to find all nth roots of a complex number.

To find the three complex cube roots of 81cos 135° + i sin 135°2, for example, look for a complex number, say r 1cos a + i sin a2, that will satisfy3r 1cos a + i sin a243 = 81cos 135° + i sin 135°2.

By De Moivre’s theorem, this equation becomes

r31cos 3a + i sin 3a2 = 81cos 135° + i sin 135°2.Set r3 = 8 and cos 3a + i sin 3a = cos 135° + i sin 135°, to satisfy this equa-tion. The first of these conditions implies that r = 2, and the second implies that

cos 3a = cos 135° and sin 3a = sin 135°.

For these equations to be satisfied, 3a must represent an angle that is coterminal with 135°. Therefore, we must have

3a = 135° + 360° # k, k any integeror a = 135° + 360°

# k3

, k any integer.

Now, let k take on the integer values 0, 1, and 2.

If k = 0, then a = 135° + 360°# 0

3= 45°.

If k = 1, then a = 135° + 360°# 1

3= 495°

3= 165°.

If k = 2, then a = 135° + 360°# 2

3= 855°

3= 285°.

In the same way, a = 405° when k = 3. But note that 405° = 45° + 360°, so sin 405° = sin 45° and cos 405° = cos 45°. Similarly, if k = 4, then a = 525°, which has the same sine and cosine values as 165°. Continuing with larger val-ues of k would repeat solutions already found. Therefore, all of the cube roots (three of them) can be found by letting k = 0, 1, and 2, respectively.

When k = 0, the root is 21cos 45° + i sin 45°2.When k = 1, the root is 21cos 165° + i sin 165°2.When k = 2, the root is 21cos 285° + i sin 285°2.

In summary, we see that 21cos 45° + i sin 45°2, 21cos 165° + i sin 165°2, and 21cos 285° + i sin 285°2 are the three cube roots of 81cos 135° + i sin 135°2.

nth Root

For a positive integer n, the complex number a + bi is an nth root of the complex number x + yi if the following holds.1a + bi 2n = x + yi

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380 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

nth Root Theorem

If n is any positive integer, r is a positive real number, and u is in degrees, then the nonzero complex number r 1cos u + i sin u2 has exactly n distinct nth roots, given by the following.

n!r 1cos A + i sin A 2 or n!r cis A,where

A =U + 360° # k

n , or A =

360° # kn

, k = 0, 1, 2, . . . , n - 1

If u is in radians, then

A =U + 2Pk

n , or A =

2Pkn

, k = 0, 1, 2, . . . , n - 1.

EXAMPLE 2 Finding Complex Roots

Find the two square roots of 4i. Write the roots in rectangular form.

SOLUTION First write 4i in trigonometric form.

4 acos p2

+ i sin p2b Trigonometric form (using radian measure)

Here r = 4 and u = p2 . The square roots have absolute value 24 = 2 and argu-ments as follows.

a =p2

2+ 2pk

2= p

4+ pk

Because there are two square roots, let k = 0 and 1.

If k = 0, then a = p4

+ p # 0 = p4

If k = 1, then a = p4

+ p # 1 = 5p4

Using these values for a, the square roots are 2 cis p4 and 2 cis 5p4 , which can

be written in rectangular form as22 + i22 and -22 - i22.■✔ Now Try Exercise 23(a).

Be careful simplifying here.

This screen confirms the results of Example 2.

EXAMPLE 3 Finding Complex Roots

Find all fourth roots of -8 + 8i23. Write the roots in rectangular form.SOLUTION -8 + 8i23 = 16 cis 120° Write in trigonometric form.Here r = 16 and u = 120°. The fourth roots of this number have absolute value 24 16 = 2 and arguments as follows.

a = 120°4

+ 360°# k

4= 30° + 90° # k

This screen shows how a calculator finds r and u for the number in Example 3.

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3818.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers

Because there are four fourth roots, let k = 0, 1, 2, and 3.

If k = 0, then a = 30° + 90° # 0 = 30°.If k = 1, then a = 30° + 90° # 1 = 120°.If k = 2, then a = 30° + 90° # 2 = 210°.If k = 3, then a = 30° + 90° # 3 = 300°.

Using these angles, the fourth roots are

2 cis 30°, 2 cis 120°, 2 cis 210°, and 2 cis 300°.

These four roots can be written in rectangular form as23 + i, -1 + i23, -23 - i, and 1 - i23.The graphs of these roots lie on a circle with center at the origin and radius 2. See Figure 12. The roots are equally spaced about the circle, 90° apart. (For con-venience, we label the real axis x and the imaginary axis y.)

EXAMPLE 4 Solving an Equation (Complex Roots)

Find all complex number solutions of x5 - 1 = 0. Graph them as vectors in the complex plane.

SOLUTION Write the equation as

x5 - 1 = 0, or x5 = 1.

Because 15 = 1, there is a real number solution, 1, and it is the only one. There are a total of five complex number solutions. To find these solutions, first write 1 in trigonometric form.

1 = 1 + 0i = 11cos 0° + i sin 0°2 Trigonometric formThe absolute value of the fifth roots is 25 1 = 1. The arguments are given by

0° + 72° # k, k = 0, 1, 2, 3, and 4.By using these arguments, we find that the fifth roots are as follows.

Real solution 11cos 0° + i sin 0°2, k = 011cos 72° + i sin 72°2, k = 111cos 144° + i sin 144°2, k = 211cos 216° + i sin 216°2, k = 311cos 288° + i sin 288°2 k = 4

–2x

–2

2 cis 30°30°

2 cis 300°

2 cis 120°

2 cis 210°

Figure 12 ■✔ Now Try Exercise 29.

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382 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

The solution set of the equation can be written as5cis 0°, cis 72°, cis 144°, cis 216°, cis 288°6.The first of these roots is the real number 1. The others cannot easily be expressed in rectangular form but can be approximated using a calculator.

The tips of the arrows representing the five fifth roots all lie on a unit circle and are equally spaced around it every 72°, as shown in Figure 13.

–1x1

–1

1(cos 0° + i sin 0°)

1(cos 144° + i sin 144°)

1(cos 72° + i sin 72°)

1(cos 216° + i sin 216°)

1(cos 288° + i sin 288°)

72°

Figure 13 ■✔ Now Try Exercise 41.

CONCEPT PREVIEW Fill in the blanks to correctly complete each problem.

8.4 Exercises

1. If z = 31cos 30° + i sin 30°2, it follows that

z 3 = 1cos + i sin 2= 1 + i 2= + i, or simply .

2. If we are given

z = 161cos 80° + i sin 80°2,then any fourth root of z has r = , and the fourth root with least positive argument has u = .

3. 3cos 6° + i sin 6°430= cos + i sin = + i

4. Based on the result of Exercise 3,

cos 6° + i sin 6° is a1n2 root of .

CONCEPT PREVIEW Answer each question.

5. How many real tenth roots of 1 exist? 6. How many nonreal complex tenth roots of 1 exist?

Find each power . Write answers in rectangular form. See Example 1.

7. 331cos 30° + i sin 30°243 8. 321cos 135° + i sin 135°244 9. 1cos 45° + i sin 45°28 10. 321cos 120° + i sin 120°243 11. 33 cis 100°43 12. 33 cis 40°4313. A23 + i B5 14. A2 - 2i23 B415. A222 - 2i22 B6 16. ¢22

2- 22

2 i≤8

17. 1-2 - 2i25 18. 1-1 + i27

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3838.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers

For each of the following, (a) find all cube roots of each complex number. Write answers in trigonometric form. (b) Graph each cube root as a vector in the complex plane. See Examples 2 and 3.

19. cos 0° + i sin 0° 20. cos 90° + i sin 90° 21. 8 cis 60°

22. 27 cis 300° 23. -8i 24. 27i

28. 2 - 2i23 29. -223 + 2i 30. 23 - i25. -64 26. 27 27. 1 + i2331. second (square) 32. fourth 33. sixth

Find and graph all specified roots of 1.

34. second (square) 35. third (cube) 36. fourth

Find and graph all specified roots of i.

40. x4 + i = 0 41. x3 - 8 = 0 42. x3 + 27 = 0

37. x3 - 1 = 0 38. x3 + 1 = 0 39. x3 + i = 0

43. x4 + 1 = 0 44. x4 + 16 = 0 45. x4 - i = 0

Find all complex number solutions of each equation. Write answers in trigonometric form. See Example 4.

46. x5 - i = 0 47. x3 - A4 + 4i23 B = 0 48. x4 - A8 + 8i23 B = 0Solve each problem.

49. Solve the cubic equationx3 = 1

by writing it as x3 - 1 = 0, factoring the left side as the difference of two cubes, and using the zero-factor property. Apply the quadratic formula as needed. Then compare the solutions to those of Exercise 37.

50. Solve the cubic equation

x3 = -27

by writing it as x3 + 27 = 0, factoring the left side as the sum of two cubes, and using the zero-factor property. Apply the quadratic formula as needed. Then com-pare the solutions to those of Exercise 42.

51. Mandelbrot Set The fractal known as the Man-delbrot set is shown in the figure. To determine whether a complex number z = a + bi is in this set, perform the following sequence of calcula-tions. Repeatedly compute

z, z2 + z, 1z2 + z22 + z,31z2 + z22 + z42 + z, . . . . In a manner analogous to the Julia set, the complex number z does not belong to the

Mandelbrot set if any of the resulting absolute values exceeds 2. Otherwise z is in the set and the point 1a, b2 should be shaded in the graph. Determine whether the following numbers belong to the Mandelbrot set. (Source: Lauwerier, H., Fractals, Princeton University Press.)

(a) z = 0 + 0i (b) z = 1 - 1i (c) z = -0.5i

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384 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

52. Basins of Attraction The fractal shown in the figure is the solution to Cayley’s problem of determining the basins of attraction for the cube roots of unity. The three cube roots of unity are

w1 = 1, w2 = - 12

+ 232

and w3 = - 12

- 232

A fractal of this type can be generated by repeatedly evaluating the function

ƒ1z2 = 2z3 + 13z2

where z is a complex number. We begin by picking z1 = a + bi and successively computing z2 = ƒ1z12, z3 = ƒ1z22, z4 = ƒ1z32, . . . . Suppose that if the resulting val-ues of ƒ1z2 approach w1, we color the pixel at 1a, b2 red. If they approach w2, we color it blue, and if they approach w3, we color it yellow. If this process continues for a large number of different z1, a fractal similar to the figure will appear. Deter-mine the appropriate color of the pixel for each value of z1. (Source: Crownover, R., Introduction to Fractals and Chaos, Jones and Bartlett Publishers.)

(a) z1 = i (b) z1 = 2 + i (c) z1 = -1 - i

53. The screens here illustrate how a pentagon can be graphed using a graphing calculator.

Note that a pentagon has five sides, and the Tstep is 3605 = 72. The display at the bottom of the graph screen indicates that one fifth root of 1 is 1 + 0i = 1. Use this technique to find all fifth roots of 1, and express the real and imaginary parts in decimal form.

−1.2

−1.8

1.2

1.8

The calculator is in parametric, degree, and connected graph modes.

54. Use the method of Exercise 53 to find the first three of the ten 10th roots of 1.

Use a calculator to find all solutions of each equation in rectangular form.

55. x2 - 3 + 2i = 0 56. x2 + 2 - i = 0

57. x5 + 2 + 3i = 0 58. x3 + 4 - 5i = 0

Relating Concepts

For individual or collaborative investigation (Exercises 59–62)

In earlier work we derived identities, or formulas, for cos 2u and sin 2u. These identities can also be derived using De Moivre’s theorem. Work Exercises 59–62 in order , to see how this is done.

59. De Moivre’s theorem states that 1cos u + i sin u22 = .60. Expand the left side of the equation in Exercise 59 as a binomial and combine

like terms to write the left side in the form a + bi.61. Use the result of Exercise 60 to obtain the double-angle formula for cosine.

62. Repeat Exercise 61, but find the double-angle formula for sine.

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3858.5 Polar Equations and Graphs

1. Find each product or quotient. Simplify the answers.

(a) 2-24 # 2-3 (b) 2-8272 2. Write each of the following in rectangular form for the complex numbers

w = 3 + 5i and z = -4 + i.

(a) w + z (and give a geometric representation)

(b) w - z (c) wz (d) wz

3. Write each of the following in rectangular form.

(a) 11 - i23 (b) i33 4. Solve 3x2 - x + 4 = 0 over the set of complex numbers.

5. Write each complex number in trigonometric (polar) form, where 0° … u 6 360°.

(a) -4i (b) 1 - i23 (c) -3 - i 6. Write each complex number in rectangular form.

(a) 41cos 60° + i sin 60°2 (b) 5 cis 130°(c) 71cos 270° + i sin 270°2 (d) 2 cis 0°

7. Write each of the following in the form specified for the complex numbers

w = 121cos 80° + i sin 80°2 and z = 31cos 50° + i sin 50°2.(a) wz (trigonometric form) (b)

(rectangular form)

8. Find the four complex fourth roots of -16. Write them in both trigonometric and rectangular forms.

Chapter 8 Quiz (Sections 8.1—8.4)

8.5 Polar Equations and Graphs

Polar Coordinate System Previously we have used the rectangular coor-dinate system to graph points and equations. In the rectangular coordinate system, each point in the plane is specified by giving two numbers 1x, y2. These repre-sent the directed distances from a pair of perpendicular axes, the x-axis and the y-axis.

Now we consider the polar coordinate system which is based on a point, called the pole, and a ray, called the polar axis. The polar axis is usually drawn in the direction of the positive x-axis, as shown in Figure 14.

■ Polar Coordinate System

■ Graphs of Polar Equations

■ Conversion from Polar to Rectangular Equations

■ Classification of Polar Equations

Pole

Polar axis

Figure 14

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386 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

In Figure 15 the pole has been placed at the origin of a rectangular coordi-nate system so that the polar axis coincides with the positive x-axis. Point P has rectangular coordinates 1x, y2. Point P can also be located by giving the directed angle u from the positive x-axis to ray OP and the directed distance r from the pole to point P. The ordered pair 1r, u2 gives the polar coordinates of point P. If r 7 0 then point P lies on the terminal side of u, and if r 6 0 then point P lies on the ray pointing in the opposite direction of the terminal side of u, a distance ( r ( from the pole.

Figure 16 shows rectangular axes superimposed on a polar coordinate grid.

P(x, y)P(r, U)

r > 0

Figure 15

(r, u)

(–r, u)

r > 0

Figure 16

Rectangular and Polar Coordinates

If a point has rectangular coordinates 1x, y2 and polar coordinates 1r, u2, then these coordinates are related as follows.

x = r cos U y = r sin U

r2 = x2 + y2 tan U =yx

, if x ≠ 0

EXAMPLE 1 Plotting Points with Polar Coordinates

Plot each point in the polar coordinate system. Then determine the rectangular coordinates of each point.

(a) P12, 30°2 (b) Q a -4, 2p3b (c) R a5, - p

SOLUTION

(a) In the point P12, 30°2, r = 2 and u = 30°, so P is located 2 units from the origin in the positive direction on a ray making a 30° angle with the polar axis, as shown in Figure 17.

We find the rectangular coordinates as follows.

x = r cos u y = r sin u Conversion equationsx = 2 cos 30° y = 2 sin 30° Substitute.

x = 2 ¢232

≤ y = 2 a 12b Evaluate the functions.

x = 23 y = 1 Multiply.The rectangular coordinates are 123, 12.

x0 1

P(2, 30°)

30°

Figure 17

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3878.5 Polar Equations and Graphs

(b) In the point Q A -4, 2p3 B , r is negative, so Q is 4 units in the opposite direction from the pole on an extension of the 2p3 ray. See Figure 18. The rectangular coordinates are

x = -4 cos 2p3

= -4 a - 12b = 2

and y = -4 sin 2p3

= -4 ¢232

≤ = -223.(c) Point R A5, - p4 B is shown in Figure 19. Because u is negative, the angle is

measured in the clockwise direction.

x = 5 cos a - p4b = 522

2 and y = 5 sin a - p

4b = - 522

■✔ Now Try Exercises 13(a), (c), 15(a), (c), and 21(a), (c).

–10

Q(–4, )32P

2p3

Figure 18

–10

R(5, – )4Pp4

–

Figure 19

While a given point in the plane can have only one pair of rectangular coordinates, this same point can have an infinite number of pairs of polar coordinates. For example, 12, 30°2 locates the same point as12, 390°2, 12, -330°2, and 1-2, 210°2.

EXAMPLE 2 Giving Alternative Forms for Coordinates of Points

Determine the following.

(a) Three other pairs of polar coordinates for the point P13, 140°2(b) Two pairs of polar coordinates for the point with rectangular coordinates 1-1, 12SOLUTION

(a) Three pairs that could be used for the point are 13, -220°2, 1-3, 320°2, and 1-3, -40°2. See Figure 20.

(b) As shown in Figure 21, the point 1-1, 12 lies in the second quadrant. Because tan u = 1-1 = -1, one possible value for u is 135°. Also,

r = 2x2 + y2 = 21-122 + 12 = 22.Two pairs of polar coordinates are A22, 135° B and A -22, 315° B .

■✔ Now Try Exercises 13(b), 15(b), 21(b), and 25.

0– 40°–220°

140°

320°

Figure 20

(–1, 1)

Figure 21

LOOKING AHEAD TO CALCULUSTechniques studied in calculus

associated with derivatives and

integrals provide methods of finding

slopes of tangent lines to polar curves,

areas bounded by such curves, and

lengths of their arcs.

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388 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Graphs of Polar Equations An equation in the variables x and y is a rectangular (or Cartesian) equation. An equation in which r and u are the vari-ables instead of x and y is a polar equation.

r = 3 sin u, r = 2 + cos u, r = u Polar equations

Although the rectangular forms of lines and circles are the ones most often encountered, they can also be defined in terms of polar coordinates. The polar equation of the line ax + by = c can be derived as follows.

Line: ax + by = c Rectangular equation of a line a1r cos u2 + b1r sin u2 = c Convert to polar coordinates. r 1a cos u + b sin u2 = c Factor out r.

r =c

a cos U + b sin U Polar equation of a line

For the circle x2 + y2 = a2, the polar equation can be found in a similar manner.

Circle: x2 + y2 = a2 Rectangular equation of a circle r2 = a2 x2 + y2 = r2

r =ta Polar equation of a circle; r can be negative in polar coordinates.

We use these forms in the next example.

This is the polar equation of

ax + by = c.

These are polar equations of x 2 + y 2 = a2.

–3

–3 3

y = x – 3 (rectangular)

r = (polar)cos U – sin U3

(a)

−10

Polar graphing mode

(b)

Figure 22

EXAMPLE 3 Finding Polar Equations of Lines and Circles

For each rectangular equation, give the equivalent polar equation and sketch its graph.

(a) y = x - 3 (b) x2 + y2 = 4SOLUTION

(a) This is the equation of a line.

y = x - 3

x - y = 3 Write in standard form ax + by = c.

r cos u - r sin u = 3 Substitute for x and y. r 1cos u - sin u2 = 3 Factor out r.

r = 3cos u - sin u Divide by cos u - sin u.

A traditional graph is shown in Figure 22(a), and a calculator graph is shown in Figure 22(b).

(b) The graph of x2 + y2 = 4 is a circle with center at the origin and radius 2.

x2 + y2 = 4 r2 = 4 x2 + y2 = r2

r = 2 or r = -2

The graphs of r = 2 and r = -2 coincide. See Figure 23 on the next page.

In polar coordinates, we may have r 6 0.

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3898.5 Polar Equations and Graphs

■✔ Now Try Exercises 37 and 39.

0 2–2

–2

x2 + y2 = 4 (rectangular) r = 2 (polar)

−4.1

−6.6

4.1

6.6

Polar graphing mode

(a) (b)

Figure 23

To graph polar equations, evaluate r for various values of u until a pattern appears, and then join the points with a smooth curve. The next four examples illustrate curves that are not usually discussed when rectangular coordinates are covered. (Using graphing calculators makes the task of graphing them quite a bit easier than using traditional point-plotting methods.)

EXAMPLE 4 Graphing a Polar Equation (Cardioid)

Graph r = 1 + cos u.

ALGEBRAIC SOLUTION

To graph this equation, find some ordered pairs as in the table. Once the pattern of values of r becomes clear, it is not necessary to find more ordered pairs. The table includes approximate values for cos u and r.

GRAPHING CALCULATOR SOLUTION

We choose degree mode and graph values of u in the interval 30°, 360°4. The screen in Figure 25(a) shows the choices needed to generate the graph in Figure 25(b).

180° 0°

90°

270°

1 2

r = 1 + cos U

Figure 24

(a)

−2.05

−3.3

2.05

3.3

Polar graphing mode

(b)

Figure 25

■✔ Now Try Exercise 47.

U cos U r = 1 + cos U U cos U r = 1 + cos U 0° 1 2 135° -0.7 0.3 30° 0.9 1.9 150° -0.9 0.1 45° 0.7 1.7 180° -1 0 60° 0.5 1.5 270° 0 1 90° 0 1 315° 0.7 1.7120° -0.5 0.5 330° 0.9 1.9

Connect the points in order—from 12, 0°2 to 11.9, 30°2 to 11.7, 45°2 and so on. See Figure 24. This curve is called a cardioid because of its heart shape. The curve has been graphed on a polar grid.

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390 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

The equation r = 3 cos 2u in Example 5 has a graph that belongs to a family of curves called roses.

r = a sin nU and r = a cos nU Equations of roses

The graph has n leaves if n is odd, and 2n leaves if n is even.

The absolute value of a determines the length of the leaves.

EXAMPLE 5 Graphing a Polar Equation (Rose)

Graph r = 3 cos 2u.

SOLUTION Because the argument is 2u, the graph requires a greater number of points than when the argument is just u. We complete the table using selected angle measures through 360° in order to see the pattern of the graph. Approxi-mate values in the table have been rounded to the nearest tenth.

U 2U cos 2U r = 3 cos 2U U 2U cos 2U r = 3 cos 2U 0° 0° 1 3 120° 240° -0.5 -1.515° 30° 0.9 2.6 135° 270° 0 030° 60° 0.5 1.5 180° 360° 1 345° 90° 0 0 225° 450° 0 060° 120° -0.5 -1.5 270° 540° -1 -375° 150° -0.9 -2.6 315° 630° 0 090° 180° -1 -3 360° 720° 1 3

Plotting these points in order gives the graph of a four-leaved rose. Note in Fig-ure 26(a) how the graph is developed with a continuous curve, beginning with the upper half of the right horizontal leaf and ending with the lower half of that leaf. As the graph is traced, the curve goes through the pole four times. This can be seen as a calculator graphs the curve. See Figure 26(b).

180° 0°

90°

270°

3Start

Finish

r = 3 cos 2U

−4.1

−6.6

4.1

6.6

Polar graphing mode

(a) (b)

Figure 26 ■✔ Now Try Exercise 51.

NOTE To sketch the graph of r = 3 cos 2u in polar coordinates, it may be helpful to first sketch the graph of y = 3 cos 2x in rectangular coordinates. The minimum and maximum values of this function may be used to deter-mine the location of the tips of the leaves, and the x-intercepts of this function may be used to determine where the polar graph passes through the pole.

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3918.5 Polar Equations and Graphs

U (radians) r = 2U

-p -6.3 p3 2.1

- p2 -3.1p

2 3.1

- p4 -1.6 p 6.3

0 03p2 9.4

6 1 2p 12.6

Radian measures have been rounded.

EXAMPLE 6 Graphing a Polar Equation (Lemniscate)

Graph r2 = cos 2u.ALGEBRAIC SOLUTION

Complete a table of ordered pairs, and sketch the graph, as in Figure 27. The point 1-1, 0°2, with r negative, may be plotted as 11, 180°2. Also, 1-0.7, 30°2 may be plotted as 10.7, 210°2, and so on.

Values of u for 45° 6 u 6 135° are not included in the table because the corresponding values of cos 2u are negative (quadrants II and III) and so do not have real square roots. Values of u greater than 180° give 2u greater than 360° and would repeat the points already found. This curve is called a lemniscate.

GRAPHING CALCULATOR SOLUTION

To graph r2 = cos 2u with a graphing calculator, first solve for r by considering both square roots. Enter the two polar equations as

r1 = 2cos 2u and r2 = -2cos 2u.See Figures 28(a) and (b).

180° 0°

90°

270°

r2 = cos 2r2 = cos 2U

Figure 27

U 0° 30° 45° 135° 150° 180°2U 0° 60° 90° 270° 300° 360°cos 2U 1 0.5 0 0 0.5 1

r = t!cos 2U {1 {0.7 0 0 {0.7 {1 Settings for the graph below (a)−1.4

−2.2

1.4

2.2

r2 r 2 = -2cos 2u is in red.

(b)

Figure 28

■✔ Now Try Exercise 53.

EXAMPLE 7 Graphing a Polar Equation (Spiral of Archimedes)

Graph r = 2u (with u measured in radians).

SOLUTION Some ordered pairs are shown in the table. Because r = 2u does not involve a trigonometric function of u, we must also consider negative values of u. The graph in Figure 29(a) on the next page is a spiral of Archimedes. Figure 29(b) shows a calculator graph of this spiral.

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392 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

■✔ Now Try Exercise 59.

4 8

r = 2U

p 0

3p2

More of the spiral is shown in this calculator graph, with -8p … u … 8p.

(a) (b)

Figure 29

−50

EXAMPLE 8 Converting a Polar Equation to a Rectangular Equation

For the equation r = 41 + sin u , write an equivalent equation in rectangular coor-dinates, and graph.

SOLUTION r = 41 + sin u Polar equation

r 11 + sin u2 = 4 Multiply by 1 + sin u. r + r sin u = 4 Distributive property

2x2 + y2 + y = 4 Let r = 2x2 + y2 and r sin u = y. 2x2 + y2 = 4 - y Subtract y. x2 + y2 = 14 - y22 Square each side. x2 + y2 = 16 - 8y + y2 Expand the right side.

x2 = -8y + 16 Subtract y2.

x2 = -81y - 22 Rectangular equationThe final equation represents a parabola and is graphed in Figure 30.

■✔ Now Try Exercise 63.

x2 = –8( y – 2)2

4–4 0

Figure 30

The conversion in Example 8 is not necessary when using a graphing cal-culator. Figure 31 shows the graph of r = 41 + sin u , graphed directly with the calculator in polar mode. ■

−10

0° … u … 360°

Figure 31

Classification of Polar Equations The table on the next page summa-rizes common polar graphs and forms of their equations. In addition to circles, lemniscates, and roses, we include limaçons. Cardioids are a special case of limaçons, where 0 ab 0 = 1.

NOTE Some other polar curves are the cissoid, kappa curve, conchoid, trisectrix, cruciform, strophoid, and lituus. Refer to older textbooks on analytic geometry or the Internet to investigate them.

Conversion from Polar to Rectangular Equations

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3938.5 Polar Equations and Graphs

Polar Graphs and Forms of Equations

Circles and Lemniscates

Circles Lemniscates

r = a cos u r = a sin u r2 = a2 sin 2u r2 = a2 cos 2u

Limaçons

r = a { b sin u or r = a { b cos u

6 1ab

= 1 1 6ab

6 2 ab

Ú 2

Rose Curves

2n leaves if n is even, n Ú 2 n leaves if n is odd

n = 2 n = 4 n = 3 n = 5

r = a sin nu r = a cos nu r = a cos nu r = a sin nu

CONCEPT PREVIEW Fill in the blank to correctly complete each sentence.

1. For the polar equation r = 3 cos u, if u = 60°, then r = .

2. For the polar equation r = 2 sin 2u, if u = 15°, then r = .

3. For the polar equation r2 = 4 sin 2u, if u = 15°, then r = .

4. For the polar equation r2 = -2 cos 2u, if u = 60°, then r = .

8.5 Exercises

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394 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

CONCEPT PREVIEW For each point given in polar coordinates, state the quadrant in which the point lies if it is graphed in a rectangular coordinate system.

5. 15, 135°2 6. 12, 60°2 7. 16, -30°2 8. 14.6, 213°2CONCEPT PREVIEW For each point given in polar coordinates, state the axis on which the point lies if it is graphed in a rectangular coordinate system. Also state whether it is on the positive portion or the negative portion of the axis. (For example, 15, 0°2 lies on the positive x-axis.)

9. 17, 360°2 10. 14, 180°2 11. 12, -90°2 12. 18, 450°2For each pair of polar coordinates, (a) plot the point, (b) give two other pairs of polar coordinates for the point, and (c) give the rectangular coordinates for the point. See Examples 1 and 2.

13. 11, 45°2 14. 13, 120°2 15. 1-2, 135°2 16. 1-4, 30°217. 15, -60°2 18. 12, -45°2 19. 1-3, -210°2 20. 1-1, -120°221. a3, 5p

3b 22. a4, 3p

2b 23. a -2, p

3b 24. a -5, 5p

28. 10, -32 29. A22, 22 B 30. A -22, 22 BFor each pair of rectangular coordinates, (a) plot the point and (b) give two pairs of polar coordinates for the point, where 0° … u 6 360°. See Example 2(b).

34. 1-2, 02 35. ¢ - 32

, - 3232

≤ 36. ¢12

, - 232

≤31. ¢232 , 32 ≤ 32. ¢ - 232 , - 12 ≤ 33. 13, 0225. 11, -12 26. 11, 12 27. 10, 32

180° 0°

90°

270°

180° 0°

90°

270°

Concept Check Match each equation with its polar graph from choices A–D.

3180° 0°

90°

270°

1180° 0°

90°

270°

43. r = 3 44. r = cos 3u 45. r = cos 2u 46. r =2

cos u + sin u

37. x - y = 4 38. x + y = -7 39. x2 + y2 = 16

40. x2 + y2 = 9 41. 2x + y = 5 42. 3x - 2y = 6

For each rectangular equation, give the equivalent polar equation and sketch its graph. See Example 3.

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3958.5 Polar Equations and Graphs

Graph each spiral of Archimedes. See Example 7.

59. r = u (Use both positive and nonpositive values.)

60. r = -4u (Use a graphing calculator in a window of 3-30, 304 by 3-30, 304, in radian mode, and u in 3-12p, 12p4.)

Solve each problem.

71. Find the polar equation of the line that passes through the points 11, 0°2 and 12, 90°2.72. Explain how to plot a point 1r, u2 in polar coordinates, if r 6 0 and u is in degrees.Concept Check The polar graphs in this section exhibit symmetry. Visualize an xy-plane superimposed on the polar coordinate system, with the pole at the origin and the polar axis on the positive x-axis. Then a polar graph may be symmetric with respect to the

x-axis (the polar axis), the y-axis A the line u = p2 B , or the origin (the pole).73. Complete the missing ordered pairs in the graphs below.

(a) y

(r, u)

–u

( , )

(b)

( , )( , )

(r, u)

–u

p – u

(c)

( , )( , )

(r, u)

p + u u

63. r = 21 - cos u 64. r =

31 - sin u

67. r = 2 sec u 68. r = -5 csc u

65. r = -2 cos u - 2 sin u 66. r =3

4 cos u - sin u

61. r = 2 sin u 62. r = 2 cos u

For each equation, find an equivalent equation in rectangular coordinates, and graph. See Example 8.

69. r = 2cos u + sin u 70. r =

22 cos u + sin u

47. r = 2 + 2 cos u 48. r = 8 + 6 cos u

51. r = 4 cos 2u 52. r = 3 cos 5u

Graph each polar equation. In Exercises 47–56, also identify the type of polar graph. See Examples 4 – 6.

53. r2 = 4 cos 2u 54. r2 = 4 sin 2u

55. r = 4 - 4 cos u 56. r = 6 - 3 cos u

57. r = 2 sin u tan u (This is a cissoid.)

58. r = cos 2ucos u

(This is a cissoid with a loop.)

49. r = 3 + cos u 50. r = 2 - cos u

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396 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

74. Based on the results in Exercise 73, fill in the blank(s) to correctly complete each sentence.

(a) The graph of r = ƒ1u2 is symmetric with respect to the polar axis if substitution of for u leads to an equivalent equation.

(b) The graph of r = ƒ1u2 is symmetric with respect to the vertical line u = p2 if substitution of for u leads to an equivalent equation.

(c) Alternatively, the graph of r = ƒ1u2 is symmetric with respect to the vertical line u = p2 if substitution of for r and for u leads to an equivalent equation.

(d) The graph of r = ƒ1u2 is symmetric with respect to the pole if substitution of for r leads to an equivalent equation.

(e) Alternatively, the graph of r = ƒ1u2 is symmetric with respect to the pole if substitution of for u leads to an equivalent equation.

(f) In general, the completed statements in parts (a)–(e) mean that the graphs of polar equations of the form r = a{ b cos u (where a may be 0) are symmetric with respect to .

(g) In general, the completed statements in parts (a)–(e) mean that the graphs of polar equations of the form r = a{ b sin u (where a may be 0) are symmetric with respect to .

Source: Karttunen, H., P. Kröger, H. Oja, M. Putannen, and K. Donners (Editors), Fundamental Astronomy, 4th edition, Springer-Verlag. Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Saunders College Publishers.

Satellite a eMercury 0.39 0.206

Venus 0.78 0.007

Earth 1.00 0.017

Mars 1.52 0.093

Jupiter 5.20 0.048

Saturn 9.54 0.056

Uranus 19.20 0.047

Neptune 30.10 0.009

Pluto 39.40 0.249

Spirals of Archimedes The graph of r = au in polar coordinates is an example of a spiral of Archimedes. With a calculator set to radian mode, use the given value of a and interval of u to graph the spiral in the window specified.

75. a = 1, 0 … u … 4p, 3-15, 154 by 3-15, 154 76. a = 2, -4p … u … 4p, 3-30, 304 by 3-30, 30477. a = 1.5, -4p … u … 4p, 3-20, 204 by 3-20, 204 78. a = -1, 0 … u … 12p, 3-40, 404 by 3-40, 404Intersection of Polar Curves Find the polar coordinates of the points of intersection of the given curves for the specified interval of u.

79. r = 4 sin u, r = 1 + 2 sin u; 0 … u 6 2p

80. r = 3, r = 2 + 2 cos u; 0° … u 6 360°

81. r = 2 + sin u, r = 2 + cos u; 0 … u 6 2p

82. r = sin 2u, r = 22 cos u; 0 … u 6 p

(Modeling) Solve each problem.

83. Orbits of Satellites The polar equation

r =a11 - e 221 + e cos u

can be used to graph the orbits of the satellites of our sun, where a is the average distance in astronomi-cal units from the sun and e is a constant called the eccentricity. The sun will be located at the pole. The table lists the values of a and e.

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3978.5 Polar Equations and Graphs

(a) Graph the orbits of the four closest satellites on the same polar grid. Choose a viewing window that results in a graph with nearly circular orbits.

(b) Plot the orbits of Earth, Jupiter, Uranus, and Pluto on the same polar grid. How does Earth’s distance from the sun compare to the others’ distances from the sun?

84. Radio Towers and Broadcasting Patterns Radio stations do not always broadcast in all directions with the same intensity. To avoid interference with an existing sta-tion to the north, a new station may be licensed to broadcast only east and west. To create an east-west signal, two radio towers are sometimes used. See the figure. Locations where the radio signal is received correspond to the interior of the curve

r2 = 40,000 cos 2u,

where the polar axis (or positive x-axis) points east.

(a) Graph r2 = 40,000 cos 2u for 0° … u … 360°, where distances are in miles. Assuming the radio towers are located near the pole, use the graph to describe the regions where the signal can be received and where the signal cannot be received.

(b) Suppose a radio signal pattern is given by the following equation. Graph this pattern and interpret the results.

r2 = 22,500 sin 2u

Relating Concepts

For individual or collaborative investigation (Exercises 85–92)

In rectangular coordinates, the graph of

ax + by = c

is a horizontal line if a = 0 or a vertical line if b = 0. Work Exercises 85–92 in order, to determine the general forms of polar equations for horizontal and vertical lines.

85. Begin with the equation y = k, whose graph is a horizontal line. Make a trigo-nometric substitution for y using r and u.

86. Solve the equation in Exercise 85 for r.

87. Rewrite the equation in Exercise 86 using the appropriate reciprocal function.

88. Sketch the graph of the equation r = 3 csc u. What is the corresponding rectan-gular equation?

89. Begin with the equation x = k, whose graph is a vertical line. Make a trigono-metric substitution for x using r and u.

90. Solve the equation in Exercise 89 for r.

91. Rewrite the equation in Exercise 90 using the appropriate reciprocal function.

92. Sketch the graph of r = 3 sec u. What is the corresponding rectangular equation?

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398 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

8.6 Parametric Equations, Graphs, and Applications

Basic Concepts We have graphed sets of ordered pairs that correspond to a function of the form y = ƒ1x2 or r = g1u2. Another way to determine a set of ordered pairs involves the equations x = ƒ1t2 and y = g1t2, where t is a real number in an interval I. Each value of t leads to a corresponding x-value and a corresponding y-value, and thus to an ordered pair 1x, y2.

■ Basic Concepts■ Parametric Graphs

and Their Rectangular Equivalents

■ The Cycloid■ Applications of

Parametric Equations

Parametric Equations of a Plane Curve

A plane curve is a set of points 1x, y2 such that x = ƒ1t2, y = g1t2, and ƒ and g are both defined on an interval I. The equations x = ƒ1t2 and y = g1t2 are parametric equations with parameter t.

Graphing calculators are capable of graphing plane curves defined by parametric equations. The calculator must be set to parametric mode. ■

Parametric Graphs and Their Rectangular Equivalents

EXAMPLE 1 Graphing a Plane Curve Defined Parametrically

Let x = t2 and y = 2t + 3, for t in 3-3, 34. Graph the set of ordered pairs 1x, y2.ALGEBRAIC SOLUTION

Make a table of corresponding values of t, x, and y over the domain of t. Plot the points as shown in Figure 32. The graph is a portion of a parabola with horizontal axis y = 3. The arrowheads indicate the direction the curve traces as t increases.

t x y

-3 9 -3-2 4 -1-1 1 1

0 0 31 1 52 4 73 9 9

–3 3 6 9–3

(9, 9)

(9, –3)

x = t 2

y = 2t + 3for t in [–3, 3]

Figure 32

GRAPHING CALCULATOR SOLUTION

We set the parameters of the TI-84 Plus as shown to obtain the graph. See Figure 33.

−4−2

Figure 33

Duplicate this graph and observe how the curve is traced. It should match Figure 32.

■✔ Now Try Exercise 9(a).

EXAMPLE 2 Finding an Equivalent Rectangular Equation

Find a rectangular equation for the plane curve of Example 1,

x = t2, y = 2t + 3, for t in 3-3, 34.SOLUTION To eliminate the parameter t, first solve either equation for t.

y = 2 t + 3 Choose the simpler equation. 2 t = y - 3 Subtract 3 and rewrite.

t =y - 3

2 Divide by 2.

This equation leads to a unique

solution for t.

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3998.6 Parametric Equations, Graphs, and Applications

Now substitute this result into the first equation to eliminate the parameter t.

x = t2

x = a y - 32b 2 Substitute for t.

x =1y - 322

4 Aab B2 = a2b2

4x = 1y - 322 Multiply by 4.This is the equation of a horizontal parabola opening to the right, which agrees with the graph given in Figure 32. Because t is in 3-3, 34, x is in 30, 94 and y is in 3-3, 94. The rectangular equation must be given with restricted domain as

4x = 1y - 322, for x in 30, 94.■✔ Now Try Exercise 9(b).

EXAMPLE 3 Graphing a Plane Curve Defined Parametrically

Graph the plane curve defined by x = 2 sin t, y = 3 cos t, for t in 30, 2p4.SOLUTION To convert to a rectangular equation, it is not productive here to solve either equation for t. Instead, we use the fact that sin2 t + cos2 t = 1 to apply another approach.

x = 2 sin t y = 3 cos t Given equations x2 = 4 sin2 t y2 = 9 cos2 t Square each side.

4= sin2 t

9= cos2 t Solve for sin2 t and cos2 t.

Now add corresponding sides of the two equations.

9= sin2 t + cos2 t

9= 1 sin2 t + cos2 t = 1

This is an equation of an ellipse. See Figure 34.

–3

2–2 0

x = 2 sin ty = 3 cos t

fort in [0, 2P]

4y2

9+ = 1

−3.1

−5

3.1

Parametric graphing modeFigure 34 ■✔ Now Try Exercise 31.

Parametric representations of a curve are not unique. In fact, there are infinitely many parametric representations of a given curve. If the curve can be described by a rectangular equation y = ƒ1x2, with domain X, then one simple parametric representation is

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400 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

EXAMPLE 4 Finding Alternative Parametric Equation Forms

Give two parametric representations for the equation of the parabola.

y = 1x - 222 + 1SOLUTION The simplest choice is to let

x = t, y = 1t - 222 + 1, for t in 1-∞, ∞2.Another choice, which leads to a simpler equation for y, is

x = t + 2, y = t2 + 1, for t in 1-∞, ∞2.■✔ Now Try Exercise 33.

NOTE Verify that another choice in Example 4 is

x = 2 + tan t, y = sec2 t, for t in A - p2 , p2 B . Other choices are possible.

ALGEBRAIC SOLUTION

There is no simple way to find a rectangular equation for the cycloid from its parametric equations. Instead, begin with a table using selected values for t in 30, 2p4. Approximate values have been rounded as necessary.

EXAMPLE 5 Graphing a Cycloid

Graph the cycloid.

x = t - sin t, y = 1 - cos t, for t in 30, 2p4

Plotting the ordered pairs 1x, y2 from the table of values leads to the portion of the graph in Figure 35 from 0 to 2p.

t 0p4

p2 p

3p2 2p

x 0 0.08 0.6 p 5.7 2py 0 0.3 1 2 1 0

P(x, y)

2pp

Figure 35

GRAPHING CALCULATOR SOLUTION

It is easier to graph a cycloid with a graphing calculator in parametric mode than with tradi-tional methods. See Figure 36.

Using a larger interval for t would show that the cycloid repeats the pattern shown here every 2p units.

■✔ Now Try Exercise 37.

Figure 36

The Cycloid The cycloid is a special case of the trochoid—a curve traced out by a point at a given distance from the center of a circle as the circle rolls along a straight line. If the given point is on the circumference of the circle, then the path traced as the circle rolls along a straight line is a cycloid, which is defined parametrically as follows.

x = at − a sin t, y = a − a cos t, for t in 1−H, H 2Other curves related to trochoids are hypotrochoids and epitrochoids,

which are traced out by a point that is a given distance from the center of a circle that rolls not on a straight line, but on the inside or outside, respectively, of another circle. The classic Spirograph toy can be used to draw these curves.

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4018.6 Parametric Equations, Graphs, and Applications

The cycloid has an interesting physical property. If a flexible cord or wire goes through points P and Q as in Figure 37, and a bead is allowed to slide due to the force of gravity without friction along this path from P to Q, the path that requires the shortest time takes the shape of the graph of an inverted cycloid.

Figure 37

v0 sin U

v0 cos U

Figure 38

Applications of Parametric Equations Parametric equations are used to simulate motion. If an object is thrown with a velocity of v feet per second at an angle u with the horizontal, then its flight can be modeled by

x = 1v cos U 2 t and y = 1v sin U 2 t − 16 t2 + h,where t is in seconds and h is the object’s initial height in feet above the ground. Here, x gives the horizontal position information and y gives the vertical posi-tion information. The term -16t2 occurs because gravity is pulling downward. See Figure 38. These equations ignore air resistance.

400

600

x3, y3 x2, y2

x1, y1

(a)

400

600

x3, y3 x2, y2

x1, y1

(b)

Figure 39

EXAMPLE 6 Simulating Motion with Parametric Equations

Three golf balls are hit simultaneously into the air at 132 ft per sec (90 mph) at angles of 30°, 50°, and 70° with the horizontal.

(a) Assuming the ground is level, determine graphically which ball travels the greatest distance. Estimate this distance.

(b) Which ball reaches the greatest height? Estimate this height.

SOLUTION

(a) Use the following parametric equations to model the flight of the golf balls.

x = 1v cos u2t and y = 1v sin u2t - 16t2 + hWrite three sets of parametric equations.

x1 = 1132 cos 30°2t, y1 = 1132 sin 30°2t - 16t2x2 = 1132 cos 50°2t, y2 = 1132 sin 50°2t - 16t2x3 = 1132 cos 70°2t, y3 = 1132 sin 70°2t - 16t2

The graphs of the three sets of parametric equations are shown in Figure 39(a), where 0 … t … 3. From the graph in Figure 39(b), where 0 … t … 9, we see that the ball hit at 50° travels the greatest distance. Using the tracing feature of the TI-84 Plus calculator, we find that this distance is about 540 ft.

(b) Again, use the tracing feature to find that the ball hit at 70° reaches the great-est height, about 240 ft.

■✔ Now Try Exercise 43.

Substitute h = 0, v = 132 ft per sec, and u = 30°, 50°, and 70°.

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402 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

A TI-84 Plus calculator allows us to view the graphing of more than one equation either sequentially or simultaneously. By choosing the latter, the three golf balls in Figure 39 can be viewed in flight at the same time. ■

EXAMPLE 7 Examining Parametric Equations of Flight

Jack launches a small rocket from a table that is 3.36 ft above the ground. Its initial velocity is 64 ft per sec, and it is launched at an angle of 30° with respect to the ground. Find the rectangular equation that models its path. What type of path does the rocket follow?

SOLUTION The path of the rocket is defined by the parametric equations

x = 164 cos 30°2t and y = 164 sin 30°2t - 16t2 + 3.36or, equivalently, x = 3223 t and y = -16t2 + 32t + 3.36.

From x = 3223 t, we solve for t to obtaint = x

3223 . Divide by 3223.Substituting for t in the other parametric equation yields the following.

y = -16t2 + 32t + 3.36

y = -16 ¢ x3223 ≤2 + 32 ¢ x3223 ≤ + 3.36 Let t = x3223 .

y = - 1192

x2 + 233

x + 3.36 Simplify.

This equation defines a parabola. The rocket follows a parabolic path.

■✔ Now Try Exercise 47(a).

ALGEBRAIC SOLUTION

The equation y = -16t2 + 32t + 3.36 tells the vertical position of the rocket at time t. We need to determine the positive value of t for which y = 0 because this value corresponds to the rocket at ground level. This yields

0 = -16t2 + 32t + 3.36.

Using the quadratic formula, the solutions are t = -0.1 or t = 2.1. Because t represents time, t = -0.1 is an unacceptable answer. Therefore, the flight time is 2.1 sec.

The rocket was in the air for 2.1 sec, so we can use t = 2.1 and the parametric equation that models the horizontal position, x = 3223 t, to obtain

x = 3223 12.12 ≈ 116.4 ft.

EXAMPLE 8 Analyzing the Path of a Projectile

Determine the total flight time and the horizontal distance traveled by the rocket in Example 7.

GRAPHING CALCULATOR SOLUTION

Figure 40 shows that when t = 2.1, the hor-izontal distance x covered is approximately 116.4 ft, which agrees with the algebraic solution.

−10

−30

150

Figure 40

■✔ Now Try Exercise 47(b).

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4038.6 Parametric Equations, Graphs, and Applications

CONCEPT PREVIEW Fill in the blank to correctly complete each sentence.

1. For the plane curve defined by

x = t2 + 1, y = 2t + 3, for t in 3-4, 44,the ordered pair that corresponds to t = -3 is .

2. For the plane curve defined by

x = -3t + 6, y = t2 - 3, for t in 3-5, 54,the ordered pair that corresponds to t = 4 is .

3. For the plane curve defined by

x = cos t, y = 2 sin t, for t in 30, 2p4,the ordered pair that corresponds to t = p3 is .

4. For the plane curve defined by

x = 2t, y = t2 + 3, for t in 10, ∞2,the ordered pair that corresponds to t = 16 is .

8.6 Exercises

CONCEPT PREVIEW Match the ordered pair from Column II with the pair of para-metric equations in Column I on whose graph the point lies. In each case, consider the given value of t.

I II

5. x = 3t + 6, y = -2t + 4; t = 2

6. x = cos t, y = sin t; t = p47. x = t, y = t2; t = 5

8. x = t2 + 3, y = t2 - 2; t = 2

A. 15, 252B. 17, 22C. 112, 02D. Q222 , 222 R

For each plane curve, (a) graph the curve, and (b) find a rectangular equation for the curve. See Examples 1 and 2.

9. x = t + 2, y = t2, for t in 3-1, 14 10. x = 2t, y = t + 1, for t in 3-2, 34

11. x = 2t , y = 3t - 4, for t in 30, 44 12. x = t2, y = 2t , for t in 30, 44

13. x = t3 + 1, y = t3 - 1, for t in 1-∞, ∞2 14. x = 2t - 1, y = t2 + 2, for t in 1-∞, ∞2

15. x = 2 sin t, y = 2 cos t, for t in 30, 2p4 16. x = 25 sin t, y = 23 cos t, for t in 30, 2p4

17. x = 3 tan t, y = 2 sec t, for t in A - p2 , p2 B 18. x = cot t, y = csc t, for t in 10, p2

19. x = sin t, y = csc t, for t in 10, p2 20. x = tan t, y = cot t, for t in A0, p2 B

21. x = t, y = 2t2 + 2 , for t in 1-∞, ∞2 22. x = 2t , y = t2 - 1, for t in 30, ∞2

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404 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Lissajous Figures The screen shown here is an example of a Lissajous figure. Such figures occur in electronics and may be used to find the frequency of an unknown voltage. Graph each Lissajous figure for t in 30, 6.54 using the window 3-6, 64 by 3-4, 44.39. x = 2 cos t, y = 3 sin 2t

40. x = 3 cos t, y = 2 sin 2t

41. x = 3 sin 4t, y = 3 cos 3t 42. x = 4 sin 4t, y = 3 sin 5t

−4

−6

23. x = 2 + sin t, y = 1 + cos t, for t in 30, 2p4 24. x = 1 + 2 sin t, y = 2 + 3 cos t, for t in 30, 2p4

27. x = t + 2, y = t - 4, for t in 1-∞, ∞2 28. x = t2 + 2, y = t2 - 4, for t in 1-∞, ∞2

25. x = t + 2, y = 1t + 2 ,

for t≠ -2 26. x = t - 3, y = 2

t - 3 , for t≠ 3

Graph each plane curve defined by the parametric equations for t in 30, 2p4. Then find a rectangular equation for the plane curve. See Example 3.

29. x = 3 cos t, y = 3 sin t 30. x = 2 cos t, y = 2 sin t

31. x = 3 sin t, y = 2 cos t 32. x = 4 sin t, y = 3 cos t

Give two parametric representations for the equation of each parabola. See Example 4.

33. y = 1x + 322 - 1 34. y = 1x + 422 + 235. y = x2 - 2x + 3 36. y = x2 - 4x + 6

Graph each cycloid defined by the given equations for t in the specified interval. See Example 5.

37. x = 2t - 2 sin t, y = 2 - 2 cos t, for t in 30, 4p4 38. x = t - sin t, y = 1 - cos t, for t in 30, 4p4

(Modeling) Do the following. See Examples 6–8.

(a) Determine parametric equations that model the path of the projectile.

(b) Determine a rectangular equation that models the path of the projectile.

43. Flight of a Model Rocket A model rocket is launched from the ground with vel-ocity 48 ft per sec at an angle of 60° with respect to the ground.

44. Flight of a Golf Ball Tyler is playing golf. He hits a golf ball from the ground at an angle of 60° with respect to the ground at velocity 150 ft per sec.

45. Flight of a Softball Sally hits a softball when it is 2 ft above the ground. The ball leaves her bat at an angle of 20° with respect to the ground at velocity 88 ft per sec.

60°

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4058.6 Parametric Equations, Graphs, and Applications

46. Flight of a Baseball Francisco hits a baseball when it is 2.5 ft above the ground. The ball leaves his bat at an angle of 29° from the horizon-tal with velocity 136 ft per sec.

29°

2.5 ft

(Modeling) Solve each problem. See Examples 7 and 8.

47. Path of a Rocket A rocket is launched from the top of an 8-ft platform. Its initial velocity is 128 ft per sec. It is launched at an angle of 60° with respect to the ground.(a) Find the rectangular equation that models its path. What type of path does the

rocket follow?

(b) Determine the total flight time, to the nearest second, and the horizontal distance the rocket travels, to the nearest foot.

48. Simulating Gravity on the Moon If an object is thrown on the moon, then the para-metric equations of flight are

x = 1v cos u2t and y = 1v sin u2t - 2.66t2 + h.Estimate, to the nearest foot, the distance a golf ball hit at 88 ft per sec (60 mph) at an angle of 45° with the horizontal travels on the moon if the moon’s surface is level.

49. Flight of a Baseball A baseball is hit from a height of 3 ft at a 60° angle above the horizontal. Its initial velocity is 64 ft per sec.

(a) Write parametric equations that model the flight of the baseball.

(b) Determine the horizontal distance, to the nearest tenth of a foot, traveled by the ball in the air. Assume that the ground is level.

(c) What is the maximum height of the baseball, to the nearest tenth of a foot? At that time, how far has the ball traveled horizontally?

(d) Would the ball clear a 5-ft-high fence that is 100 ft from the batter?

50. Path of a Projectile A projectile has been launched from the ground with initial velocity 88 ft per sec. The parametric equations

x = 82.7t and y = -16t2 + 30.1t

model the path of the projectile, where t is in seconds.

(a) Approximate the angle u that the projectile makes with the horizontal at the launch, to the nearest tenth of a degree.

(b) Write parametric equations for the path using the cosine and sine functions.

Work each problem.

51. Give two parametric representations of the parabola y = a1x - h22 + k.52. Give a parametric representation of the rectangular equation x

a2 -y 2

b2 = 1.

53. Give a parametric representation of the rectangular equation x2

a2 +y 2

b2 = 1.

54. The spiral of Archimedes has polar equation r = a u, where r2 = x2 + y2. Show that a parametric representation of the spiral of Archimedes is

x = au cos u, y = au sin u, for u in 1-∞, ∞2.55. Show that the hyperbolic spiral r u = a, where r2 = x2 + y2, is given parametri-

cally by

x = a cos uu

, y = a sin uu

, for u in 1-∞, 02 ´ 10, ∞2.

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406 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

56. The parametric equations x = cos t, y = sin t, for t in 30, 2p4 and the parametric equations x = cos t, y = -sin t, for t in 30, 2p4 both have the unit circle as their graph. However, in one case the circle is traced out clockwise (as t moves from 0 to 2p), and in the other case the circle is traced out counterclockwise. For which pair of equations is the circle traced out in the clockwise direction?

Concept Check Consider the parametric equations x = ƒ1t2, y = g1t2, for t in 3a, b4, with c 7 0, d 7 0.

57. How is the graph affected if the equation x = ƒ1t2 is replaced by x = c + ƒ1t2?58. How is the graph affected if the equation y = g1t2 is replaced by y = d + g1t2?

Key Terms

8.1 imaginary unitcomplex number real part imaginary part nonreal complex

number pure imaginary

number standard form complex

conjugates

8.2 resultant real axis

imaginary axis complex plane rectangular form of a

complex number trigonometric (polar)

form of a complex number

absolute value (modulus) argument

8.4 nth root of a complex number

8.5 polar coordinate systempole polar axis polar coordinates rectangular

(Cartesian) equation polar equation cardioid polar grid

rose curve lemniscate spiral of

Archimedes limaçon

8.6 plane curve parametric

equations of a plane curve

parameter cycloid

Quick ReviewConcepts Examples

3 - 4i

Real Imaginary part part

Complex Numbers

Imaginary Unit i

i = !−1, and therefore i2 = −1.Complex NumberIf a and b are real numbers, then any number of the form a + bi is a complex number. In the complex number a + bi, a is the real part and b is the imaginary part.

8.1

Chapter 8 Test Prep

New Symbols

i imaginary unita + bi complex number

cis U cos u + i sin u

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407

Concepts Examples

CHAPTER 8 Test Prep 407

Simplify. 2-4 = 2i2-12 = i212 = i24 # 3 = 2i23 12 + 3i2 + 13 + i2 - 12 - i2 = 12 + 3 - 22 + 13 + 1 + 12i= 3 + 5i 16 + i213 - 2i2

= 18 - 12i + 3i - 2i2 FOIL method = 118 + 22 + 1-12 + 32i i2 = -1 = 20 - 9i

3 + i1 + i =

13 + i211 - i211 + i211 - i2 = 3 - 3i + i - i21 - i2 = 4 - 2i

212 - i22

= 2 - i

Trigonometric (Polar) Form of Complex NumbersLet the complex number x + yi correspond to the vector with direction angle u and magnitude r.

x = r cos U y = r sin U

r = !x2 + y2 tan U = yx

, if x ≠ 0

The trigonometric (polar) form of the expression x + yi is

r 1cos U + i sin U 2 or r cis U.

Write 21cos 60° + i sin 60°2 in rectangular form.21cos 60° + i sin 60°2

= 2 ¢12

+ i # 232

≤ = 1 + i23

Write -22 + i22 in trigonometric form.r = 4A -22 B2 + A22 B2 = 2tan u = -1 and u is in quadrant II, so u = 180° - 45° = 135°.

-22 + i22 = 2 cis 135°.RealImaginary

√2

–√2

Meaning of !−aIf a 7 0, then !−a = i !a.Adding and Subtracting Complex NumbersAdd or subtract the real parts, and add or subtract the imagi-nary parts.

Multiplying and Dividing Complex NumbersMultiply complex numbers as with binomials, and use the fact that i2 = -1.

Divide complex numbers by multiplying the numerator and denominator by the complex conjugate of the denominator.

Trigonometric (Polar) Form of Complex Numbers8.2

The Product and Quotient Theorems

Product and Quotient TheoremsIf r11cos u1 + i sin u12 and r21cos u2 + i sin u22 are any two complex numbers, then the following hold.3r1 1cos U1 + i sin U1 2 4 # 3r2 1cos U2 + i sin U2 2 4

= r1r2 3cos 1U1 + U2 2 + i sin 1U1 + U2 2 4and

r1 1cos U1 + i sin U1 2r2 1cos U2 + i sin U2 2 =

r1r2

3cos 1U1 − U2 2 + i sin 1U1 − U2 2 4 ,where r 2 1cos u2 + i sin u22 ≠ 0

8.3

Let z1 = 41cos 135° + i sin 135°2and z2 = 21cos 45° + i sin 45°2. z1z2 = 81cos 180° + i sin 180°2 4 # 2 = 8; = 81-1 + i # 02 135° + 45° = 180° = -8

z1z2

= 21cos 90° + i sin 90°2 42 = 2; = 210 + i # 12 135° - 45° = 90° = 2i

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408 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Concepts Examples

Polar Equations and Graphs

Rectangular and Polar CoordinatesIf a point has rectangular coordinates 1x, y2 and polar coor-dinates 1r, u2, then these coordinates are related as follows.

x = r cos U y = r sin U r2 = x2 + y2 tan U = y

x , if x ≠ 0

8.5

De Moivre’s Theorem; Powers and Roots of Complex Numbers8.4

De Moivre’s Theorem3r 1cos U + i sin U 2 4n = rn 1cos nU + i sin nU 2nth Root TheoremIf n is any positive integer, r is a positive real number, and u is in degrees, then the nonzero complex number r 1cos u + i sin u2 has exactly n distinct nth roots, given by the following.

n!r 1cos A + i sin A 2 , or n!r cis A,where

A =U + 360° # k

n , k = 0, 1, 2, . . . , n - 1.

If u is in radians, then

A =U + 2Pk

n , k = 0, 1, 2, . . . , n - 1.

Let z = 41cos 180° + i sin 180°2. Find z3 and the square roots of z.341cos 180° + i sin 180°243 Find z3.

= 431cos 3 # 180° + i sin 3 # 180°2 = 641cos 540° + i sin 540°2 = 641-1 + i # 02 = -64

For the given z, r = 4 and u = 180°. Its square roots are24 acos 180°2

+ i sin 180°2b

= 210 + i # 12 = 2i

and 24 acos 180° + 360°2

+ i sin 180° + 360°2

b = 210 + i1-122 = -2i.

Find the rectangular coordinates for the point 15, 60°2 in polar coordinates.

x = 5 cos 60° = 5 a12b = 5

y = 5 sin 60° = 5 ¢232

≤ = 5232

The rectangular coordinates are Q52 , 5232 R .

Polar Equations and Graphs

r = a cos Ur = a sin U

f Circles r2 = a2 sin 2Ur2 = a2 cos 2U

f Lemniscatesr = a t b sin Ur = a t b cos U f Limaçons r = a sin nUr = a cos nU f Rose curves

Find polar coordinates for 1-1, -12 in rectangular coordinates.

–1

r = 21-122 + 1-122 = 22 tan u = 1 and u is in quadrant III, so u = 225°. One pair of polar coordinates for 1-1, -12 is A22, 225° B .

Graph r = 4 cos 2u.

180° 0°

90°

270°

r = 4 cos 2U

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409CHAPTER 8 Review Exercises

Concepts Examples

Parametric Equations, Graphs, and Applications8.6

Parametric Equations of a Plane CurveA plane curve is a set of points 1x, y2 such that x = ƒ1t2, y = g1t2, and ƒ and g are both defined on an interval I. The equations

x = ƒ1t2 and y = g1t2are parametric equations with parameter t .

Flight of an ObjectIf an object has initial velocity v and initial height h, and trav-els such that its initial angle of elevation is u, then its flight after t seconds can be modeled by the following parametric equations.

x = 1v cos U 2 t and y = 1v sin U 2 t − 16 t2 + h

Graph x = 2 - sin t, y = cos t - 1, for 0 … t … 2p.

−3.1

−5

3.1

Joe kicks a football from the ground at an angle of 45° with a velocity of 48 ft per sec. Give the parametric equa-tions that model the path of the football and the distance it travels before hitting the ground.

x = 148 cos 45°2t = 2422 t y = 148 sin 45°2t - 16t2 = 2422 t - 16t2

When the ball hits the ground, y = 0.

2422 t - 16t2 = 0 Substitute y = 0. 8t A322 - 2t B = 0 Factor.t = 0 or t = 322

2 Zero-factor property

(Reject)

The distance it travels is x = 2422 Q3222 R = 72 ft.

Write each number as the product of a real number and i.

1. 2-9 2. 2-12 Chapter 8 Review Exercises

Solve each equation over the set of complex numbers.

3. x2 = -81 4. x12x + 32 = -4Perform each operation. Write answers in standard form.

5. 11 - i2 - 13 + 4i2 + 2i 6. 12 - 5i2 + 19 - 10i2 - 3 7. 16 - 5i2 + 12 + 7i2 - 13 - 2i2 8. 14 - 2i2 - 16 + 5i2 - 13 - i2 9. 13 + 5i218 - i2 10. 14 - i215 + 2i2 11. 12 + 6i2212. 16 - 3i22 13. 11 - i23 14. 12 + i2315.

25 - 19i5 + 3i 16.

2 - 5i1 + i

17. 2 + i1 - 5i

18. 3 + 2i

i19. i53 20. i-41

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410 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

Concept Check The complex number z, where z = x + yi, can be graphed in the plane as 1x, y2. Describe the graph of all complex numbers z satisfying the given conditions.41. The imaginary part of z is the negative of the real part of z.

42. The absolute value of z is 2.

Perform each operation. Write answers in rectangular form.

21. 351cos 90° + i sin 90°24361cos 180° + i sin 180°2422. 33 cis 135°432 cis 105°4 23. 21cos 60° + i sin 60°2

81cos 300° + i sin 300°224.

4 cis 270°2 cis 90°

25. A23 + i B326. 12 - 2i25 27. 1cos 100° + i sin 100°2628. Concept Check The vector representing a real number will lie on the -axis in

the complex plane.

Graph each complex number.

29. 5i 30. -4 + 2i 31. 3 - 3i2332. Find the sum of 7 + 3i and -2 + i. Graph both complex numbers and their resultant.

Write each complex number in its alternative form, using a calculator to approximate answers to four decimal places as necessary.

Rectangular Form Trigonometric Form

33. -2 + 2i

34. 31cos 90° + i sin 90°235. 21cos 225° + i sin 225°236. -4 + 4i2337. 1 - i

38. 4 cis 240°

39. -4i

40. 7 cis 310°

Find all roots as indicated. Write answers in trigonometric form.

43. the cube roots of 1 - i 44. the fifth roots of -2 + 2i

45. Concept Check How many real sixth roots does -64 have?

46. Concept Check How many real fifth roots does -32 have?

Find all complex number solutions. Write answers in trigonometric form.

47. x4 + 16 = 0 48. x3 + 125 = 0 49. x2 + i = 0

50. Convert 15, 315°2 to rectangular coordinates.51. Convert A -1, 23 B to polar coordinates, with 0° … u 6 360° and r 7 0.52. Concept Check Describe the graph of r = k for k 7 0.

Identify and graph each polar equation for u in 30°, 360°2.53. r = 4 cos u 54. r = -1 + cos u

55. r = 2 sin 4u 56. r = 22 cos u - sin u

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411

Find an equivalent equation in rectangular coordinates.

57. r = 31 + cos u

58. r = sin u + cos u 59. r = 2

Find an equivalent equation in polar coordinates.

60. y = x 61. y = x2 62. x2 + y2 = 25

Identify the geometric symmetry (A, B, or C) that each graph will possess.

A. symmetry with respect to the origin

B. symmetry with respect to the y-axis

C. symmetry with respect to the x-axis

63. Whenever 1r, u2 is on the graph, so is 1-r, -u2.64. Whenever 1r, u2 is on the graph, so is 1-r, u2.65. Whenever 1r, u2 is on the graph, so is 1r, -u2.66. Whenever 1r, u2 is on the graph, so is 1r, p - u2.Find a polar equation having the given graph.

67.

0 2

2 x = 2

68.

0 2

2 y = 2

69.

x + 2y = 4

70.

0 2

2 x2 + y2 = 4

71. Graph the plane curve defined by the parametric equations x = t + cos t, y = sin t, for t in 30, 2p4.

72. Show that the distance between 1r1, u12 and 1r2, u22 in polar coordinates is given byd = 2r1 2 + r 2 2 - 2r1r 2 cos1u1 - u22.

Find a rectangular equation for each plane curve with the given parametric equations.

73. x = 2t - 1, y = 2t , for t in 31, ∞274. x = 3t + 2, y = t - 1, for t in 3-5, 5475. x = 5 tan t, y = 3 sec t, for t in a - p

2 , p

76. x = t2 + 5, y = 1t2 + 1 , for t in 1-∞, ∞2

77. x = cos 2t, y = sin t, for t in 1-p, p278. Give a pair of parametric equations whose graph is the circle having center 13, 42

and passing through the origin.

CHAPTER 8 Review Exercises

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412 CHAPTER 8 Complex Numbers, Polar Equations, and Parametric Equations

1. Find each product or quotient. Simplify the answers.

(a) 2-8 # 2-6 (b) 2-228 (c) 2-202-180 2. For the complex numbers w = 2 - 4i and z = 5 + i, find each of the following in

rectangular form.

(a) w + z (and give a geometric representation) (b) w - z (c) wz (d) wz

3. Express each of the following in rectangular form.

(a) i15 (b) 11 + i22 4. Solve 2x2 - x + 4 = 0 over the set of complex numbers. 5. Write each complex number in trigonometric (polar) form, where 0° … u 6 360°.

(a) 3i (b) 1 + 2i (c) -1 - i23 6. Write each complex number in rectangular form.

(a) 31cos 30° + i sin 30°2 (b) 4 cis 40° (c) 31cos 90° + i sin 90°2 7. For the complex numbers w = 81cos 40° + i sin 40°2 and z = 21cos 10° + i sin 10°2, find each of the following in the form specified.

(a) wz (trigonometric form) (b) wz

(rectangular form) (c) z3 (rectangular form)

8. Find the four complex fourth roots of -16i. Write answers in trigonometric form. 9. Convert the given rectangular coordinates to polar coordinates. Give two pairs of

polar coordinates for each point.

(a) 10, 52 (b) 1-2, -22 10. Convert the given polar coordinates to rectangular coordinates.

(a) 13, 315°2 (b) 1-4, 90°2

Chapter 8 Test

79. (Modeling) Flight of a Baseball A batter hits a baseball when it is 3.2 ft above the ground. It leaves the bat with velocity 118 ft per sec at an angle of 27° with respect to the ground.

(a) Determine parametric equations that model the path of the baseball.

(b) Determine a rectangular equation that models the path of the baseball.

80. Mandelbrot Set Consider the complex number z = 1 + i. Compute the value of z2 + z, and show that its absolute value exceeds 2, indicating that 1 + i is not in the Mandelbrot set.

Identify and graph each polar equation for u in 30°, 360°2.11. r = 1 - cos u 12. r = 3 cos 3u13. Convert each polar equation to a rectangular equation, and sketch its graph.

(a) r = 42 sin u - cos u (b) r = 6

Graph each pair of parametric equations.

14. x = 4t - 3, y = t2, for t in 3-3, 44 15. x = 2 cos 2t, y = 2 sin 2t, for t in 30, 2p416. Julia Set Consider the complex number z = -1 + i. Compute the value of z2 - 1,

and show that its absolute value exceeds 2, indicating that -1 + i is not in the Julia set.

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Appendices

A Equations and Inequalities■ Basic Terminology of

Equations■ Linear Equations■ Quadratic Equations■ Inequalities■ Linear Inequalities

and Interval Notation■ Three-Part Inequalities

Basic Terminology of Equations An equation is a statement that two expressions are equal.

x + 2 = 9, 11x = 5x + 6x, x2 - 2x - 1 = 0 Equations

To solve an equation means to find all numbers that make the equation a true statement. These numbers are the solutions, or roots, of the equation. A number that is a solution of an equation is said to satisfy the equation, and the solutions of an equation make up its solution set. Equations with the same solution set are equivalent equations. For example,

x = 4, x + 1 = 5, and 6x + 3 = 27 are equivalent equations

because they have the same solution set, 546. However, the equationsx2 = 9 and x = 3 are not equivalent

because the first has solution set 5-3, 36 while the solution set of the second is 536.One way to solve an equation is to rewrite it as a series of simpler equiva-

lent equations using the addition and multiplication properties of equality. Let a, b, and c represent real numbers.

If a = b, then a + c = b + c.

If a = b and c 3 0, then ac = bc.

These properties can be extended: The same number may be subtracted from each side of an equation, and each side may be divided by the same nonzero number, without changing the solution set.

Linear Equations We use the properties of equality to solve linear equations.

Linear Equation in One Variable

A linear equation in one variable is an equation that can be written in the form

ax + b = 0,

where a and b are real numbers and a ≠ 0.

A linear equation is a first-degree equation because the greatest degree of the variable is 1.

413

3x + 12 = 0, 34

x = 12, 0.51x + 32 = 2x - 6 Linear equations1x + 2 = 5, 1x

= -8, x2 + 3x + 0.2 = 0 Nonlinear equations

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414 APPENDIX A Equations and Inequalities

EXAMPLE 1 Solving a Linear Equation

Solve 312x - 42 = 7 - 1x + 52.SOLUTION 312x - 42 = 7 - 1x + 52

6x - 12 = 7 - x - 5 Distributive property 6x - 12 = 2 - x Combine like terms. 6x - 12 + x = 2 - x + x Add x to each side. 7x - 12 = 2 Combine like terms. 7x - 12 + 12 = 2 + 12 Add 12 to each side. 7x = 14 Combine like terms.

7x7

= 147

Divide each side by 7.

x = 2

CHECK

A check of the solution is recommended.

312x - 42 = 7 - 1x + 52 Original equation 312 # 2 - 42≟ 7 - 12 + 52 Let x = 2.

314 - 42≟ 7 - 172 Work inside the parentheses. 0 = 0 ✓ True

The solution set is 526 . ■✔ Now Try Exercise 9.EXAMPLE 2 Solving a Linear Equation with Fractions

Solve 2x + 4

3+ 1

2 x = 1

4 x - 7

3 .

SOLUTION 2x + 4

3+ 1

2 x = 1

4 x - 7

12a 2x + 43

+ 12

xb = 12a 14

x - 73b Multiply by 12, the LCD of the fractions.

12a 2 x + 43b + 12a 1

2 xb = 12a 1

4 xb - 12a 7

3b Distributive property

412 x + 42 + 6 x = 3x - 28 Multiply. 8 x + 16 + 6 x = 3x - 28 Distributive property 14x + 16 = 3x - 28 Combine like terms. 11x = -44 Subtract 3x. Subtract 16. x = -4 Divide each side by 11.

Distribute to all terms within the parentheses.

CHECK 2 x + 4

3+ 1

2 x = 1

4 x - 7

3 Original equation

21-42 + 4

3+ 1

2 1-42≟ 1

4 1-42 - 7

3 Let x = -4.

- 103

= - 103

✓ True

The solution set is 5-46. ■✔ Now Try Exercise 11.M09_LHSD7642_11_AIE_App_pp413-449.indd 414 25/11/15 11:58 am

415APPENDIX A Equations and Inequalities 415

An equation satisfied by every number that is a meaningful replacement for the variable is an identity.

31x + 12 = 3x + 3 IdentityAn equation that is satisfied by some numbers but not others is a conditional equation.

2 x = 4 Conditional equation

The equations in Examples 1 and 2 are conditional equations. An equation that has no solution is a contradiction.

x = x + 1 Contradiction

EXAMPLE 3 Identifying Types of Equations

Determine whether each equation is an identity, a conditional equation, or a contradiction. Give the solution set.

(a) -21x + 42 + 3x = x - 8 (b) 5 x - 4 = 11 (c) 313x - 12 = 9x + 7SOLUTION

(a) -21x + 42 + 3x = x - 8 -2 x - 8 + 3x = x - 8 Distributive property x - 8 = x - 8 Combine like terms. 0 = 0 Subtract x. Add 8. When a true statement such as 0 = 0 results, the equation is an identity, and the solution set is {all real numbers}.

(b) 5 x - 4 = 11 5 x = 15 Add 4 to each side. x = 3 Divide each side by 5. This is a conditional equation, and its solution set is 536.(c) 313x - 12 = 9x + 7 9x - 3 = 9x + 7 Distributive property -3 = 7 Subtract 9x.

When a false statement such as -3 = 7 results, the equation is a contradic-tion, and the solution set is the empty set, or null set, symbolized ∅.

■✔ Now Try Exercises 21, 23, and 25.

Quadratic Equation in One Variable

An equation that can be written in the form

ax2 + bx + c = 0,

where a, b, and c are real numbers with a ≠ 0, is a quadratic equation. The given form is called standard form.

Quadratic Equations A quadratic equation is defined as follows.

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416 APPENDIX A Equations and Inequalities

A quadratic equation is a second-degree equation—that is, an equation with a squared variable term and no terms of greater degree.

x2 = 25, 4x2 + 4x - 5 = 0, 3x2 = 4x - 8 Quadratic equations

When the expression ax2 + bx + c in a quadratic equation is easily factorable over the real numbers, it is efficient to factor and then apply the following zero-factor property.

If a and b are complex numbers with ab = 0, then a = 0 or b = 0 or both equal zero.

6 a 13b 2 + 7 a 1

3b ≟ 3 Let x = 13 .

+ 73

≟ 3

3 = 3 ✓ True

Don’t factor out x here.

CHECK 6x2 + 7x = 3 Original equation

EXAMPLE 4 Using the Zero-Factor Property

Solve 6x2 + 7x = 3.

SOLUTION 6x2 + 7x = 3 6x2 + 7x - 3 = 0 Standard form

13x - 1212x + 32 = 0 Factor. 3x - 1 = 0 or 2x + 3 = 0 Zero-factor property

3x = 1 or 2x = -3 Solve each equation.

x = 13

or x = - 32

6 a - 32b 2 + 7 a - 3

2b ≟ 3 Let x = - 32 .

544

- 212≟ 3

3 = 3 ✓ True

Both values check because true statements result. The solution set is 513 , - 326. ■✔ Now Try Exercise 33.

A quadratic equation written in the form x2 = k, where k is a constant, can be solved using the square root property.

If x2 = k, then x = !k or x = −!k.That is, the solution set of x2 = k isE !k, −!k F , which may be abbreviated Et!k F .

EXAMPLE 5 Using the Square Root Property

Solve each quadratic equation.(a) x2 = 17 (b) 1x - 422 = 12SOLUTION

(a) x2 = 17 x = {217 Square root property

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417APPENDIX A Equations and Inequalities

(b) 1x - 422 = 12 x - 4 = {212 Generalized square root property x = 4 { 212 Add 4. x = 4 { 223 112 = 14 # 3 = 213 CHECK 1x - 422 = 12 Original equation

A4 + 223 - 4 B2≟ 12 Let x = 4 + 213. A223 B2≟ 12

22 # A 23 B2≟ 12 12 = 12 ✓ True

A4 - 223 - 4 B2≟ 12 Let x = 4 - 213. A -223 B2≟ 12

1-222 # A 23 B2≟ 12 12 = 12 ✓ True

The solution set is E4{ 223 F . ■✔ Now Try Exercises 43 and 47. Any quadratic equation can be solved by the quadratic formula, which says

that the solutions of the quadratic equation ax2 + bx + c = 0, where a ≠ 0, are given by

x =−b t !b2 − 4 ac

2 a . This formula is derived

in algebra courses.

EXAMPLE 6 Using the Quadratic Formula

Solve x2 - 4x = -2.

SOLUTION x2 - 4x + 2 = 0 Write in standard form. Here a = 1, b = -4, and c = 2.

x = -b { 2b2 - 4ac2a

Quadratic formula

x =-1-42 { 21-422 - 4112122

2112 Substitute a = 1, b = -4, and c = 2. x = 4 { 216 - 8

2 Simplify.

x = 4 { 2222

116 - 8 = 18 = 14 # 2 = 212 x =

2 A 2{ 22 B2

Factor out 2 in the numerator.

x = 2 { 22 Lowest termsThe solution set is E2{ 22 F . ■✔ Now Try Exercise 55.

The fraction bar extends under -b.

Factor first, then divide.

Inequalities An inequality says that one expression is greater than, greater than or equal to, less than, or less than or equal to another. As with equa-tions, a value of the variable for which the inequality is true is a solution of the inequality, and the set of all solutions is the solution set of the inequality. Two inequalities with the same solution set are equivalent.

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418 APPENDIX A Equations and Inequalities

Linear Inequality in One Variable

A linear inequality in one variable is an inequality that can be written in the form

ax + b + 0,*where a and b are real numbers and a ≠ 0.*The symbol 7 can be replaced with 6 , … , or Ú .

EXAMPLE 7 Solving a Linear Inequality

Solve -3x + 5 7 -7.

SOLUTION -3x + 5 7 -7 -3x + 5 - 5 7 -7 - 5 Subtract 5. -3x 7 -12 Combine like terms.

-3x-3 6

-12-3

Divide by -3. Reverse the direction of the inequality symbol when multiplying or dividing by a negative number.

x 6 4

Don’t forget to reverse the

inequality symbol here.

Inequalities are solved with the properties of inequality, which are similar to the properties of equality. Let a, b, and c represent real numbers.

1. If a * b, then a + c * b + c.2. If a * b and if c + 0, then ac * bc. 3. If a * b and if c * 0, then ac + bc. Replacing 6 with 7 , … , or Ú results in similar properties. (Restrictions on c remain the same.)

Multiplication may be replaced by division in Properties 2 and 3. Always remem-ber to reverse the direction of the inequality symbol when multiplying or dividing by a negative number.

0 4

Figure 1

The original inequality -3x + 5 7 -7 is satisfied by any real number less than 4. The solution set can be written using set-builder notation as5x ' x 6 46, Set-builder notationwhich is read “the set of all x such that x is less than 4.”

The solution set 5x 0 x 6 46 is an example of an interval. Using interval notation, we write it as 1-∞, 42. Interval notationThe symbol -∞ does not represent an actual number. Rather, it is used to show that the interval includes all real numbers less than 4. The interval 1-∞, 42 is an example of an open interval because the endpoint, 4, is not part of the interval. An interval that includes both its endpoints is a closed interval. A square bracket indicates that a number is part of an interval, and a parenthesis indicates that a number is not part of an interval.

The solution set 1-∞, 42 is graphed in Figure 1. ■✔ Now Try Exercise 73.

Linear Inequalities and Interval Notation The definition of a linear inequality is similar to the definition of a linear equation.

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419APPENDIX A Equations and Inequalities

Summary of Types of Intervals (Assume that a * b.)

Type of Interval Set

Interval Notation Graph

Open interval e

5x ' x 7 a65x 'a 6 x 6 b65x ' x 6 b6

1a, ∞21a, b21-∞, b2

Other intervals g 5x ' x Ú a65x 'a 6 x … b65x 'a … x 6 b65x ' x … b6

3a, ∞21a, b43a, b21-∞, b4Closed interval 5x 'a … x … b6 3a, b4Disjoint interval 5x ' x 6 a or x 7 b6 1-∞, a2 ´ 1b, ∞2All real numbers 5x ' x is a real number6 1-∞, ∞2

a b

EXAMPLE 8 Solving a Three-Part Inequality

Solve -2 6 5 + 3x 6 20. Give the solution set in interval notation.

SOLUTION -2 6 5 + 3x 6 20 -2 - 5 6 5 + 3x - 5 6 20 - 5 Subtract 5 from each part. -7 6 3x 6 15 Combine like terms in each part.

-73

6 3x3

6 153

Divide each part by 3.

- 73

6 x 6 5

The solution set, graphed in Figure 2, is the interval A - 73 , 5 B .■✔ Now Try Exercise 83.

5073

–

Figure 2

Three-Part Inequalities The inequality -2 6 5 + 3x 6 20 says that 5 + 3x is between -2 and 20. This inequality is solved using an extension of the properties of inequality given earlier, working with all three expressions at the same time.

Concept Check Fill in the blank to correctly complete each sentence.

1. A(n) is a statement that two expressions are equal.

2. To an equation means to find all numbers that make the equation a true statement.

Appendix A Exercises

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420 APPENDIX A Equations and Inequalities

Solve each equation. See Examples 1 and 2.

7. 5 x + 4 = 3x - 4 8. 9x + 11 = 7x + 1

9. 613x - 12 = 8 - 110 x - 142 10. 41-2 x + 12 = 6 - 12 x - 4211.

x - 2 x + 43

= 53

12. 74

+ 15

x - 32

= 45

13. 3x + 5 - 51x + 12 = 6 x + 7 14. 51x + 32 + 4 x - 3 = -12 x - 42 + 215. 23x - 14 + 2 x2 + 34 = 2 x + 2 16. 432 x - 13 - x2 + 54 = -6 x - 2817. 0.2 x - 0.5 = 0.1x + 7 18. 0.01x + 3.1 = 2.03x - 2.96

19. -412 x - 62 + 8 x = 5 x + 24 + x 20. -813x + 42 + 6 x = 41x - 82 + 4 x

3. A linear equation is a(n) because the greatest degree of the variable is 1.

4. A(n) is an equation satisfied by every number that is a meaningful replace-ment for the variable.

5. A(n) is an equation that has no solution.

6. Concept Check Which one is not a linear equation?

A. 5 x + 71x - 12 = -3x B. 9x2 - 4 x + 3 = 0 C. 7x + 8 x = 13x D. 0.04 x - 0.08 x = 0.40

Determine whether each equation is an identity, a conditional equation, or a contradic-tion. Give the solution set. See Example 3.

21. 412 x + 72 = 2 x + 22 + 312 x + 22 22. 12

16 x + 202 = x + 4 + 21x + 3223. 21x - 82 = 3x - 16 24. -81x + 52 = -8 x - 51x + 8225. 41x + 72 = 21x + 122 + 21x + 12 26. -612 x + 12 - 31x - 42 = -15 x + 1Concept Check Use choices A–D to answer each question.

A. 3x2 - 17x - 6 = 0 B. 12x + 522 = 7C. x2 + x = 12 D. 13x - 121x - 72 = 027. Which equation is set up for direct use of the zero-factor property? Solve it.

28. Which equation is set up for direct use of the square root property? Solve it.

29. Which one or more of these equations can be solved using the quadratic formula?

30. Only one of the equations is set up so that the values of a, b, and c can be deter-mined immediately. Which one is it? Solve it.

Solve each equation using the zero-factor property. See Example 4.

31. x2 - 5x + 6 = 0 32. x2 + 2x - 8 = 0 33. 5x2 - 3x - 2 = 0

34. 2x2 - x - 15 = 0 35. -4x2 + x = -3 36. -6x2 + 7x = -10

37. x2 - 100 = 0 38. x2 - 64 = 0 39. 4x2 - 4x + 1 = 0

40. 9x2 - 12x + 4 = 0 41. 25x2 + 30x + 9 = 0 42. 36x2 + 60x + 25 = 0

Solve each equation using the square root property. See Example 5.

43. x2 = 16 44. x2 = 121 45. x2 - 27 = 0

46. x2 - 48 = 0 47. 13x - 122 = 12 48. 14x + 122 = 20

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421APPENDIX A Equations and Inequalities

Solve each equation using the quadratic formula. See Example 6.

49. x2 - 4x + 3 = 0 50. x2 - 7x + 12 = 0 51. 2x2 - x - 28 = 0

52. 4x2 - 3x - 10 = 0 53. x2 - 2x - 2 = 0 54. x2 - 10x + 18 = 0

55. x2 - 6x = -7 56. x2 - 4x = -1 57. x2 - x - 1 = 0

58. x2 - 3x - 2 = 0 59. -2x2 + 4x + 3 = 0 60. -3x2 + 6x + 5 = 0

Concept Check Match the inequality in each exercise in Column I with its equivalent interval notation in Column II.

61. x 6 -6

62. x … 6

63. -2 6 x … 6

64. x2 Ú 0

65. x Ú -6

66. 6 … x

67.

68.

69.

70.

–2 0 6

0 8

–3 30

–6 0

A. 1-2, 64B. 3-2, 62C. 1-∞, -64D. 36, ∞2E. 1-∞, -32 ´ 13, ∞2F. 1-∞, -62G. 10, 82H. 1-∞, ∞2I. 3-6, ∞2J. 1-∞, 64

71. Explain how to determine whether to use a parenthesis or a square bracket when writing the solution set of a linear inequality in interval notation.

72. Concept Check The three-part inequality a 6 x 6 b means “a is less than x and x is less than b.” Which inequality is not satisfied by some real number x?

A. -3 6 x 6 10 B. 0 6 x 6 6

C. -3 6 x 6 -1 D. -8 6 x 6 -10

Solve each inequality. Give the solution set in interval notation. See Example 7.

73. -2x + 8 … 16 74. -3x - 8 … 7

75. -2x - 2 … 1 + x 76. -4x + 3 Ú -2 + x

77. 31x + 52 + 1 Ú 5 + 3x 78. 6x - 12x + 32 Ú 4x - 579. 8x - 3x + 2 6 21x + 72 80. 2 - 4x + 51x - 12 6 -61x - 2281.

4x + 7-3 … 2x + 5 82.

2x - 5-8 … 1 - x

Solve each inequality. Give the solution set in interval notation. See Example 8.

83. -5 6 5 + 2x 6 11 84. -7 6 2 + 3x 6 5

87. -11 7 -3x + 1 7 -17 88. 2 7 -6x + 3 7 -3

85. 10 … 2x + 4 … 16 86. -6 … 6x + 3 … 21

89. -4 … x + 12

… 5 90. -5 … x - 33

… 1

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422 APPENDIX B Graphs of Equations

■ The Rectangular Coordinate System

■ Equations in Two Variables

■ Circles

The Rectangular Coordinate System Each real number corresponds to a point on a number line. This idea is extended to ordered pairs of real numbers by using two perpendicular number lines, one horizontal and one vertical, that intersect at their zero-points. The point of intersection is the origin. The horizontal line is the x-axis, and the vertical line is the y-axis. See Figure 1.

The x-axis and y-axis together make up a rectangular coordinate system, or Cartesian coordinate system (named for one of its coinventors, René Descartes. The other coinventor was Pierre de Fermat). The plane into which the coordinate system is introduced is the coordinate plane, or xy-plane. See Figure 1. The x-axis and y-axis divide the plane into four regions, or quadrants, labeled as shown. The points on the x-axis or the y-axis belong to no quadrant.

Each point P in the xy-plane corresponds to a unique ordered pair 1a, b2 of real numbers. The point P corresponding to the ordered pair 1a, b2 often is written P1a, b2 as in Figure 1 and referred to as

“the point 1a, b2.” The numbers a and b are the coordinates of point P.

B Graphs of Equations

Rectangular (Cartesian) Coordinate System

Figure 1

ba x-axis

y-axis

QuadrantII

QuadrantI

QuadrantIII

QuadrantIV

P(a, b)

Plotting Points

Figure 2

B(–5, 6) A(3, 4)

E(–3, 0)

D(4, –3)C(–2, –4)

4 units

3 units

Equations in Two Variables Ordered pairs are used to express the solu-tions of equations in two variables. When an ordered pair represents the solution of an equation with the variables x and y, the x-value is written first. For example, we say that 11, 22 is a solution of 2x - y = 0.Substituting 1 for x and 2 for y in the equation gives a true statement.

2x - y = 0 2112 - 2≟ 0 Let x = 1 and y = 2.

0 = 0 ✓ True

To locate on the xy-plane the point corresponding to the ordered pair 13, 42, for example, start at the origin, move 3 units in the positive x-direction, and then move 4 units in the positive y-direction. See Figure 2. Point A corresponds to the ordered pair 13, 42.

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423APPENDIX B Graphs of Equations

EXAMPLE 1 Finding Ordered-Pair Solutions of Equations

For each equation, find at least three ordered pairs that are solutions.

(a) y = 4x - 1 (b) x = 2y - 1 (c) y = x2 - 4SOLUTION

(a) Choose any real number for x or y, and substitute in the equation to obtain the corresponding value of the other variable. For example, let x = -2 and then let y = 3.

y = 4x - 1 y = 41-22 - 1 Let x = -2. y = -8 - 1 Multiply. y = -9 Subtract.

y = 4x - 1 3 = 4x - 1 Let y = 3. 4 = 4x Add 1. 1 = x Divide by 4.

This gives the ordered pairs 1-2, -92 and 11, 32. Verify that the ordered pair 10, -12 is also a solution.

(b) x = 2y - 1 Given equation 1 = 2y - 1 Let x = 1. 1 = y - 1 Square each side. 2 = y Add 1.

One ordered pair is 11, 22. Verify that the ordered pairs 10, 12 and 12, 52 are also solutions of the equation.

y = x2 - 4

and determine the corresponding y-values.

x y

-2 0-1 -3

0 -41 -32 0

1-222 - 4 = 4 - 4 = 0 1-122 - 4 = 1 - 4 = -3

02 - 4 = -4 12 - 4 = -3 22 - 4 = 0

Five ordered pairs are 1-2, 02, 1-1, -32, 10, -42, (1, -32, and 12, 02.■✔ Now Try Exercises 15(a), 19(a), and 21(a).

The graph of an equation is found by plotting ordered pairs that are solutions of the equation. The intercepts of the graph are good points to plot first. An x-intercept is a point where the graph intersects the x-axis, and a y-intercept is a point where the graph intersects the y-axis. In other words, the x-intercept is represented by an ordered pair with y-coordinate 0, and the y-intercept is represented by an ordered pair with x-coordinate 0.

A general algebraic approach for graphing an equation using intercepts and point-plotting follows.

y-intercept

(0, y)

(x, 0)

x-intercept

Intercepts

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424 APPENDIX B Graphs of Equations

Step 2 We use the intercepts and the other ordered pairs found in Exam-ple 1(a): 1-2, -92 and 11, 32.

Step 3 Plot the four ordered pairs from Steps 1 and 2 as shown in Figure 3.

Step 4 Join the points plotted in Step 3 with a straight line. This line, also shown in Figure 3, is the graph of the equation y = 4x - 1.

Figure 3

–2 2x

–6

–3

–9

y = 4x – 1

x = √y – 1

Figure 4

*Intercepts are sometimes defined as numbers, such as x-intercept 14 and y-intercept -1. In this text, we define them as ordered pairs, such as A14 , 0 B and 10, -12.

EXAMPLE 2 Graphing Equations

Graph each of the equations here, from Example 1.

(a) y = 4x - 1 (b) x = 2y - 1 (c) y = x2 - 4SOLUTION

(a) Step 1 Let y = 0 to find the x-intercept, and let x = 0 to find the y-intercept.

The intercepts are A14 , 0 B and 10, -12.* y = 4x - 1 0 = 4x - 1 Let y = 0. 1 = 4x

= x

y = 4x - 1 y = 4102 - 1 Let x = 0. y = 0 - 1 y = -1

Graphing an Equation by Point Plotting

Step 1 Find the intercepts.

Step 2 Find as many additional ordered pairs as needed.

Step 3 Plot the ordered pairs from Steps 1 and 2.

Step 4 Join the points from Step 3 with a smooth line or curve.

(b) For x = 2y - 1, the y-intercept 10, 12 was found in Example 1(b). Solvex = 20 - 1 Let y = 0.

to find the x-intercept. When y = 0, the quantity under the radical symbol is negative, so there is no x-intercept. In fact, y - 1 must be greater than or equal to 0, so y must be greater than or equal to 1.

We plot the ordered pairs 10, 12, 11, 22, and 12, 52 from Example 1(b) and join the points with a smooth curve as in Figure 4. To confirm the direc-tion the curve will take as x increases, we find another solution, 13, 102.

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425APPENDIX B Graphs of Equations

Figure 5

y = x2 – 4

–2 2

–4

(c) In Example 1(c), we made a table of five ordered pairs that satisfy the equa-tion y = x2 - 4.1-2, 02, 1-1, -32, 10, -42, 11, -32, 12, 02

x-intercept y-intercept x-intercept

Plotting the points and joining them with a smooth curve gives the graph in Figure 5. This curve is called a parabola.

■✔ Now Try Exercises 15(b), 19(b), and 21(b).

Center-Radius Form of the Equation of a Circle

A circle with center 1h, k2 and radius r has equation1x − h 2 2 + 1 y − k 2 2 = r2,which is the center-radius form of the equation of the circle. As a special case, a circle with center 10, 02 and radius r has the following equation.

x2 + y2 = r2

EXAMPLE 3 Finding the Center-Radius Form

Find the center-radius form of the equation of each circle described.

(a) center 1-3, 42, radius 6 (b) center 10, 02, radius 3SOLUTION

(a) 1x - h22 + 1y - k22 = r2 Center-radius form 3x - 1-3242 + 1y - 422 = 62 Substitute. Let 1h, k2 = 1-3, 42 and r = 6. 1x + 322 + 1y - 422 = 36 Simplify.Be careful with signs here.

(b) The center is the origin and r = 3.

x2 + y2 = r2 Special case of the center-radius form x2 + y2 = 32 Let r = 3. x2 + y2 = 9 Apply the exponent.

✔ Now Try Exercises 35(a) and 41(a).

(x, y)

(h, k)

Figure 6

Circles By definition, a circle is the set of all points in a plane that lie a given distance from a given point. The given distance is the radius of the circle, and the given point is the center.

We can find the equation of a circle from its definition using the distance formula. Suppose that the point 1h, k2 is the center and the circle has radius r, where r 7 0. Let 1x, y2 represent any point on the circle. See Figure 6.

21x2 - x122 + 1y2 - y122 = d Distance formula 21x - h22 + 1y - k22 = r 1x1, y12 = 1h, k2, 1x2, y22 = 1x, y2, and d = r

1x − h 2 2 + 1 y − k 2 2 = r2 Square each side.

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426 APPENDIX B Graphs of Equations

EXAMPLE 4 Graphing Circles

Graph each circle discussed in Example 3.

(a) 1x + 322 + 1y - 422 = 36 (b) x2 + y2 = 9SOLUTION

(a) Writing the given equation in center-radius form3x - 1-3242 + 1y - 422 = 62 gives 1-3, 42 as the center and 6 as the radius. See Figure 7.

Figure 7

(–3, 4)

(x + 3)2 + (y – 4)2 = 36

(x, y)

–9 3–2

Figure 8

(0, 0)

x2 + y2 = 9

(x, y)

–3 3

–3

(b) The graph with center 10, 02 and radius 3 is shown in Figure 8.✔ Now Try Exercises 35(b) and 41(b).

Concept Check Fill in the blank to correctly complete each sentence.

1. The point 1-1, 32 lies in quadrant in the rectangular coordinate system. 2. The point 14, 2 lies on the graph of the equation y = 3x - 6. 3. Any point that lies on the x-axis has y-coordinate equal to .

4. The y-intercept of the graph of y = -2x + 6 is .

5. The x-intercept of the graph of 2x + 5y = 10 is .

6. Give three ordered pairs from the table.

Appendix B Exercises

x y

2 -5-1 7

3 -95 -176 -21

Concept Check Graph the points on a coordinate system and identify the quadrant or axis for each point.

7. 13, 22 8. 1-7, 62 9. 1-7, -42 10. 18, -52 11. 10, 52 12. 1-8, 02 13. 14.5, 72 14. 1-7.5, 82For each equation, (a) give a table with at least three ordered pairs that are solutions, and (b) graph the equation. See Examples 1 and 2.

15. y = 12

x - 2 16. y = - 12

x + 2 17. 2x + 3y = 5

18. 3x - 2y = 6 19. y = x2 20. y = x2 + 2

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427APPENDIX B Graphs of Equations

Concept Check Fill in the blank(s) to correctly complete each sentence.

27. The circle with equation x2 + y2 = 49 has center with coordinates and radius equal to .

28. The circle with center 13, 62 and radius 4 has equation . 29. The graph of 1x - 422 + 1y + 722 = 9 has center with coordinates . 30. The graph of x2 + 1y - 522 = 9 has center with coordinates .Concept Check Match each equation in Column I with its graph in Column II.

31. 1x - 322 + 1y - 222 = 25 32. 1x - 322 + 1y + 222 = 25 33. 1x + 322 + 1y - 222 = 25 34. 1x + 322 + 1y + 222 = 25

–2–3

3–2

–3

In the following exercises, (a) find the center-radius form of the equation of each circle described, and (b) graph it. See Examples 3 and 4.

35. center 10, 02, radius 6 36. center 10, 02, radius 9 37. center 12, 02, radius 6 38. center 13, 02, radius 3 39. center 10, 42, radius 4 40. center 10, -32, radius 7 41. center 1-2, 52, radius 4 42. center 14, 32, radius 5 43. center 15, -42, radius 7 44. center 1-3, -22, radius 645. center A22, 22 B , radius 22 46. center A -23, -23 B , radius 23Connecting Graphs with Equations Use each graph to determine an equation of the circle in center-radius form.

47.

(1, 1) (5, 1)

(3, –1)

(3, 3)

48.

(–1, –5)

(–4, –2) (2, –2)

(–1, 1)

21. y = 2x - 3 22. y = 2x - 3 23. y = 0 x - 2 024. y = - 0 x + 4 0 25. y = x3 26. y = -x3

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428 APPENDIX C Functions

C Functions

Relations and Functions In algebra, we use ordered pairs to represent related quantities. For example, 13, $10.502 might indicate that we pay $10.50 for 3 gallons of gas. The amount we pay depends on the number of gallons pumped, so the amount (in dollars) is called the dependent variable, and the number of gallons pumped is called the independent variable.

Generalizing, if the value of the second component y depends on the value of the first component x, then y is the dependent variable and x is the indepen-dent variable.

Independent variable Dependent variable1x, y2A set of ordered pairs such as 513, 10.502, 18, 28.002, 110, 35.0026 is a

relation. A function is a special kind of relation.

■ Relations and Functions

■ Domain and Range■ Determining

Whether Relations Are Functions

■ Function Notation■ Increasing,

Decreasing, and Constant Functions

Relation and Function

A relation is a set of ordered pairs. A function is a relation in which, for each distinct value of the first component of the ordered pairs, there is exactly one value of the second component.

EXAMPLE 1 Deciding Whether Relations Define Functions

Decide whether each relation defines a function.

F = 511, 22, 1-2, 42, 13, 426 G = 511, 12, 11, 22, 11, 32, 12, 326 H = 51-4, 12, 1-2, 12, 1-2, 026

SOLUTION Relation F is a function because for each different x-value there is exactly one y-value. We can show this correspondence as follows.51, -2, 36 x-values of F

52, 4, 46 y-values of FAs the correspondence below shows, relation G is not a function because

one first component corresponds to more than one second component.51, 26 x-values of G51, 2, 36 y-values of G

In relation H the last two ordered pairs have the same x-value paired with two different y-values (-2 is paired with both 1 and 0), so H is a relation but not a function. In a function, no two ordered pairs can have the same first compo-nent and different second components.

Different y-values

H = 51-4, 12, 1-2, 12, 1-2, 026 Not a functionSame x-value ■✔ Now Try Exercises 1 and 3.

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429APPENDIX C Functions

Figure 1

1–2

Fx-values y-values

F is a function.

–4

–2

H is not a function.

x-values y-values

Relations and functions can also be expressed as a correspondence or mapping from one set to another, as shown in Figure 1 for function F and relation H from Example 1. The arrow from 1 to 2 indicates that the ordered pair 11, 22 belongs to F— each first component is paired with exactly one second compo-nent. In the mapping for relation H, which is not a function, the first component -2 is paired with two different second compo-nents, 1 and 0.

Because relations and functions are sets of ordered pairs, we can represent them using tables and graphs. A table and graph for function F are shown in Figure 2.

Finally, we can describe a relation or function using a rule that tells how to determine the dependent variable for a specific value of the independent vari-able. The rule may be given in words: for instance, “the dependent variable is twice the independent variable.” Usually the rule is an equation, such as the one below.

Dependent variable y = 2x Independent variable

Figure 2

(3, 4)(–2, 4)

(1, 2)

Graph of F

x y

1 2-2 4 3 4

In a function, there is exactly one value of the dependent variable, the sec-ond component, for each value of the independent variable, the first component.

Domain and Range

For every relation consisting of a set of ordered pairs 1x, y2, there are two important sets of elements.

The set of all values of the independent variable 1x2 is the domain.The set of all values of the dependent variable 1y2 is the range.

SOLUTION

(a) The domain is the set of x-values, 53, 4, 66. The range is the set of y-values, 5-1, 2, 5, 86. This relation is not a function because the same x-value, 4, is paired with two different y-values, 2 and 5.

(b) The domain is 54, 6, 7, -36 and the range is 5100, 200, 3006. This map-ping defines a function. Each x-value corresponds to exactly one y-value.

EXAMPLE 2 Finding Domains and Ranges of Relations

Give the domain and range of each relation. Tell whether the relation defines a function.

(a) 513, -12, 14, 22, 14, 52, 16, 826(b) 4 100

62007300–3

-5 20 25 2

Domain and Range We consider two important concepts concerning relations.

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430 APPENDIX C Functions

(c) This relation, represented by a table, is a set of ordered pairs. The domain is the set of x-values 5-5, 0, 56, and the range is the set of y-values 526. The table defines a function because each different x-value corresponds to exactly one y-value (even though it is the same y-value).

■✔ Now Try Exercises 9, 11, and 13.

EXAMPLE 3 Finding Domains and Ranges from Graphs

Give the domain and range of each relation.

(a)

0(0, –1)

(4, –3)

(–1, 1)(1, 2)

(b)

0–4

–6

Range

Domain

(c)

(d)

0 2

–3

SOLUTION

(a) The domain is the set of x-values, 5-1, 0, 1, 46. The range is the set of y-values, 5-3, -1, 1, 26.

(b) The x-values of the points on the graph include all numbers between -4 and 4, inclusive. The y-values include all numbers between -6 and 6, inclusive.

The domain is 3-4, 44. The range is 3-6, 64. Use interval notation.(c) The arrowheads indicate that the line extends indefinitely left and right, as

well as up and down. Therefore, both the domain and the range include all real numbers, which is written 1-∞, ∞2. Interval notation for the

set of all real numbers

(d) The arrowheads indicate that the graph extends indefinitely left and right, as well as upward. The domain is 1-∞, ∞2. Because there is a least y-value, -3, the range includes all numbers greater than or equal to -3, written 3-3, ∞2.

■✔ Now Try Exercise 19.

Vertical Line Test

If every vertical line intersects the graph of a relation in no more than one point, then the relation is a function.

Determining Whether Relations Are Functions Because each value of x leads to only one value of y in a function, any vertical line must intersect the graph in at most one point. This is the vertical line test for a function.

x y

-5 20 25 2

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431APPENDIX C Functions

The graph in Figure 3(a) represents a function because each vertical line intersects the graph in no more than one point. The graph in Figure 3(b) is not the graph of a function because there exists a vertical line that intersects the graph in more than one point.

EXAMPLE 4 Using the Vertical Line Test

Use the vertical line test to determine whether each relation graphed in Example 3 is a function.

SOLUTION We repeat each graph from Example 3, this time with vertical lines drawn through the graphs.

(a)

(4, –3)

(–1, 1)(1, 2)

(0, –1)

(b)

0–4

–6

(c)

(d)

0 2

–3

The graphs of the relations in parts (a), (c), and (d) pass the vertical line test because every vertical line intersects each graph no more than once. Thus, these graphs represent functions.

The graph of the relation in part (b) fails the vertical line test because the same x-value corresponds to two different y-values. Therefore, it is not the graph of a function.

■✔ Now Try Exercises 15 and 17.

(a)

x1 x2 x3

y2y1

This is the graph of a function.Each x-value corresponds

to only one y-value.

x10

y1 (x1, y1)

(x1, y2)

This is not the graph of a function.The same x-value corresponds to

two different y-values. (b)

Figure 3

The vertical line test is a simple method for identifying a function defined by a graph. Deciding whether a relation defined by an equation or an inequality is a function, as well as determining the domain and range, is more difficult. The next example gives some hints that may help.

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432 APPENDIX C Functions

EXAMPLE 5 Identifying Functions, Domains, and Ranges

Decide whether each relation defines y as a function of x, and give the domain and range.

(a) y = x + 4 (b) y = 22x - 1 (c) y2 = x (d) y = 5x - 1

SOLUTION

(a) In the defining equation (or rule), y = x + 4, y is always found by adding 4 to x. Thus, each value of x corresponds to just one value of y, and the relation defines a function. The variable x can represent any real number, so the domain is5x ' x is a real number6, or 1-∞, ∞2.

Because y is always 4 more than x, y also may be any real number, and so the range is 1-∞, ∞2.

(b) For any choice of x in the domain of y = 22x - 1, there is exactly one cor-responding value for y (the radical is a nonnegative number), so this equation defines a function. The equation involves a square root, so the quantity under the radical sign cannot be negative.

2x - 1 Ú 0 Solve the inequality. 2x Ú 1 Add 1.

x Ú 12

Divide by 2.

The domain of the function is C 12 , ∞ B. Because the radical must represent a nonnegative number, as x takes values greater than or equal to 12 , the range is 5y ' y Ú 06, or 30, ∞2. See Figure 4.

(c) The ordered pairs 116, 42 and 116, -42 both satisfy the equation y2 = x. There exists at least one value of x —for example, 16—that corresponds to two values of y, 4 and -4, so this equation does not define a function. Because x is equal to the square of y, the values of x must always be nonnegative. The domain of the relation is 30, ∞2. Any real number can be squared, so the range of the relation is 1-∞, ∞2. See Figure 5.

(d) Given any value of x in the domain of

y = 5x - 1 ,

we find y by subtracting 1 from x, and then dividing the result into 5. This process produces exactly one value of y for each value in the domain, so this equation defines a function.

The domain of y = 5x - 1 includes all real numbers except those that make the denominator 0. We find these numbers by setting the denominator equal to 0 and solving for x.

x - 1 = 0 x = 1 Add 1.

Thus, the domain includes all real numbers except 1, written as the inter-val 1-∞, 12 ´ 11, ∞2. Values of y can be positive or negative, but never 0, because a fraction cannot equal 0 unless its numerator is 0. Therefore, the range is the interval 1-∞, 02 ´ 10, ∞2, as shown in Figure 6.

■✔ Now Try Exercises 23, 25, and 29.

Figure 4

2 4 6

Domain

Range

y = √2x – 1

4 8 12 16

–4

–2

y2 = x

Domain

Range

Figure 5

Figure 6

–4

2 6

–5

Range

Domain

y = 5x – 1

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433APPENDIX C Functions

Function Notation When a function ƒ is defined with a rule or an equation using x and y for the independent and dependent variables, we say, “y is a function of x” to emphasize that y depends on x. We use the notation

y = ƒ 1x 2 ,called function notation, to express this and read ƒ1x2 as “ƒ of x,” or “ƒ at x.” The letter ƒ is the name given to this function.

For example, if y = 3x - 5, we can name the function ƒ and write

ƒ1x2 = 3x - 5.Note that ƒ 1x 2 is just another name for the dependent variable y. For example, if y = ƒ1x2 = 3x - 5 and x = 2, then we find y, or ƒ122, by replacing x with 2.

ƒ122 = 3 # 2 - 5 Let x = 2. ƒ122 = 1 Multiply, and then subtract.

The statement “In the function ƒ, if x = 2, then y = 1” represents the ordered pair 12, 12 and is abbreviated with function notation as follows.

ƒ122 = 1The symbol ƒ122 is read “ƒ of 2” or “ƒ at 2.”

Function notation can be illustrated as follows.

Name of the function Defining expression $1%1&

y = ƒ1x2 = 3x - 5Value of the function Name of the independent variable

EXAMPLE 6 Using Function Notation

Let ƒ1x2 = -x2 + 5x - 3 and g1x2 = 2x + 3. Find each of the following.(a) ƒ122 (b) ƒ1q2 (c) g1a + 12SOLUTION

(a) ƒ1x2 = -x2 + 5x - 3 ƒ122 = -22 + 5 # 2 - 3 Replace x with 2. ƒ122 = -4 + 10 - 3 Apply the exponent and multiply. ƒ122 = 3 Add and subtract.

Thus, ƒ122 = 3, and the ordered pair 12, 32 belongs to ƒ.(b) ƒ1x2 = -x2 + 5x - 3

ƒ1q2 = -q2 + 5q - 3 Replace x with q.(c) g1x2 = 2x + 3

g1a + 12 = 21a + 12 + 3 Replace x with a + 1. g1a + 12 = 2a + 2 + 3 Distributive property g1a + 12 = 2a + 5 Add.

■✔ Now Try Exercises 35, 43, and 49.

Functions can be evaluated in a variety of ways, as shown in Example 7.

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434 APPENDIX C Functions

EXAMPLE 7 Using Function Notation

For each function, find ƒ132.(a) ƒ1x2 = 3x - 7 (b) ƒ = 51-3, 52, 10, 32, 13, 12, 16, -126(c)

–23

fDomain Range

(d)

2 4

y = f(x)

SOLUTION

(a) ƒ1x2 = 3x - 7 ƒ132 = 3132 - 7 Replace x with 3. ƒ132 = 2 Simplify.ƒ132 = 2 indicates that the ordered pair 13, 22 belongs to ƒ.

(b) For ƒ = 51-3, 52, 10, 32, 13, 12, 16, -126, we want ƒ132, the y-value of the ordered pair where x = 3. As indicated by the ordered pair 13, 12, when x = 3, y = 1, so ƒ132 = 1.

(d) To evaluate ƒ132 using the graph, find 3 on the x-axis. See Figure 8. Then move up until the graph of ƒ is reached. Moving horizontally to the y-axis gives 4 for the corresponding y-value. Thus, ƒ132 = 4.

■✔ Now Try Exercises 51, 53, and 55.

Figure 7

–23

fDomain Range

Figure 8

2 43

y = f(x)

Increasing, Decreasing, and Constant Functions Informally speak-ing, a function increases over an open interval of its domain if its graph rises from left to right on the interval. It decreases over an open interval of its domain if its graph falls from left to right on the interval. It is constant over an open interval of its domain if its graph is horizontal on the interval.

For example, consider Figure 9.

The function increases over the open interval 1-2, 12 because the y-values continue to get larger for x-values in that interval.

The function is constant over the open inter-val 11, 42 because the y-values are always 5 for all x-values there.

The function decreases over the open interval 14, 62 because in that interval the y-values continuously get smaller.

The intervals refer to the x-values where the y-values either increase, decrease, or are con-stant.

The formal definitions of these concepts follow.

Figure 9

–2 1 2 4 6

y isincreasing.

y isdecreasing.y is

constant.

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435APPENDIX C Functions

Increasing, Decreasing, and Constant Functions

Suppose that a function ƒ is defined over an open interval I and x1 and x2 are in I.

(a) ƒ increases over I if, whenever x1 6 x2, ƒ1x12 6 ƒ1x22.(b) ƒ decreases over I if, whenever x1 6 x2, ƒ1x12 7 ƒ1x22.(c) ƒ is constant over I if, for every x1 and x2, ƒ1x12 = ƒ1x22.

f (x1)

f (x2)

x1 x20

Whenever x1 < x2, and f (x1) < f (x2), f is increasing.

y = f(x)

xf (x2)

f (x1)

x1 x20

Whenever x1 < x2, and f (x1) > f (x2), f is decreasing.

y = f(x)

f(x1) = f (x2)

x1 x20

For every x1 and x2, if f (x1) = f (x2), then f is constant.

y = f(x)

(a) (b) (c)

Figure 10

Figure 10 illustrates these ideas.

Figure 11

8(1, 8)

(–2, –1)

1 3

–2

–3 –1

SOLUTION We observe the domain and ask, “What is happening to the y-values as the x-values are getting larger?” Moving from left to right on the graph, we see the following:

On the open interval 1-∞, -22, the y-values are decreasing.On the open interval 1-2, 12, the y-values are increasing.On the open interval 11, ∞2, the y-values are constant (and equal to 8).

Therefore, the function is decreasing on 1-∞, -22, increasing on 1-2, 12, and constant on 11, ∞2.

■✔ Now Try Exercise 65.

NOTE To decide whether a function is increasing, decreasing, or con-stant over an interval, ask yourself, “What does y do as x goes from left to right?” Our definition of increasing, decreasing, and constant function behavior applies to open intervals of the domain, not to individual points.

EXAMPLE 8 Determining Open Intervals of a Domain

Figure 11 shows the graph of a function. Determine the largest open intervals of the domain over which the function is increasing, decreasing, or constant.

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436 APPENDIX C Functions

Appendix C Exercises

Decide whether each relation defines a function. See Example 1.

1. 515, 12, 13, 22, 14, 92, 17, 826 2. 518, 02, 15, 72, 19, 32, 13, 826 3. 512, 42, 10, 22, 12, 626 4. 519, -22, 1-3, 52, 19, 126 5. 51-3, 12, 14, 12, 1-2, 726 6. 51-12, 52, 1-10, 32, 18, 326 7. x y

3 -47 -4

10 -4

8. x y

-4 220 224 22

Decide whether each relation defines a function, and give the domain and range. See Examples 1– 4.

9. 511, 12, 11, -12, 10, 02, 12, 42, 12, -426 10. 512, 52, 13, 72, 13, 92, 15, 112611. 2

511173

12. 1235

10151927

14. x y

0 01 -12 -2

13. x y

0 0-1 1-2 2

15.

x0 2

16.

–3

17.

18.

–44

–3

y 19.

–3

–4

y 20.

–2 2

Decide whether each relation defines y as a function of x. Give the domain and range. See Example 5.

21. y = x2 22. y = x3 23. x = y6

24. x = y4 25. y = 2x - 5 26. y = -6x + 4

27. y = 2x 28. y = -2x 29. y = 24x + 1 30. y = 27 - 2x 31. y = 2

(Video) TRIGONOMETRY TRICK/SHORTCUT FOR JEE/NDA/NA/CETs/AIRFORCE/RAILWAYS/BANKING/SSC-CGL

x - 3 32. y =-7

x - 5

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437APPENDIX C Functions

33. Concept Check Choose the correct answer: For function ƒ, the notation ƒ132 meansA. the variable ƒ times 3, or 3ƒ.

B. the value of the dependent variable when the independent variable is 3.

C. the value of the independent variable when the dependent variable is 3.

D. ƒ equals 3.

34. Concept Check Give an example of a function from everyday life. (Hint: Fill in the blanks: depends on , so is a function of .)

Let ƒ1x2 = -3x + 4 and g1x2 = -x2 + 4x + 1. Find each of the following. Simplify if necessary. See Example 6.

35. ƒ102 36. ƒ1-32 37. g1-22 38. g110239. ƒa1

3b 40. ƒa - 7

3b 41. ga1

2b 42. ga - 1

43. ƒ1p2 44. g1k2 45. ƒ1-x2 46. g1-x247. ƒ1x + 22 48. ƒ1a + 42 49. ƒ12m - 32 50. ƒ13t - 22For each function, find (a) ƒ122 and (b) ƒ1-12. See Example 7.51. ƒ = 51-1, 32, 14, 72, 10, 62, 12, 226 52. ƒ = 512, 52, 13, 92, 1-1, 112, 15, 32653.

–1235

10151927

f 54.

–13

55.

2–2 4

y = f(x)

56.

y = f(x)

57.

0–2

4321–2x

–1

58.

0–2

4321–2 –1x

Use the graph of y = ƒ1x2 to find each function value: (a) ƒ1-22, (b) ƒ102, (c) ƒ112, and (d) ƒ142. See Example 7(d).

59.

–4–2

4321–2x

–1

60.

–4–2

4321–2x

–1

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438 APPENDIX D Graphing Techniques

Determine the largest open intervals of the domain over which each function is (a) increasing, (b) decreasing, and (c) constant. See Example 8.

61. 62. 63.

64. 65. 66.

(0, –2)

(–2, –4)

0 2–2

–4

(–1, 3)(–3, 1)

0 2–1–2

1–3

y(–2, 4)

(2, –4)

0 2–2

–4

(–3, –1)

(3, 1)

0 3–3

–1

(–1, –3) (1, –3)(0, –2)

0 2–2x

(–2, 0) (2, 0)

(0, –4)

D Graphing Techniques

Graphing techniques presented in this section show how to graph functions that are defined by altering the equation of a basic function.

■ Stretching and Shrinking

■ Reflecting■ Symmetry■ Translations

Stretching and Shrinking We begin by considering how the graphs of y = aƒ1x2 and y = ƒ1ax2 compare to the graph of y = ƒ1x2, where a 7 0.

EXAMPLE 1 Stretching or Shrinking Graphs

Graph each function.

(a) g1x2 = 2 0 x 0 (b) h1x2 = 12

0 x 0 (c) k1x2 = 0 2x 0SOLUTION

(a) Comparing the tables of values for ƒ1x2 = 0 x 0 and g1x2 = 2 0 x 0 in Figure 1 on the next page, we see that for corresponding x-values, the y-values of g are each twice those of ƒ. The graph of ƒ1x2 = 0 x 0 is vertically stretched. The graph of g1x2, shown in blue in Figure 1, is narrower than that of ƒ1x2, shown in red for comparison.

NOTE Recall from algebra that 'a ' is the absolute value of a number a.

∣a ∣ = b a if a is positive or 0−a if a is negative

Thus, '2 ' = '2 ' and ' -2 ' = '2 ' . We use absolute value functions to illustrate many

of the graphing techniques in this section.

–2

Graph of the absolutevalue function

y = u x u0

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439APPENDIX D Graphing Techniques

–3 0

6g(x) = 2!x!

f (x) = !x!

Figure 1

x ƒ 1x 2 = ∣x ∣ g 1x 2 = 2 ∣x ∣-2 2 4-1 1 2

0 0 01 1 22 2 4

(b) The graph of h1x2 = 12 0 x 0 is also the same general shape as that of ƒ1x2, but here the coefficient 12 is between 0 and 1 and causes a vertical shrink. The graph of h1x2 is wider than the graph of ƒ1x2, as we see by comparing the tables of values. See Figure 2.

42–2–4 0

h(x) = !x!

f (x) = !x!

Figure 2

x ƒ 1x 2 = ∣x ∣ h 1x 2 = 12 ∣x ∣-2 2 1-1 1 12

0 0 0

1 1 122 2 1

Vertical Stretching or Shrinking of the Graph of a Function

Suppose that a 7 0. If a point 1x, y2 lies on the graph of y = ƒ1x2, then the point 1x, ay2 lies on the graph of y = aƒ 1x 2 .(a) If a 7 1, then the graph of y = aƒ1x2 is a vertical stretching of the

graph of y = ƒ1x2.(b) If 0 6 a 6 1, then the graph of y = aƒ1x2 is a vertical shrinking of the

graph of y = ƒ1x2.Figure 3 shows graphical interpretations of vertical stretching and shrinking.

In both cases, the x-intercepts of the graph remain the same but the y-intercepts are affected.

(x, y)

(x, ay)y = af(x), a > 1

y = f(x)

Vertical stretchinga > 1

(x, y)

(x, ay)

y = af(x), 0 < a < 1

y = f(x)

Vertical shrinking0 < a < 1

Figure 3

(c) Use the property of absolute value that states 0 ab 0 = 0 a 0 # 0 b 0 to rewrite 0 2x 0 .k1x2 = 0 2x 0 = 0 2 0 # 0 x 0 = 2 0 x 0

Therefore, the graph of k1x2 = 0 2x 0 is the same as the graph of g1x2 = 2 0 x 0 in part (a). This is a horizontal shrink of the graph of ƒ1x2 = 0 x 0 . See Figure 1.

■✔ Now Try Exercises 13 and 15.

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440 APPENDIX D Graphing Techniques

Horizontal Stretching or Shrinking of the Graph of a Function

Suppose that a 7 0. If a point 1x, y2 lies on the graph of y = ƒ1x2, then the point A xa , y B lies on the graph of y = ƒ 1ax 2 .(a) If 0 6 a 6 1, then the graph of y = ƒ1ax2 is a horizontal stretching of

the graph of y = ƒ1x2.(b) If a 7 1, then the graph of y = ƒ1ax2 is a horizontal shrinking of the

graph of y = ƒ1x2.

Graphs of functions can also be stretched and shrunk horizontally.

See Figure 4 for graphical interpretations of horizontal stretching and shrinking. In both cases, the y-intercept remains the same but the x-intercepts are affected.

(x, y)

y = f(ax), 0 < a < 1

y = f(x)

Horizontal stretching0 < a < 1

( , y)xa

(x, y)

y = f(ax), a > 1

y = f(x)

Horizontal shrinkinga > 1

( , y)xa

Figure 4

Reflecting Forming the mirror image of a graph across a line is called reflecting the graph across the line.

EXAMPLE 2 Reflecting Graphs across Axes

Graph each function.

(a) g1x2 = -2x (b) h1x2 = 2-xSOLUTION

(a) The tables of values for g1x2 = -2x and ƒ1x2 = 2x are shown with their graphs in Figure 5. As the tables suggest, every y-value of the graph of g1x2 = -2x is the negative of the corresponding y-value of ƒ1x2 = 2x. This has the effect of reflecting the graph across the x-axis.

–4 –2 2 4

–2

f(x) = √x

g(x) = –√x

Figure 5

x ƒ 1x 2 = !x g 1x 2 = −!x0 0 01 1 -14 2 -2

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441APPENDIX D Graphing Techniques

(b) The domain of h1x2 = 2-x is 1-∞, 04, while the domain of ƒ1x2 = 2x is 30, ∞2. Choosing x-values for h1x2 that are negatives of those used for ƒ1x2, we see that corresponding y-values are the same. The graph of h is a reflec-tion of the graph of ƒ across the y-axis. See Figure 6.

x ƒ 1x 2 = !x h 1x 2 = !−x-4 undefined 2-1 undefined 1

0 0 01 1 undefined4 2 undefined

4–4 0

f(x) = √xh(x) = √–x

Figure 6

■✔ Now Try Exercises 23 and 29.

Reflecting across an Axis

The graph of y = −ƒ 1x 2 is the same as the graph of y = ƒ1x2 reflected across the x-axis. (If a point 1x, y2 lies on the graph of y = ƒ1x2, then 1x, -y2 lies on this reflection.)The graph of y = ƒ 1−x 2 is the same as the graph of y = ƒ1x2 reflected across the y-axis. (If a point 1x, y2 lies on the graph of y = ƒ1x2, then 1-x, y2 lies on this reflection.)

The graphs in Example 2 suggest the following generalizations.

Symmetry The graph of ƒ shown in Figure 7(a) is cut in half by the y-axis, with each half the mirror image of the other half. Such a graph is symmetric with respect to the y-axis. In general, for a graph to be symmetric with respect to the y-axis, the point 1−x, y 2 must be on the graph whenever the point 1x, y 2 is on the graph.

(–x, y) (x, y)

y = f(x)

y-axis symmetry

(a)

(x, y)

(x, –y)

x-axis symmetry

(b)Figure 7

(a) (b)

Similarly, if the graph in Figure 7(b) were folded in half along the x-axis, the portion at the top would exactly match the portion at the bottom. Such a graph is symmetric with respect to the x-axis. In general, for a graph to be symmetric with respect to the x-axis, the point 1x, −y 2 must be on the graph whenever the point 1x, y 2 is on the graph.

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442 APPENDIX D Graphing Techniques

y-axis symmetry

2–2 0

y = x2 + 4

Figure 8

–3–2

x = y2 – 3

x-axis symmetry

Figure 9

Symmetry with Respect to an Axis

The graph of an equation is symmetric with respect to the y-axis if the replacement of x with -x results in an equivalent equation.The graph of an equation is symmetric with respect to the x-axis if the replacement of y with -y results in an equivalent equation.

EXAMPLE 3 Testing for Symmetry with Respect to an Axis

Test for symmetry with respect to the x-axis and the y-axis.

(a) y = x2 + 4 (b) x = y2 - 3 (c) x2 + y2 = 16 (d) 2x + y = 4

SOLUTION

(a) In y = x2 + 4, replace x with -x.

y = x2 + 4 y = 1-x22 + 4 The result is equivalent to the original equation. y = x2 + 4

Use parentheses around -x.

Thus the graph, shown in Figure 8, is symmetric with respect to the y-axis. The y-axis cuts the graph in half, with the halves being mirror images.

Now replace y with -y to test for symmetry with respect to the x-axis.

y = x2 + 4 -y = x2 + 4 The result is not equivalent to the original equation.

y = -x2 - 4Multiply

by -1.

The graph is not symmetric with respect to the x-axis. See Figure 8.

(b) In x = y2 - 3, replace y with -y.

x = 1-y22 - 3 = y2 - 3 Same as the original equation The graph is symmetric with respect to the x-axis, as shown in Figure 9. It

is not symmetric with respect to the y-axis.

(c) Substitute -x for x and then -y for y in x2 + y2 = 16.1-x22 + y2 = 16 and x2 + 1-y22 = 16 Both simplify to the original equation,

x2 + y2 = 16.

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443APPENDIX D Graphing Techniques

–4

x2 + y2 = 16

x-axis andy-axis symmetry

Figure 10

No x-axis ory-axis symmetry

Figure 11

Another kind of symmetry occurs when a graph can be rotated 180° about the origin, with the result coinciding exactly with the original graph. Symmetry of this type is symmetry with respect to the origin. In general, for a graph to be symmetric with respect to the origin, the point 1−x, −y 2 is on the graph whenever the point 1x, y 2 is on the graph.

Figure 12 shows two such graphs.

(–x, –y)

(x, y)

(–x, –y)

(x, y)

Origin symmetry

(–x, –y)

(x, y)

Origin symmetry

Figure 12

Symmetry with Respect to the Origin

The graph of an equation is symmetric with respect to the origin if the replacement of both x with -x and y with -y at the same time results in an equivalent equation.

2x + y = 4 21-x2 + y = 4 Not equivalent

-2x + y = 4

2x + y = 4 2x + 1-y2 = 4 Not equivalent

2x - y = 4

The graph is not symmetric with respect to either axis. See Figure 11.

■✔ Now Try Exercise 35.

(d) In 2x + y = 4, replace x with -x and then replace y with -y.

The graph, a circle of radius 4 centered at the origin, is symmetric with respect to both axes. See Figure 10.

EXAMPLE 4 Testing for Symmetry with Respect to the Origin

Determine whether the graph of each equation is symmetric with respect to the origin.

(a) x2 + y2 = 16 (b) y = x3

SOLUTION

(a) Replace x with -x and y with -y.

x2 + y2 = 161-x22 + 1-y22 = 16 Equivalentx2 + y2 = 16

Use parentheses around -x and -y.

The graph, which is the circle shown in Figure 10 in Example 3(c), is sym-metric with respect to the origin.

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444 APPENDIX D Graphing Techniques

(b) In y = x3, replace x with -x and y with -y.

y = x3

-y = 1-x23 Equivalent

-y = -x3

y = x3

The graph, which is that of the cubing function, is symmetric with respect to the origin and is shown in Figure 13.

■✔ Now Try Exercise 39.

–2 –1

–8

y = x3

Origin symmetry

Figure 13Notice the following important concepts regarding symmetry:

A graph symmetric with respect to both the x- and y-axes is automatically symmetric with respect to the origin. (See Figure 10.)

A graph symmetric with respect to the origin need not be symmetric with respect to either axis. (See Figure 13.)

Of the three types of symmetry—with respect to the x-axis, with respect to the y-axis, and with respect to the origin—a graph possessing any two types must also exhibit the third type of symmetry.

A graph symmetric with respect to the x-axis does not represent a function. (See Figures 9 and 10.)

Translations The next examples show the results of horizontal and verti-cal shifts, or translations, of the graph of ƒ1x2 = 0 x 0 .

EXAMPLE 5 Translating a Graph Vertically

Graph g1x2 = 0 x 0 - 4.SOLUTION Comparing the table shown with Figure 14, we see that for corre-sponding x-values, the y-values of g are each 4 less than those for ƒ. The graph of g1x2 = 0 x 0 - 4 is the same as that of ƒ1x2 = 0 x 0 , but translated 4 units down. The lowest point is at 10, -42. The graph is symmetric with respect to the y-axis and is therefore the graph of an even function.

(0, –4)

–4 40

g(x) = |x| – 4

f(x) = |x|

Figure 14

x ƒ 1x 2 = ∣x ∣ g 1x 2 = ∣x ∣ − 4-4 4 0-1 1 -3 0 0 -4 1 1 -3 4 4 0

■✔ Now Try Exercise 51.

The graphs in Example 5 suggest the following generalization.

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445APPENDIX D Graphing Techniques

Vertical Translations

Given a function g defined by g 1x 2 = ƒ 1x 2 + c, where c is a real number:For every point 1x, y2 on the graph of ƒ, there will be a corresponding point 1x, y + c2 on the graph of g.The graph of g will be the same as the graph of ƒ, but translated c units up if c is positive or 0 c 0 units down if c is negative.

The graph of g is a vertical translation of the graph of ƒ. See Figure 15.

y = f(x) + 2

Verticaltranslation3 units down

Verticaltranslation2 units up

Original graphy = f(x) – 3

y = f(x)

–1–2

–3–4

Figure 15

EXAMPLE 6 Translating a Graph Horizontally

Graph g1x2 = 0 x - 4 0 .SOLUTION Comparing the tables of values given with Figure 16 shows that for corresponding y-values, the x-values of g are each 4 more than those for ƒ. The graph of g1x2 = 0 x - 4 0 is the same as that of ƒ1x2 = 0 x 0 , but translated 4 units to the right. The lowest point is at 14, 02. As suggested by the graphs in Figure 16, this graph is symmetric with respect to the line x = 4.

x ƒ 1x 2 = ∣x ∣ g 1x 2 = ∣x − 4 ∣-2 2 6

0 0 42 2 24 4 06 6 2 x

(4, 0) 80

g(x) = !x – 4!

f (x) = !x!

Figure 16

■✔ Now Try Exercise 49.

The graphs in Example 6 suggest the following generalization.

Horizontal Translations

Given a function g defined by g 1x 2 = ƒ 1x − c 2 , where c is a real number:For every point 1x, y2 on the graph of ƒ, there will be a corresponding point 1x + c, y2 on the graph of g.The graph of g will be the same as the graph of ƒ, but translated c units to the right if c is positive or 0 c 0 units to the left if c is negative.

The graph of g is a horizontal translation of the graph of ƒ. See Figure 17.

–3 –1 1 2 3 4

Horizontaltranslation2 units tothe lefty = f (x + 2)

Horizontaltranslation3 units tothe right

Originalgraphy = f (x)

y = f(x – 3)

Figure 17

CAUTION Errors frequently occur when horizontal shifts are involved. Find the value that causes the expression x - h to equal 0, as shown below.

F 1x 2 = 1x − 5 2 2Because +5 causes x - 5 to equal 0, the graph of F1x2 illustrates a shift of

5 units to the right.

F 1x 2 = 1x + 5 2 2Because −5 causes x + 5 to equal 0, the graph of F1x2 illustrates a shift of

5 units to the left.

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446 APPENDIX D Graphing Techniques

EXAMPLE 7 Using More Than One Transformation

Graph each function.

(a) ƒ1x2 = - 0 x + 3 0 + 1 (b) h1x2 = 0 2x - 4 0 (c) g1x2 = - 12

x2 + 4

SOLUTION

(a) To graph ƒ1x2 = - 0 x + 3 0 + 1, the lowest point on the graph of y = 0 x 0 is translated 3 units to the left and 1 unit up. The graph opens down because of the negative sign in front of the absolute value expression, making the lowest point now the highest point on the graph, as shown in Figure 18. The graph is symmetric with respect to the line x = -3.

(b) To determine the horizontal translation, factor out 2.

h1x2 = 0 2x - 4 0 h1x2 = 0 21x - 22 0 Factor out 2. h1x2 = 0 2 0 # 0 x - 2 0 0 ab 0 = 0 a 0 # 0 b 0 h1x2 = 2 0 x - 2 0 0 2 0 = 2

The graph of h is the graph of y = 0 x 0 translated 2 units to the right, and vertically stretched by a factor of 2. Horizontal shrinking gives the same appearance as vertical stretching for this function. See Figure 19.

(c) The graph of g1x2 = - 12 x2 + 4 has the same shape as that of y = x2, but it is wider (that is, shrunken vertically), reflected across the x-axis because the coefficient - 12 is negative, and then translated 4 units up. See Figure 20.

–2–3–6 0

f(x) = –⏐x + 3⏐ + 1

Figure 18

y = !x!

h(x) = !2x – 4! = 2!x – 2!

2 40

Figure 19

–3 3

–3

g(x) = – x2 + 412

Figure 20

■✔ Now Try Exercises 55, 57, and 65.

Concept Check Fill in the blank(s) to correctly complete each sentence.

1. To graph the function ƒ1x2 = x2 - 3, shift the graph of y = x2 down units. 2. To graph the function ƒ1x2 = x2 + 5, shift the graph of y = x2 up units. 3. The graph of ƒ1x2 = 1x + 422 is obtained by shifting the graph of y = x2 to the

4 units.

4. The graph of ƒ1x2 = 1x - 722 is obtained by shifting the graph of y = x2 to the 7 units.

5. The graph of ƒ1x2 = -2x is a reflection of the graph of y = 2x across the -axis.

Appendix D Exercises

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447APPENDIX D Graphing Techniques

6. The graph of ƒ1x2 = 2-x is a reflection of the graph of y = 2x across the -axis.

7. To obtain the graph of ƒ1x2 = 1x + 223 - 3, shift the graph of y = x3 to the left units and down units.

8. To obtain the graph of ƒ1x2 = 1x - 323 + 6, shift the graph of y = x3 to the right units and up units.

11. Match each equation with the sketch of its graph in A–I.

(a) y = x2 + 2 (b) y = x2 - 2 (c) y = 1x + 222(d) y = 1x - 222 (e) y = 2x2 (f) y = -x2(g) y = 1x - 222 + 1 (h) y = 1x + 222 + 1 (i) y = 1x + 222 - 1A.

0–2

0 2

(–2, 1)0

0–2

(2, 1)0

(–2, –1)0

10. Match each equation in Column I with a description of its graph from Column II as it relates to the graph of y = 23 x.

(a) y = 423 x(b) y = -23 x(c) y = 23 -x(d) y = 23 x - 4(e) y = 23 x - 4

A. a translation 4 units to the right

B. a translation 4 units down

C. a reflection across the x-axis

D. a reflection across the y-axis

E. a vertical stretching by a factor of 4

(a) y = 1x - 722(b) y = x2 - 7(c) y = 7x2

(d) y = 1x + 722(e) y = x2 + 7

A. a translation 7 units to the left

B. a translation 7 units to the right

C. a translation 7 units up

D. a translation 7 units down

E. a vertical stretching by a factor of 7

Concept Check Work each matching problem.

9. Match each equation in Column I with a description of its graph from Column II as it relates to the graph of y = x2.

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448 APPENDIX D Graphing Techniques

Graph each function. See Examples 1 and 2.

13. ƒ1x2 = 3 0 x 0 14. ƒ1x2 = 4 0 x 0 15. ƒ1x2 = 23

0 x 016. ƒ1x2 = 3

4 0 x 0 17. g1x2 = 2x2 18. g1x2 = 3x2

19. g1x2 = 12

x2 20. g1x2 = 13

x2 21. ƒ1x2 = - 12

22. ƒ1x2 = - 13

x2 23. ƒ1x2 = -3 0 x 0 24. ƒ1x2 = -2 0 x 025. h1x2 = ` - 1

2 x ` 26. h1x2 = ` - 1

3 x ` 27. h1x2 = 24x

28. h1x2 = 29x 29. ƒ1x2 = -2-x 30. ƒ1x2 = - 0 -x 0

12. Match each equation with the sketch of its graph in A–I.

(a) y = 0 x - 2 0 (b) y = 0 x 0 - 2 (c) y = 0 x 0 + 2(d) y = 2 0 x 0 (e) y = - 0 x 0 (f ) y = 0 -x 0(g) y = -2 0 x 0 (h) y = 0 x - 2 0 + 2 (i) y = 0 x + 2 0 - 2

0–2 2

0–2

(–2, –2)–2

0–2

–2

0–1

–1

–2

0 (2, 0)

–2 2

–2

0–2 2

(0, 2)

0 2

(2, 2)

Concept Check Plot each point, and then plot the points that are symmetric to the given point with respect to the (a) x-axis, (b) y-axis, and (c) origin.

31. 15, -32 32. 1-6, 12 33. 1-4, -22 34. 1-8, 02Without graphing, determine whether each equation has a graph that is symmetric with respect to the x-axis, the y-axis, the origin, or none of these. See Examples 3 and 4.

35. y = x2 + 5 36. y = 2x4 - 3

37. x2 + y2 = 12 38. y2 - x2 = -6

39. y = -4x3 + x 40. y = x3 - x

41. y = x2 - x + 8 42. y = x + 15

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449APPENDIX D Graphing Techniques

Graph each function. See Examples 5–7.

43. ƒ1x2 = x2 - 1 44. ƒ1x2 = x2 - 2 45. ƒ1x2 = x2 + 246. ƒ1x2 = x2 + 3 47. g1x2 = 1x - 422 48. g1x2 = 1x - 22249. g1x2 = 1x + 222 50. g1x2 = 1x + 322 51. g1x2 = 0 x 0 - 152. g1x2 = 0 x + 3 0 + 2 53. h1x2 = -1x + 123 54. h1x2 = -1x - 12355. h1x2 = 2x2 - 1 56. h1x2 = 3x2 - 2 57. ƒ1x2 = 21x - 222 - 458. ƒ1x2 = -31x - 222 + 1 59. ƒ1x2 = 2x + 2 60. ƒ1x2 = 2x - 361. ƒ1x2 = -2x 62. ƒ1x2 = 2x - 2 63. ƒ1x2 = 22x + 164. ƒ1x2 = 32x - 2 65. g1x2 = 1

2 x3 - 4 66. g1x2 = 1

2 x3 + 2

Connecting Graphs with Equations Each of the following graphs is obtained from the graph of ƒ1x2 = 0 x 0 or g1x2 = 1x by applying several of the transformations discussed in this section. Describe the transformations and give an equation for the graph.

67.

1 4

68.

69.

70.

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A-1

Answers to Selected Exercises

To The StudentIn this section we provide the answers that we think most students will obtain when they work the exercises using the methods explained in the text. If your answer does not look exactly like the one given here, it is not necessarily wrong. In many cases there are equivalent forms of the answer. For example, if the answer section shows 34 and your answer is 0.75, you have obtained the correct answer but written it in a different (yet equivalent) form. Unless the directions specify otherwise, 0.75 is just as valid an answer as 34 . (In answers with radicals, we give rationalized denominators when appropriate.) In general, if your answer does not agree with the one given in the text, see whether it can be transformed into the other form. If it can, then it is equivalent to the cor-rect answer. If you still have doubts, talk with your instructor.

Chapter 1 Trigonometric Functions1.1 Exercises

1. 1360 3. 180° 5. 90° 7. 1

60 9. 55° 15′ 11. (a) 60°(b) 150° 13. (a) 45° (b) 135° 15. (a) 36° (b) 126° 17. (a) 89° (b) 179° 19. (a) 75° 40′ (b) 165° 40′21. (a) 69° 49′ 30″ (b) 159° 49′ 30″ 23. 70°; 110°25. 30°; 60° 27. 40°; 140° 29. 107°; 73° 31. 69°; 21°33. 150° 35. 7° 30′ 37. 130° 39. 83° 59′ 41. 179° 19′ 43. -23° 49′ 45. 38° 32′ 47. 60° 34′49. 17° 01′ 49″ 51. 30° 27′ 53. 35.5° 55. 112.25°57. -60.2° 59. 20.91° 61. 91.598° 63. 274.316°65. 39° 15′ 00″ 67. 126° 45′ 36″ 69. -18° 30′ 54″71. 31° 25′ 47″ 73. 89° 54′ 01″ 75. 178° 35′ 58″77. 392° 79. 386° 30′ 81. 320° 83. 234° 30′ 85. 1°87. 359° 89. 179° 91. 130° 93. 240° 95. 120°

In Exercises 97 and 99, answers may vary.97. 450°, 810°; -270°, -630° 99. 360°, 720°; -360°, -720° 101. 30° + n # 360° 103. 135° + n # 360° 105. -90° + n # 360° 107. 0° + n # 360°, or n # 360° 109. 0° and 360° are coterminal angles.

Angles other than those given are possible in Exer-cises 111–121.111.

75°

435°; –285°;quadrant I

113.

174°

534°; –186°;quadrant II

115.

300°

660°; –60°;quadrant IV

117.

–61°

299°; –421°;quadrant IV

119.

90°

450°; –270°;no quadrant

121.

–90°

270°; –450°;no quadrant

123. 34 125. 1800° 127. 12.5 rotations per hr129. 4 sec

1.2 Exercises1. 180° 3. three 5. Answers are given in numerical order: 49°; 49°; 131°; 131°; 49°; 49°; 131° 7. A and P; B and Q; C and R; AC and PR; BC and QR; AB and PQ 9. A and C; E and D; ABE and CBD; EB and DB; AB and CB; AE and CD 11. 51°; 51° 13. 50°; 60°; 70° 15. 60°; 60°; 60° 17. 45°; 75°; 120° 19. 49°; 49°21. 48°; 132° 23. 91° 25. 2° 29′ 27. 25.4° 29. 22° 29′ 34″ 31. no 33. right; scalene35. acute; equilateral 37. right; scalene 39. right; isosceles 41. obtuse; scalene 43. acute; isosceles45. Angles 1, 2, and 3 form a straight angle on line m and, therefore, sum to 180°. It follows that the sum of the measures of the angles of triangle PQR is 180° because the angles marked 1 are alternate interior angles whose mea-sures are equal, as are the angles marked 2.47. Q = 42°; B = R = 48° 49. B = 106°; A = M = 44°51. X = M = 52° 53. a = 20; b = 15 55. a = 6; b = 7.5 57. x = 6 59. 30 m 61. 500 m; 700 m63. 112.5 ft 65. x = 110 67. c ≈ 111.169. (a) 236,000 mi (b) no 71. (a) 2900 mi (b) no73. (a) 14 (b) 30 arc degrees

Chapter 1 Quiz[1.1] 1. (a) 71° (b) 161° 2. 65°; 115° 3. 26°; 64°[1.2] 4. 20°; 24°; 136° 5. 130°; 50°[1.1] 6. (a) 77.2025° (b) 22° 01′ 30″ 7. (a) 50°(b) 300° (c) 170° (d) 417° 8. 1800° [1.2] 9. 10 ft10. (a) x = 12; y = 10 (b) x = 5

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A-2 Answers to Selected Exercises

51.

0 x

(1, –2)1

–2

2x + y = 0, x ≥ 0U

- 2255 ; 255 ; -2; - 12 ; 25; - 252

53.

0 x

(–1, 6) 6–6x – y = 0, x ≤ 0

–1

623737 ; - 23737 ; -6; - 16 ; -237; 2376In Exercises 11–29 and 51–61, we give, in order, sine, cosine, tangent, cotangent, secant, and cosecant.11.

0 x

(5, –12)–12

- 1213 ; 5

13 ; - 125 ;

- 512 ; 135 ; -

1312

13.

(3, 4)4

45 ; 35 ;

43 ;

34 ; 53 ;

15.

(–8, 15)15

–5

–15

1517 ; - 8

17 ; - 158 ;

- 815 ; - 178 ;

1715

17.

(–7, –24)

–30

–20

–10

–20 –10–30 10

- 2425 ; - 7

25 ; 247 ;

724 ; - 257 ; -

2524

19.

(0, 2)2

–2–2

1; 0; undefined; 0; undefined; 1

21.

(–4, 0)

–4

0; -1; 0; undefined; -1; undefined

23.

(0, –4)

–4

-1; 0; undefined; 0; undefined; -1

25.

–1

(1, Ë

232 ; 12 ; 23; 233 ; 2; 2233

27.

–1–1

(Ë

2, Ë

2)1

222 ; 222 ; 1; 1; 22; 22

29.

–3

3(–2Ë

3, –2)

- 12 ; - 232 ;

233 ;

23; - 2233 ; -231. negative 33. negative 35. positive 37. positive 39. negative 41. positive 43. negative 45. positive 47. positive 49. positive

55.

(–7, –4)

–7

–4

–4x + 7y = 0, x ≤ 0U

- 426565 ; - 726565 ; 47 ; 74 ; -

2657 ; -

2654

57.

(2, –2)–2

x + y = 0, x ≥ 0

U 2

–2

- 222 ; 222 ; -1; -1; 22; -22

59.

–2

2–2

–Ë

3x + y = 0, x ≤ 0

(–1, –Ë

- 232 ; - 12 ; 23; 233 ; -2; - 2233

61.

(0, 1)

–1

x = 0, y ≥ 0

1–1

1; 0; undefined; 0; undefined; 1

63. 0 65. 0 67. -1 69. 1 71. undefined 73. -1 75. 0 77. undefined 79. 1 81. -1 83. 0 85. -387. -3 89. 5 91. 1 93. 0 95. 0 97. 1 99. 0101. 0 103. undefined 105. 0 107. undefined109. They are equal. 111. They are negatives of each other. 113. 0.940; 0.342 115. 35°117. decrease; increase

1.4 Exercises1. cos u; sec u 3. sin u; csc u 5. possible 7. impossible 9. possible 11. 32 13. -

73 15.

17. - 25 19. 222 21. -0.4 23. 0.8 25. Because

-1 … cos u … 1, it is not possible that cos u = 32 .27. All are positive. 29. Tangent and cotangent are posi-tive. All others are negative. 31. Sine and cosecant are positive. All others are negative. 33. Cosine and secant are positive. All others are negative. 35. Sine and cosecant are positive. All others are negative. 37. All are positive.39. I, II 41. I 43. II 45. I 47. III 49. III, IV51. cos u and sec u are reciprocal functions, and sin u and csc u are reciprocal functions. The pairs have the same sign for each quadrant. 53. impossible 55. possible57. possible 59. impossible 61. possible 63. possible

65. - 45 67. - 252 69. -

233 71. 1.05

1.3 Exercises1. hypotenuse 3. same 5. positive; negative

7. 322 9. - 222

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A-3Answers to Selected Exercises

[1.2] 12. 10 23 ft, or 10 ft, 8 in. 13. x = 8; y = 6[1.3] 14.

0 x

(2, –7)

–7

In Exercises 73–83, we give, in order, sine, cosine, tangent, cotangent, secant, and cosecant.

73. 1517 ; - 8

17 ; - 158 ; -

815 ; -

178 ;

1715

75. 257 ; 22117 ; 25522 ; 22555 ; 721122 ; 725577. 826767 ; 220167 ; 8233 ; 238 ; 22013 ; 267879. 226 ; - 2346 ; - 21717 ; - 217 ; - 323417 ; 32281. 2154 ; - 14 ; - 215; - 21515 ; -4; 42151583. 1; 0; undefined; 0; undefined; 1 87. This statement is false. For example, sin 180° + cos 180° = 0 + 1-12 = -1≠ 1. 89. negative 91. positive 93. negative 95. negative 97. positive 99. negative 101. positive 103. negative 105. 2° 107. 3° 109. Quadrant II is the only quadrant in which the cosine is negative and the sine is positive.

Chapter 1 Review Exercises1. complement: 55°; supplement: 145° 3. 186° 5. 9360°7. 119.134° 9. 275° 06′ 02″ 11. 40°; 60°; 80°13. 105°; 105° 15. 0.25 km 17. N = 12°; R = 82°; M = 86° 19. p = 7; q = 7 21. k = 14 23. 12 ft

In Exercises 25–43, we give, in order, sine, cosine, tangent, cotangent, secant, and cosecant.

25. - 232 ; 12 ; -23; - 233 ; 2; - 223327. - 45 ;

35 ; -

43 ; -

34 ;

53 ; -

29. 1517 ; - 8

17 ; - 158 ; -

815 ; -

178 ;

1715

31. - 12 ; 232 ; -

233 ; -23 ; 2233 ; -2

33.

0 x

(3, 5)

5x – 3y = 0, x ≥ 0

523434 ;

323434 ;

53 ;

35 ; 234

3 ; 234

35.

0 x

(5, –12)

5 10–5–10–15

12x + 5y = 0, x ≥ 0U

- 1213 ; 5

13 ; - 125 ;

- 512 ; 135 ; -

1312

37. -1; 0; undefined; 0; undefined; -1

39. - 2398 ; - 58 ; 2395 ; 523939 ; - 85 ; - 82393941. 2255 ; - 255 ; -2; - 12 ; -25 ; 25243. - 35 ;

45 ; -

34 ; -

43 ;

54 ; -

45. (a) impossible (b) possible (c) impossible47. 40 yd 49. 9500 ft

Chapter 1 Test[1.1] 1. 23°; 113° 2. 145°; 35° 3. 20°; 70°[1.2] 4. 130°; 130° 5. 110°; 110° 6. 20°; 30°; 130°7. 60°; 40°; 100° [1.1] 8. 74.31° 9. 45° 12′ 09″10. (a) 30° (b) 280° (c) 90° 11. 2700°

sin u = - 725353 ; cos u = 225353 ; tan u = - 72 ; cot u = -

27 ;

sec u = 2532 ; csc u = - 253715.

(0, –2)

sin u = -1; cos u = 0; tan u is undefined; cot u = 0; sec u is undefined; csc u = -1

16.

(–4, –3)

–3

–4

3x – 4y = 0, x ≤ 0

sin u = - 35 ; cos u = - 45 ;

tan u = 34 ; cot u =43 ;

sec u = - 54 ; csc u = - 53

17. row 1: 1, 0, undefined, 0, undefined, 1; row 2: 0, 1, 0, undefined, 1, undefined; row 3: -1, 0, undefined, 0, undefined, -1 18. cosecant and cotangent[1.4] 19. (a) I (b) III, IV (c) III 20. (a) impossible

(b) possible (c) possible 21. sec u = - 127

22. cos u = - 22107 ; tan u = - 321020 ; cot u = - 22103 ; sec u = - 721020 ; csc u = 73Chapter 2 Acute Angles and Right Triangles2.1 Exercises1. C 3. B 5. EIn Exercises 7 and 9, we give, in order, sine, cosine, and tangent.

7. 2129 ; 2029 ;

2120 9.

np ;

mp ;

In Exercises 11–19, we give, in order, the unknown side, sine, cosine, tangent, cotangent, secant, and cosecant.

11. c = 13; 1213 ; 5

13 ; 125 ;

512 ;

135 ;

1312

13. b = 213 ; 2137 ; 67 ; 2136 ; 621313 ; 76 ; 72131315. b = 291 ; 29110 ; 310 ; 2913 ; 329191 ; 103 ; 102919117. b = 23 ; 232 ; 12 ; 23 ; 233 ; 2; 223319. a = 221 ; 25 ; 2215 ; 222121 ; 2212 ; 522121 ; 5221. sin 60° 23. sec 30° 25. csc 51° 27. cos 51.3° 29. csc175° - u2 31. 40° 33. 20° 35. 12° 37. 35°39. 18° 41. true 43. false 45. true 47. true

49. 233 51. 12 53. 2233 55. 22 57. 222 59. 161. 232 63. 23 65. 60° 67. y = 233 x 69. 60°

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A-4 Answers to Selected Exercises

23. cot 77° ≈ 0.230868 25. tan 4.72° ≈ 0.08256627. cos 51° ≈ 0.629320 29. 55.845496°31. 16.166641° 33. 38.491580° 35. 68.673241°37. 45.526434° 39. 12.227282° 41. The calculator is not in degree mode. 43. 56° 45. 1 47. 1 49. 0 51. false53. true 55. false 57. false 59. true 61. true63. 68°; 112° 65. 44°; 316° 67. 51°; 231° 69. 70 lb71. -2.9° 73. 2500 lb 75. A 2200-lb car on a 2° uphill grade has greater grade resistance. 77. 703 ft79. R would decrease; 644 ft, 1559 ft81. (a) 2 * 108 m per sec (b) 2 * 108 m per sec83. 48.7° 85. 155 ft 87. Negative values of u require greater distances for slowing down than positive values.89. A: 69 mph; B: 66 mph 91. 550 ft

Chapter 2 Quiz[2.1] 1. sin A = 35 ; cos A =

45 ; tan A =

34 ; cot A =

43 ;

sec A = 54 ; csc A =53

U sin U cos U tan U cot U sec U csc U

30° 12 232

233 23 2233 2

45° 222 222 1 1 22 2260°

232

12 23 233 2 2233

3. w = 18 ; x = 1823 ; y = 18 ; z = 18224. ' = 3x2 sin u

[2.2] In Exercises 5–7, we give, in order, sine, cosine, tangent, cotangent, secant, and cosecant.

5. 222 ; - 222 ; -1; -1; -22 ; 226. - 12 ; -

232 ;

233 ; 23 ; - 2233 ; -2

7. - 232 ; 12 ; -23 ; - 233 ; 2; - 22338. 60°; 120° 9. 135°; 225° [2.3] 10. 0.67301311. -1.181763 12. 69.497888° 13. 24.777233°[2.1–2.3] 14. false 15. true

2.4 Exercises1. B 3. A 5. C 7. 23.825 to 23.835 9. 8958.5 to 8959.5 11. 0.05

Note to student: While most of the measures resulting from solving triangles in this chapter are approxima-tions, for convenience we use = rather than ? .13. B = 53° 40′; a = 571 m; b = 777 m15. M = 38.8°; n = 154 m; p = 198 m17. A = 47.9108°; c = 84.816 cm; a = 62.942 cm19. A = 37° 40′ ; B = 52° 20′ ; c = 20.5 ft 21. No. Given three angles (the two acute angles and the right angle), there are infinitely many similar triangles satisfying the conditions.

71. (a) 60° (b) k (c) k23 (d) 2; 23 ; 30°; 60°73. x = 9232 ; y = 92 ; z = 3232 ; w = 32375. p = 15 ; r = 1522 ; q = 526 ; t = 102677. ' = s

2 79. Q222 , 222 R ; 45°81.

45° x

y 82. P

45° x

2√2

83. the legs; A 222 , 222 B 84. A 1 ,23 B2.2 Exercises1. negative; III; 60°; - 232 3. positive; III; 30°; 2335. C 7. A 9. D 11. 233 ; 23 13. 232 ; 233 ; 223315. -1; -1 17. - 232 ; - 2233In Exercises 19–35, we give, in order, sine, cosine, tangent, cotangent, secant, and cosecant.

19. - 232 ; 12 ; -23 ; - 233 ; 2; - 223321. 222 ; 222 ; 1; 1; 22 ; 2223. 232 ; - 12 ; -23 ; - 233 ; -2; 223325. - 12 ; -

232 ;

233 ; 23 ; - 2233 ; -2

27. - 222 ; - 222 ; 1; 1; -22 ; -2229. 232 ; 12 ; 23 ; 233 ; 2; 223331. - 12 ; -

232 ;

233 ; 23 ; - 2233 ; -2

33. 12 ; - 232 ; -

233 ; -23 ; - 2233 ; 2

35. - 232 ; 12 ; -23 ; - 233 ; 2; - 2233 37. - 222 39. - 232 41. -22 43. -1 45. 1 47. 234 49. 72 51. - 2912 53. false; 0≠

23 + 12 55. false;

12 ≠ 23

57. true 59. false; 0≠ 22 61. 30°; 150° 63. 120°; 300°65. 45°; 315° 67. 210°; 330° 69. 30°; 210° 71. 225°; 315°

73. A -323 , 3 B 75. yes 77. positive 79. positive81. negative 83. When an integer multiple of 360° is added to u, the resulting angle is coterminal with u. The sine values of coterminal angles are equal. 85. 0.9 87. 45°; 225°

2.3 Exercises1. J 3. E 5. D 7. H 9. G In Exercises 11– 27, the number of decimal places may vary depending on the calculator used. We show six places.11. 0.625243 13. 1.027349 15. 15.055723 17. 0.740805 19. 1.483014 21. tan 23.4° ≈ 0.432739

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A-5Answers to Selected Exercises

39. false; 1.4088321≠ 1 41. true 43. No, this will result in an angle having tangent equal to 25. The function tan-1 is not the reciprocal of the tangent (cotangent) but is, rather, the inverse tangent function. To find cot 25°, the student must find the reciprocal of tan 25°.45. B = 31° 30′ ; a = 638 ; b = 39147. B = 50.28°; a = 32.38 m; c = 50.66 m 49. 137 ft51. 73.7 ft 53. 18.75 cm 55. 1200 m 57. 140 mi59. One possible answer: Find the value of x.

61. (a) 716 mi (b) 1104 mi

Chapter 2 Test[2.1] 1. sin A = 1213 ; cos A =

513 ; tan A =

125 ; cot A =

512 ;

sec A = 135 ; csc A =1312 2. x = 4 ; y = 42