Trigonometric Identities,Inverses, and...

Precalculus—

TrigonometricIdentities, Inverses,and EquationsCHAPTER OUTLINE

7.1 Fundamental Identities and Families of Identities 654

7.2 More on Verifying Identities 661

7.3 The Sum and Difference Identities 669

7.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities 680

7.5 The Inverse Trig Functions and Their Applications 695

7.6 Solving Basic Trig Equations 711

7.7 General Trig Equations and Applications 721

CHAPTER CONNECTIONS

This chapter will unify much of what we’velearned so far , and lead us to some intriguing,sophisticated, and surprising applications oftrigonometr y. Defining the trig functions helpeus study a number of new r elationships notpossible using algebra alone. Their graphs gaveus insights into how the functions wer e relatedto each other , and enabled a study of periodicphenomena. W e will now use identities tosimplify complex expr essions and show howtrig functions often work together to modelnatural events. One such “event” is a river’sseasonal dischar ge rate, which tends to begreater during the annual snow melt. In thischapter, we’ll lear n how to pr edict thedischarge rate during specific months of thyear, infor mation of gr eat value to fisheriesoceanographers, and other scientists.

� This application appears as Exer cises 53 and 54 in Section 7.7

Trigonometric equations, identities, and substitutions also play a vital role in a study of calculus, helping tosimplify complex expressions, or rewrite an expression in a form more suitable for the tools of calculus.These connections are explored in the Connections to Calculus feature following Chapter 7.

Connectionsto Calculus 653

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654 CHAPTER 7 Trigonometric Identities, Inverses, and Equations 7–000

Precalculus—

7.1 Fundamental Identities and Families of Identities

In this section, we begin laying the foundation necessary to work with identitiessuccessfully. The cornerstone of this effort is a healthy respect for the fundamentalidentities and vital role they play. Students are strongly encouraged to do more thanmemorize them—they should be internalized, meaning they must become a naturaland instinctive part of your core mathematical knowledge.

A. Fundamental Identities and Identity Families

An identity is an equation that is true for all elements in the domain. In trigonometry,some identities result directly from the way the functions are defined. For instance,the reciprocal relationships we first saw in Section 6.2 followed directly from their def-initions. We call identities of this type fundamental identities. Successfully workingwith other identities will depend a great deal on your mastery of these fundamentaltypes. For convenience, the definitions of the trig functions are reviewed here, followedby the fundamental identities that result.

Given point P(x, y) is on the terminal side of angle in standard position, withthe distance from the origin to (x, y), we have

Fundamental Trigonometric Identities

Reciprocal Ratio Pythagorean identities identities identities

EXAMPLE 1 � Proving a Fundamental IdentityUse the coordinate definitions of the trigonometric functions to prove the identity

.

Solution � We begin with the left-hand side.

substitute for , for

square terms

add terms

Noting that implies we have

substitute x 2 � y 2 for r 2, simplfy

Now try Exercises 7 through 10 �

�r2

r2 � 1

r2 � x2 � y2,r � 2x2 � y2

�x2 � y2

r2

�x2

r2 �y2

r2

sin �y

rcos �

xr

cos2� � sin2� � axrb2

� ayrb2

cos2� � sin2� � 1

cot2� � 1 � csc2�cot � �cos �

sin �tan � �

1

cot �

1 � tan2� � sec2�tan � �sec �

csc �cos � �

1

sec �

cos2� � sin2� � 1tan � �sin �

cos �sin � �

1

csc �

cot � �xy; y � 0csc � �

ry; y � 0sec � �

rx; x � 0

tan � �yx; x � 0sin � �

yr

cos � �xr

r � 2x2 � y2�

Precalculus—

WORTHY OF NOTEThe Pythagorean identities areused extensively in future courses.See Example 1 of the Connectionsto Calculus feature at the end ofthis chapter.

LEARNING OBJECTIVESIn Section 7.1 you will see

how we can:

A. Use fundamentalidentities to helpunderstand and recognizeidentity “families”

B. Verify other identitiesusing the fundamentalidentities and basicalgebra skills

C. Use fundamentalidentities to express agiven trig function interms of the other five

654 7–2

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7–3 Section 7.1 Fundamental Identities and Families of Identities 655

Precalculus—

The fundamental identities seem to naturally separate themselves into the threegroups or families listed, with each group having additional relationships that canbe inferred from the definitions. For instance, since is the reciprocal of must be the reciprocal of Similar statements can be maderegarding and as well as and . Recognizing these additional“family members” enlarges the number of identities you can work with, and willhelp you use them more effectively. In particular, since they are reciprocals:

See Exercises 11 and 12.

EXAMPLE 2 � Identifying Families of IdentitiesStarting with use algebra to write four additional identities thatbelong to the Pythagorean family.

Solution � original identity

1) subtract cos2

2) take square root

original identity

3) subtract sin2

4) take square root

For the identities involving a radical, the choice of sign will depend on thequadrant of the terminal side.

Now try Exercises 13 and 14 �

The fact that each new equation in Example 2 represents an identity gives us moreoptions when attempting to verify or prove more complex identities. For instance,since we can replace with or replace with any time they occur in an expression. Note there are many other membersof this family, since similar steps can be performed on the other Pythagorean identities.In fact, each of the fundamental identities can be similarly rewritten and there are a va-riety of exercises at the end of this section for practice.

B. Verifying an Identity Using Algebra

Note that we cannot prove an equation is an identity by repeatedly substituting inputvalues and obtaining a true equation. This would be an infinite exercise and we mighteasily miss a value or even a range of values for which the equation is false. Instead weattempt to rewrite one side of the equation until we obtain a match with the other side,so there can be no doubt. As hinted at earlier, this is done using basic algebra skillscombined with the fundamental identities and the substitution principle. For now we’llfocus on verifying identities by using algebra. In Section 7.2 we’ll introduce someguidelines and ideas that will help you verify a wider range of identities.

cos2�,1 � sin2�1 � sin2�,cos2�cos2� � 1 � sin2�,

cos � � �21 � sin2�

� cos2� � 1 � sin2�

cos2� � sin2� � 1 sin � � �21 � cos2�

� sin2� � 1 � cos2�

cos2 � � sin2 � � 1

cos2� � sin2� � 1,

tan � cot � � 1.sin � csc � � 1, cos � sec � � 1, and

cot �tan �sec �cos �sin �.csc �, csc �

sin �

A. You’ve just seen how

we can use fundamental

identities to help understand

and recognize identity

“families”

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Precalculus—

EXAMPLE 3 � Using Algebra to Help Verify an IdentityUse the distributive property to verify that is anidentity.

Solution � Use the distributive property to simplify the left-hand side.

distribute

substitute 1 for sin csc

1 � sin2 � cos2

Since we were able to transform the left-hand side into a duplicate of the right,there can be no doubt the original equation is an identity.


Often we must factor an expression, rather than multiply, to begin the verificationprocess.

EXAMPLE 4 � Using Algebra to Help Verify an IdentityVerify that is an identity.

Solution � The left side is as simple as it gets. The terms on the right side have a commonfactor and we begin there.

factor out

substitute for

power property of exponents


Examples 3 and 4 show you can begin the verification process on either the left orright side of the equation, whichever seems more convenient. Example 5 shows howthe special products and/or can be used in the verification process.

EXAMPLE 5 � Using a Special Product to Help V erify an IdentityUse a special product and fundamental identities to verify that

is an identity.

Solution � Begin by squaring the left-hand side, in hopes of using a Pythagorean identity.

binomial square

rewrite terms

substitute 1 for


Another common method used to verify identities is simplification by combining

terms, using the model For the right-hand

side immediately becomes , which gives See Exercises 39

through 44.

1

cos �� sec �.

sin2� � cos2�

cos �

sec � �sin2�

cos �� cos �,

A

B�

C

D�

AD � BC

BD.

cos2� � sin2� � 1 � 2 sin � cos � � cos2� � sin2� � 2 sin � cos �

1sin � � cos �22 � sin2� � 2 sin � cos � � cos2�

1sin � � cos �22 � 1 � 2 sin � cos �

1A � B22 � A2 � 2AB � B21A � B2 1A � B2 � A2 � B2

cot � tan � � 1 � 12 � 1 � 1cot � tan �22

sec2� � 1tan2� � cot2� tan2�

cot2� cot2� sec2� � cot2� � cot2� 1sec2� � 12

1 � cot2� sec2� � cot2�

�� cos2�

�� 1 � sin2�

sin � 1csc � � sin �2 � sin � csc � � sin2�

sin �1csc � � sin �2 � cos2�

B. You’ve just seen how we

can verify other identities using

the fundamental identities and

basic algebra skills

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Precalculus—

C. Writing One Function in Terms of Another

Any one of the six trigonometric functions can be written in terms of any of the otherfunctions using fundamental identities. The process involved offers practice in work-ing with identities, highlights how each function is related to the other, and has practi-cal applications in verifying more complex identities.

EXAMPLE 6 � Writing One Trig Function in Terms of AnotherWrite the function cos in terms of the tangent function.

Solution � Begin by noting these functions share “common ground” via , since

Starting with ,

Pythagorean identity

square roots

We can now substitute for in

substitute for

Note we have written in terms of the tangent function.


Example 6 also reminds us of a very important point—the sign we choose for thefinal answer is dependent on the terminal side of the angle. If the terminal side is in QIor QIV we chose the positive sign since in those quadrants. If the angle ter-minates in QII or QIII, the final answer is negative since in those quadrants.

Similar to our work in Section 6.7, given the value of and the quadrant of ,the fundamental identities enable us to find the value of the other five functions at . Infact, this is generally true for any given trig function and angle .

EXAMPLE 7 � Using a Known Value and Quadrant Analysis to F ind Other Function Values

Given with the terminal side of in QIV, find the value of the other

five functions of . Use a calculator to check your answer.

Solution � The function value follows immediately, since cotangent and tangent

are reciprocals. The value of can be found using


substitute for

square , substitute for 1

combine terms

take square rootssec � � �41

9

sec2� �1681

81

8181

�409

�81

81�

1600

81

tan ��409

� 1 � a�40

9b2

sec2� � 1 � tan2 �

sec2� � 1 � tan2�.sec �

tan � � �40

9

�

�cot � ��9

40

��

�cot �cos � 6 0

cos � 7 0

cos �

sec ��21 � tan2�cos � �1

�21 � tan2�

cos � �1

sec �.sec ��21 � tan2�

sec � � �21 � tan2�

sec2� � 1 � tan2�

sec2�sec2� � 1 � tan2� and cos � �1

sec �.

sec �

�

WORTHY OF NOTEAlthough identities are valid whereboth expressions are defined, thisdoes not preclude a difference inthe domains of each function. Forexample, the result of Example 6 isindeed an identity, even though the

left side is defined at while the

right side is not.

�

2

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Precalculus—

Since is positive for a terminal side in QIV, we have

This automatically gives (reciprocal identities), and we find

using or the ratio identity (verify).

This result and another reciprocal identity gives us our final value,

Check � As in Example 9 of Section 6.7, we find using (TAN�1) 40 9, which shows (Figure 7.1). Since the terminal side of is in

QIV, one possible value for is . Note in Figure 7.1, the (ANS)feature was used to compute , which we then stored as X. In Figure 7.2, we verify that

, , and .


Figure 7.1 Figure 7.2

sin � � �40

41cos � �

9

41tan � � �

40

9

�

(–)2nd2� � �r��r � 1.3495ENTER)

�TAN2nd�r

csc � � �41

40.

tan � �sin �

cos �sin2 � � 1 � cos2�sin � � �

40

41

cos � �9

41

sec � �41

9.sec �

C. You’ve just seen how

we can use fundamental

identities to express a given

trig function in terms of the

other five

2. The three fundamental reciprocal identities are, , and . From these, we can infer three

additional reciprocal relationships: ,, and .cot � � 1/sec � � 1/

csc � � 1/tan � � 1/

cos � � 1/sin � � 1/

� CONCEPTS AND VOCABULAR Y

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

7.1 EXERCISES

1. Three fundamental ratio identities are

, and cot � �?

sin �.tan � �

?

cos �, tan � �

?

csc �

4. An is an equation that is true for allelements in the . To show an equation is anidentity, we employ basic algebra skills combinedwith the identities and the substitutionprinciple.

3. Starting with the Pythagorean identity, the identity

can be derived by dividing both sides by .Alternatively, dividing both sides of this equation by

, we obtain the identity .sin2 �

1 � tan2 � � sec2 �cos2 � � sin2 � � 16. Name at least four algebraic skills that are used

with the fundamental identities in order to rewrite atrigonometric expression. Use algebra to quicklyrewrite 1sin � � cos �22.

5. Use the pattern to add the

following terms, and comment on this processversus “finding a common denominator.” cos �

sin ��

sin �

sec �

A

B�

C

D�

AD � BC

BD

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Precalculus—

32.

Verify the equation is an identity using special productsand fundamental identities.

33.

34.

35.

36.

37.

38.

Verify the equation is an identity using fundamental

identities and to combine terms.

39.

40.

41.

42.

43.

44.

Write the given function entirely in terms of the secondfunction indicated.

45. in terms of

46. in terms of

47. in terms of

48. in terms of

49. in terms of

50. in terms of csc �cot �

sin �cot �

sin �sec �

cot �sec �

sec �tan �

sin �tan �

csc �

cos ��

sec �

csc �� cot �

sec �

sin ��

csc �

sec �� tan �

cot �

sec ��

cos �

sin ��

cos � � 1

tan �

tan �

csc ��

sin �

cos ��

sin � � 1

cot �

sec �

1�

tan2�

sec �� cos �

cos2�

sin ��

sin �

1� csc �

AB

�CD

�AD � BC

BD

1sec � � tan �2 1sec � � tan �2csc �

� sin �

1csc � � cot �2 1csc � � cot �2tan �

� cot �

1sec � � 12 1sec � � 12 � tan2�

11 � sin �2 11 � sin �2 � cos2�

11 � tan �22sec �

� sec � � 2 sin �

1sin � � cos �22cos �

� sec � � 2 sin �

cos � cot � � cos �

cot � � cot2�� sin �

� DEVELOPING YOUR SKILLS

Use the definitions of the trigonometric functions toprove the following fundamental identities.

7. 8.

9. 10.

Starting with the ratio identity given, use substitutionand fundamental identities to write four new identitiesbelonging to the ratio family. Answers may vary.

11. 12.

Starting with the Pythagorean identity given, usealgebra to write four additional identities belonging tothe Pythagorean family. Answers may vary.

13. 14.

Verify the equation is an identity using multiplicationand fundamental identities.

15. 16.

17. 18.

19.

20.

21.

22.

23.

24.

Verify the equation is an identity using factoring andfundamental identities.

25.

26.

27.

28.

29.

30.

31.sin � tan � � sin �

tan � � tan2�� cos �

1 � cos �

sin � � cos � sin �� csc �

1 � sin �

cos � � cos � sin �� sec �

sin � cos � � cos �

sin � � sin2�� cot �

sin � cos � � sin �

cos � � cos2�� tan �

sin2� cot2� � sin2� � 1

tan2� csc2� � tan2� � 1

cot � 1sec � � tan �2 � csc � � 1

tan � 1csc � � cot �2 � sec � � 1

cot � 1tan � � cot �2 � csc2�

sin � 1csc � � sin �2 � cos2�

tan � 1cot � � tan �2 � sec2�

cos � 1sec � � cos �2 � sin2�

csc2� tan2� � sec2�sec2� cot2� � csc2�

cos � tan � � sin �sin � cot � � cos �

cot2� � 1 � csc2�1 � tan2� � sec2�

cot � �cos �

sin �tan � �

sin �

cos �

cot � �cos �

sin �tan � �

sin �

cos �

cot2� � 1 � csc2�1 � tan2� � sec2�

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Precalculus—

56. with in QII

57. with in QIII

58. with in QIV

59. with in QII

60. with in QIV�cot � � �11

2

�sec � � �9

7

�cos � �23

25

�sin � � �7

13

�csc � �7x

For the given trig function and the quadrant in which terminates, state the value of the other five trigfunctions. Use a calculator to verify your answers.

�f (�)

62. Area of a regular polygon:

The area of a regular polygon is given by theformula shown, where n represents the number ofsides and x is the length of each side.

a. Rewrite the formula in terms of a single trigfunction.

b. Verify the formula for a square with sides of8 m given the point (2, 2) is on the terminalside of a angle in standard position.45°

A � anx2

4b cos1180�

n 2sin1180�

n 2

� WORKING WITH FORMULAS

61. The versine function:

For centuries, the haversine formula has been usedin navigation to calculate the nautical distancebetween any two points on the surface of the Earth.One part of the formula requires the calculation ofV, where is half the difference of latitudesbetween the two points. Use a fundamental identityto express V in terms of cosine.

�

V � 2 sin2�

69. Show can be factored into.

70. Show can be factoredinto .

71. Angle of intersection: At their point of intersection,the angle between any two nonparallel linessatisfies the relationship

where m1 and m2 represent theslopes of the two lines. Rewrite the equation interms of a single trig function.

72. Angle of intersection: Use the result of Exercise 71to find the tangent of the angle between the lines

and .

73. Angle of intersection: Use the result of Exercise 71to find the tangent of the angle between the lines

and .Y2 � �2x � 7Y1 � 3x � 1

Y2 �7

3x � 1Y1 �

2

5x � 3

sin � � m1m2sin �,1m2 � m12cos � �

�

21csc � � 122 1csc � � 12 1csc � � 122 cot2� csc � � 212 cot2�

�111 � sin �2 11 � sin �22cos2� sin � � cos2�

� APPLICATIONS

Writing a given expression in an alternative form is askill used at all levels of mathematics. In addition tostandard factoring skills, it is often helpful todecompose a power into smaller powers (as in writingA3 as ).

63. Show that cos3 can be written as .

64. Show that tan3 can be written as .

65. Show that can be written as.

66. Show that can be written as .

67. Show can be factored into.

68. Show can be factoredinto .11 � cos �2 11 � cos �2 12 cos � � 1322 sin2� cos � � 13 sin2 �

1sec � � 42 1sec � � 12 1sec � � 12sec � � 4 tan2�tan2�

cot �1csc2� � 12cot3�

tan �1sec2�2tan � � tan3�

tan �1sec2� � 12�

cos �11 � sin2�2�

A # A2

51. with in QII

52. with in QII

53. with in QIII

54. with in QIV

55. with in QI�cot � �x

5

�sec � �5

3

�tan � �15

8

�sin � �12

37

�cos � � �20

29

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7–9 Section 7.2 More on Verifying Identities 661

Precalculus—

� EXTENDING THE CONCEPT

74. The word tangent literally means “to touch,” which in mathematics we take tomean touches in only and exactly one point. In the figure, the circle has a radiusof 1 and the vertical line is “tangent” to the circle at the x-axis. The figure can beused to verify the Pythagorean identity for sine and cosine, as well as the ratioidentity for tangent. Discuss/Explain how.

75. Simplify using factoring andfundamental identities.

sin ��2 sin4� � 13 sin3� � 2 sin2� � 13

� MAINTAINING YOUR SKILLS

78. (4.2) Use the rational zeroes theorem and other“tools” to find all zeroes of the function

.

79. (6.3) Use a reference rectangle and the rule offourths to sketch the graph of for t in[0, ).2�

y � 2 sin12t2

f 1x2 � 2x4 � 9x3 � 4x2 � 36x � 16

x

y

cos �

tan �

sin �

�

1

Exercise 74

7.2 More on Verifying Identities

In Section 7.1, our primary goal was to illustrate how basic algebra skills can be usedto rewrite trigonometric expressions and verify simple identities. In this section, we’llsharpen and refine these skills so they can be applied more generally, as we develop theability to verify a much wider range of identities.

A. Identities Due to Symmetry

The symmetry of the unit circle and the wrappingfunction presented in Chapter 6’s Reinforcing BasicSkills feature, lead us directly to the final group offundamental identities. Given , consider thepoints on the unit circle associated with t and , asshown in Figure 7.3. From our definitions of the trigfunctions, and , and we recog-nize sine is an odd function: . The re-maining identities due to symmetry can similarly bedeveloped, and are shown here with the completefamily of fundamental identities.

sin1�t2 � �sin tsin1�t2 � �ysin t � y

�tt 7 0


how we can:

A. Identify and use identitiesdue to symmetry

B. Verify general identities

C. Use counterexamples andcontradictions to show anequation is not an identity

Figure 7.3

x

1

�t�y

t

(x, y)

(x, �y)

y

76. (5.5) Solve for x:

77. (6.6) Standing 265 ft from the base of theStrastosphere Tower in Las Vegas, Nevada, theangle of elevation to the top of the tower is about

. Approximate the height of the tower to thenearest foot.77°

2351 �2500

1 � e�1.015x

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Precalculus—

Fundamental Trigonometric Identities

Reciprocal Ratio Pythagorean Identities due identities identities identities to symmetry

EXAMPLE 1 � Using Symmetry to Verify an IdentityVerify the identity: (1 � tan t)2 � sec2t � 2 tan(�t)

Solution � Begin by squaring the left-hand side, in hopes of using a Pythagorean identity.

binomial square

rewrite terms

substitute sec2t for

At this point, we appear to be off by two signs, but quickly recall that tangent is anodd function and �tan t � tan(�t). By writing sec2t � 2 tan t as sec2t � 2(�tan t),we can complete the verification.

rewrite expression to obtain

substitute for


B. Verifying Identities

We’re now ready to put these ideas and the ideas from Section 7.1 to work for us. Whenverifying identities we attempt to mold, change, or rewrite one side of the equality until we obtain a match with the other side. What follows is a collection of the ideasand methods we’ve observed so far, which we’ll call the Guidelines for Verifying Iden-tities. But remember, there really is no right place to start. Think things over for a mo-ment, then attempt a substitution, simplification, or operation and see where it leads. Ifyou hit a dead end, that’s okay! Just back up and try something else.

Guidelines for Verifying Identities

1. As a general rule, work on only one side of the identity.• We cannot assume the equation is true, so properties of equality cannot be

applied.• We verify the identity by changing the form of one side until we get a match

with the other.2. Work with the more complex side, as it is easier to reduce/simplify than to

“build.”3. If an expression contains more than one term, it is often helpful to combine

terms using .

4. Converting all functions to sines and cosines can be helpful.5. Apply other algebra skills as appropriate: distribute, factor, multiply by a

conjugate, and so on.6. Know the fundamental identities inside out, upside down, and

backward—they are the key!

A

B�

C

D�

AD � BC

BD

�tan ttan 1�t 2 � sec2t � 2 tan1�t 2�tan t � sec2t � 21�tan t2

1 � tan2t � sec2t � 2 tan t � 1 � tan2t � 2 tan t

11 � tan t22 � 1 � 2 tan t � tan2t

tan1�t2 � �tan tcot2t � 1 � csc2tcot t �cos t

sin ttan t �

1

cot t

cos1�t2 � cos t1 � tan2t � sec2ttan t �sec t

csc tcos t �

1

sec t

sin1�t2 � �sin tcos2t � sin2t � 1tan t �sin t

cos tsin t �

1

csc t

WORTHY OF NOTEThe identities due to symmetry aresometimes referred to as theeven/odd properties. Theseproperties can help express thecofunction identities in shifted form.For example,

� cos at ��

2b.

� cos c�at ��

2b d

sin t � cos a�

2� tb

WORTHY OF NOTEWhen verifying identities, it isactually permissible to work on each side of the equalityindependently, in the effort tocreate a “match.” But properties of equality can never be used,since we cannot assume anequality exists.


we can identify and use

identities due to symmetry

cob19537_ch07_662-669.qxd 1/26/11 8:58 AM Page 662

substitute for cot x,

for sec x, for csc x1

sin x1

cos x

cos xsin x

multiply numerator anddenominator by cos x sin x(the LCD for all fractions)

Note how these ideas are employed in Examples 2 through 5, particularly the fre-quent use of fundamental identities.

EXAMPLE 2 � Verifying an IdentityVerify the identity:

Solution � As a general rule, the side with the greater number of terms or the side withrational terms is considered “more complex,” so we begin with the right-hand side.

substitute for

decompose rational term

substitute for

factor out

substitute for


In the first step of Example 2, we converted all functions to sines and cosines. Dueto their use in the ratio identities, this often leads to compound fractions that we’ll needto simplify to complete a proof.

EXAMPLE 3 � Verifying an Identity by Simplifying Compound Fractions

Verify the identity

Solution � Beginning with the left-hand side, we’ll use the reciprocal and ratio identities toexpress all functions in terms of sines and cosines.

distribute

simplify

factor out cos x in numerator

simplify


� cos x

�cos x 1sin x � cos x21sin x � cos x2

�cos x sin x � cos 2x

sin x � cos x

�

112 1cos x sin x2 � acos x

sin xb1cos x sin x2

a 1cos x

b1 cos x sin x2 � a 1

sin xb1cos x sin x2

�

a1 �cos x

sin xb1cos x sin x2

a 1cos x

�1

sin xb1cos x sin x2

1 � cot x

sec x � csc x�

1 �cos x

sin x

1cos x

�1

sin x

1 � cot x

sec x � csc x� cos x

sec2 � � 1tan2� � sin2� tan2�

sin2� � sin2� 1sec2� � 121

cos2�sec2� � sin2� sec2� � sin2�

�sin2�

1# 1

cos2�� sin2�

tan2�sin2�

cos2� tan2� � sin2� �

sin2�

cos2�� sin2�

sin2� tan2� � tan2� � sin2�.


Precalculus—

cob19537_ch07_662-669.qxd 1/26/11 8:59 AM Page 663

Examples 2 and 3 involved factoring out a common expression. Just as often, we’llneed to multiply numerators and denominators by a common expression, as in Example 4.

EXAMPLE 4 � Verifying an Identity by Multiplying Conjugates

Verify the identity: .

Solution � Both sides of the identity have a single term and one is really no more complexthan the other. As a matter of choice we begin with the left side. Noting thedenominator on the left has the term sec t, with a corresponding term of tan2t onthe right, we reason that multiplication by a conjugate might be productive.

multiply numerator and denominator by the conjugate

distribute:

substitute for

multiply above and below by �1


Example 4 highlights the need to be very familiar with families of identities. To re-place , we had to use , not simply tan2t, since the related Pythagoreanidentity is .

As noted in the Guidelines, combining rational terms is often helpful. At this point,

students are encouraged to work with the pattern as a means of

combing rational terms quickly and efficiently.

EXAMPLE 5 � Verifying an Identity by Combining Terms

Verify the identity: .

Solution � We begin with the left-hand side.

combine terms:

simplify numerator, substitute tan x for


Identities come in an infinite variety and it would be impossible to illustrate allvariations. The general ideas and skills presented should prepare you to verify any ofthose given in the Exercise Set, as well as those you encounter in your future studies.See Exercises 35 through 58.

sin xcos x

�tan2x � cos2x

tan x

�11 � tan2x2 � 11 � cos2x2

asin x

1b a 1

cos xb

AB

�CD

�AD � BC

BD sec x

sin x�

sin xsec x

�sec2x � sin2x

sin x sec x

sec x

sin x�

sin xsec x

�tan2x � cos2x

tan x

A

B�

C

D�

AD � BC

BD

1 � tan2t � sec2t�tan2t1 � sec2t

�1 � cos t

tan2t

1 � sec2t�tan2t�

cos t � 1

�tan2t

cos t sec t � 1, 1A � B 2 1A � B 2 � A2 � B 2�cos t � 1

1 � sec2t

cos t

1 � sec t� a cos t

1 � sec tb a1 � sec t

1 � sec tb

cos t

1 � sec t�

1 � cos t

tan2t


Precalculus—

substitute for sec2x,

for sin2x, for sec x1

cos x1 � cos2x

1 � tan2x

B. You’ve just seen how

we can verify general identities

11 � tan2t � sec2t 1 1 � sec2t � �tan2t 2

cob19537_ch07_662-669.qxd 1/26/11 8:59 AM Page 664

C. Showing an Equation Is Not an Identity

To show an equation is not an identity, we need only find a single value for which thefunctions involved are defined but the equation is false. This can often be done by trialand error, or even by inspection. To illustrate the process, we’ll use two common mis-conceptions that arise in working with identities.

EXAMPLE 6 � Showing an Equation Is Not an IdentityShow the equations given are not identities.

a. b.

Solution � a. The assumption here seems to be that we can factor out the coefficient fromthe argument. By inspection we note the amplitude of sin(2x) is whilethe amplitude of 2 sin x is This means they cannot possibly be equal forall values of x, although they are equal for integer multiples of . For instance,substituting for x shows the left- and right-hand sides of the equation areequal (see Figure 7.4). However, Figure 7.5 shows the two sides of the equation

are not equal when . This equation is not an identity.x ��

6

��

A � 2.A � 1,

cos1� � �2 � cos � � cos �sin12x2 � 2 sin x


Precalculus—


b. The assumption here is that we can distribute function values. This is similar tosaying , a statement obviously false for all values except

Here we’ll substitute convenient values to prove the equation false,

namely, and

substitute for and for

simplify

result is false


Many times, a graphical test can be used to help determine if an equation is an iden-tity. While not fool-proof, seeing if the graphs appear identical can either suggest the iden-tity is true, or definitely show it is not. When testing identities, it helps to the left-handside of the equation as Y1 on the screen, andthe right-hand side as Y2. We can then test whetheran identity relationship might exist by graphingboth relations, and noting whether two graphs or asingle graph appears on the screen.

Consider the equation .After entering and

, we obtain the graph shown inFigure 7.6 using the calculator’s 7:ZTrigfeature.

ZOOM

Y2 � sin1X22 Y1 � 1 � cos1X22 sin2x1 � cos2x �

GRAPH

Y=

ENTER

�1 � 0

cos � � �12

2�12

2

��

4�

3�

4 cos a3�

4�

�

4b � cos a3�

4b � cos a�

4b

� ��

4.� �

3�

4

x � 0.1x � 4 � 1x � 2

�2�

�4

4

2�

Figure 7.6

cob19537_ch07_662-669.qxd 1/26/11 5:21 PM Page 665

Since only one graph appears in this window, it seems that an identity relationshiplikely exists between Y1 and Y2. This can be further supported (but still not proven) byentering to vertically translate the graph of Y2 up 1 unit (be sure todeactivate Y2) and observing the (see Figure 7.7). Of course the calculator’sTABLE feature could also be used on Y1 and Y2.

To graphically investigate the equation , andand note the existence of two distinct graphs (Figure 7.8). The resulting

graphs confirm our solution to Example 6(a): —even thoughfor integer multiples of (the points of intersection). See Exercises

65 through 68.�sin12x2 � 2 sin x

2 sin xsin12x2 �2 sin1X2Y2 �

Y1 � sin12X2GRAPH2 sin xsin12x2 �

GRAPH

Y3 � Y2 � 1


Precalculus—

�2�

�4

4

2�

Figure 7.7


we can use counterexamples

and contradictions to show an

equation is not an identity

�2�

�4

4

2�

Figure 7.8

2. To show an equation is not an identity, we mustfind at least input value where bothsides of the equation are defined, but results in a

equation.



7.2 EXERCISES

1. The identities tan x andcot x are examples of

due to .cos1�x2 cot x � cos x

�sin x tan x � sin1�x2

10.

11.

12.1 � cos1�x2

sin x � cos x sin 1�x2 � csc x

1 � sin1�x2cos x � cos1�x2 sin x

� sec x

1 � cot21�x2 � csc2x


Verify that the following equations are identities.

7.

8.

9. sin21�x2 � cos2x � 1

1sec x � 12 3sec1�x2 � 1 4 � tan2 x

11 � sin x2 31 � sin1�x2 4 � cos2x

3. To verify an identity, always begin with the moreexpression, since it is easier tothan to .

4. Converting all terms to functions of andmay help verify an identity.

5. Discuss/Explain why you must not add, subtract,multiply, or divide both sides of the equation whenverifying identities.

6. Discuss/Explain the difference between operatingon both sides of an equation (see Exercise 5) andworking on each side independently.

cob19537_ch07_662-669.qxd 1/26/11 8:59 AM Page 666


Precalculus—

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.tan x

1 � sec x�

tan x

1 � sec x� 2 csc x

cot x

1 � csc x�

cot x

1 � csc x� 2 sec x

cos x

1 � cos x�

cos x

1 � cos x� �2 cot2x

sin x

1 � sin x�

sin x

1 � sin x� �2 tan2x

1

cos2x�

1

sin2x� csc2x sec2x

csc xcos x

�cos xcsc x

�cot2x � sin2x

cot x

1 � cos x

sin x�

sin x

1 � cos x

1 � sin xcos x

�cos x

1 � sin x

sin �

1 � cos �� csc � � cot �

cos �

1 � sin �� sec � � tan �

1sec x � cos x

� cot x csc x

1

csc x � sin x� tan x sec x

cos x � sec xsec x

� �sin2x

sin x � csc xcsc x

� �cos2x

csc x

cot x � tan x� cos x

sec x

cot x � tan x� sin x

sin x

cot x� sec x � cos x

cos x

tan x� csc x � sin x

cot x cos x � csc x � sin x

tan x � cot x � sec x csc x

sin2x cot2x � 1 � sin2x

cos2x tan2x � 1 � cos2x35. 36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49. 50.

51.

52.

53.

54.

55.

56.

57.

58.sin4x � cos4x

sin3x � cos3x�

sin x � cos x

1 � sin x cos x

sin4x � cos4x

sin3x � cos3x�

sin x � cos x

1 � sin x cos x

cos x

sin x�

sin xcos x

�sec xcsc x

�csc x � sin x

cos x

cos x

sin x�

sin xcos x

�csc xsec x

�sec x � cos x

sin x

1csc x � cot x22 �1cos x � 122

sin2x

1sec x � tan x22 �1sin x � 122

cos2x

sin4x � cos4x

sin2x� 2 � csc2x

cos4x � sin4x

cos2x� 2 � sec2x

csc4x � cot4x

csc2x � cot2x� 1

sec4x � tan4x

sec2x � tan2x� 1

tan x

cot x � tan x� 1 � cos2x

cot x

cot x � tan x� 1 � sin2x

cot x � tan x

cot2x � tan2x� sin x cos x

tan2x � cot2x

tan x � cot x� csc x sec x

1 � cot x

1 � cot x�

sin x � cos x

sin x � cos x

cos x � sin x

1 � tan x�

cos x � sin x

1 � tan x

1 � cos x

1 � cos x� 1csc x � cot x22

1 � sin x

1 � sin x� 1tan x � sec x22

sin x tan x � cos x � sec x

cos x cot x � sin x � csc x

cos2x 1tan2x � sec2x2 � �cos2x

sin2x 1cot2x � csc2x2 � �sin2x

csc2x

1 � tan2x� cot2x

sec2x

1 � cot2x� tan2x

cob19537_ch07_662-669.qxd 1/26/11 8:59 AM Page 667

Show that the following equations are not identities.

59.

60.

61.

62.

63.

64. cos2� � sin2� � �1

tan a�

4b �

tan �

tan 4

tan 12�2 � 2 tan �

cos 12�2 � 2 cos �

cos a�

4b � cos � � cos a�

4� �b

sin a� ��

3b � sin � � sin a�

3b

Determine which of the following are not identities byusing a calculator to compare the graphs of the left- andright-hand sides of each equation.

65.

66.

67.

68.cos x

1 � sin x� sec x � tan x

cos x

1 � sin x�

1 � sin xcos x

cos12x2 � 1 � 2 sin2x

1 � sin2�

cos �� cos �


Precalculus—


69. Distance to top of movie screen:

At a theater, the optimum viewing angle depends on a number offactors, like the height of the screen, the incline of the auditorium, thelocation of a seat, the height of your eyes while seated, and so on. Oneof the measures needed to find the “best” seat is the distance from youreyes to the top of the screen. For a theater with the dimensions shown,this distance is given by the formula here (x is the diagonal distancefrom the horizontal floor to your seat). (a) Show the formula isequivalent to (b) Find thedistance d if and you are sitting in the eighth row with therows spaced 3 ft apart.

70. The area of triangle ABC:

If one side and three angles of a triangle are known, its area can be computed using thisformula, where side c is opposite angle C. Find the area of the triangle shown in thediagram.

A �c2 sin A sin B

2 sin C

� � 18°d2 � 800 � 40x 1cos � � sin �2 � x2.

d2 � 120 � x cos �22 � 120 � x sin �22

72. Pythagorean theorem: For the triangle shown, (a) find an expression for the area of the triangle interms of cot x and cos x, then determine its area

given (b) Show the expression you found in

part (a) is equivalent to

and recompute thearea using thisexpression. Did theanswers match?

A �1

2 1csc x � sin x2

x ��

6.

� APPLICATIONS

71. Pythagorean theorem: For the triangle shown, (a) find an expression for the length of thehypotenuse in terms of tan x and cot x, thendetermine the length of the hypotenuse when

rad. (b) Show the expression you found inpart (a) is equivalent to andrecompute the length ofthe hypotenuse using thisexpression. Did theanswers match?

h � 1csc x sec xx � 1.5

h√cot x

√tan xcos x

cot x

Exercise 72

d

D

�

�

3 ft

(not to scale)

3 ft

20 ft

20 ft

x

BA

C

45�

20 cm60�

75�

cob19537_ch07_662-669.qxd 1/26/11 5:21 PM Page 668

7–17 Section 7.3 The Sum and Difference Identities 669

Precalculus—

73. Viewing distance: Referring to Exercise 69, find aformula for D—the distance from a patron’s eyesto the bottom of the movie screen. Simplify theresult using a Pythagorean identity, then find thevalue of D for the same seat described in part (b) ofExercise 69.

74. Viewing angle: Referring to Exercises 69 and 73,once d and D are known, the viewing angle (theangle subtended by the movie screen and theviewer’s eyes) can be found using the formula

Find the value of cos �cos � �d2 � D2 � 202

2dD.

�

for the same seat described in part (b) of Exercise 69.

75. Intensity of light: In a study of the luminousintensity of light, the expression

can occur.

Simplify the equation for the moment

76. Intensity of light: Referring to Exercise 75, findthe angle given and � � 60°.I1 � I2�

I1 � I2.

sin � �I1 cos �

21I1cos �22 � 1I2sin �22

78. Use factoring to show the equation is an identity:sin4x � 2 sin2x cos2x � cos4x � 1.


77. Verify the identity sin6x � cos6x

sin4x � cos4x�1� sin2x cos2x.

81. (6.6) Use anappropriate trig ratio tofind the length of thebridge needed to crossthe lake shown in thefigure.

82. (2.3) Graph usingtransformations of atoolbox function:f 1x2 � �2�x � 3� � 6


79. (4.4) Graph the rational function given.

80. (6.2) Verify that is a point on the unit

circle, then state the values of sin t, cos t, and tan tassociated with this point.

a27

4,

3

4b

h1x2 �x � 1

x2 � 4

d

62�

400 ydExercise 82

7.3 The Sum and Difference Identities

The sum and difference formulas for sine and cosine have a long and ancient history.Originally developed to help study the motion of celestial bodies, they were used cen-turies later to develop more complex concepts, such as the derivatives of the trig func-tions, complex number theory, and the study of wave motion in different mediums.These identities are also used to find exact results (in radical form) for many nonstan-dard angles, a result of great importance to the ancient astronomers and still of notablemathematical significance today.

A. The Sum and Difference Identities for Cosine

On the unit circle with center C, consider the point A on the terminal side of angle and point B on the terminal side of angle as shown in Figure 7.9. Since thecoordinates of A and B are and respectively. Using the 1cos �, sin �2,1cos �, sin �2 r � 1,�,

�,


how we can:

A. Develop and use sum anddifference identities forcosine

B. Use the cofunctionidentities to develop thesum and differenceidentities for sine andtangent

C. Use the sum anddifference identities toverify other identities

cob19537_ch07_662-669.qxd 1/26/11 8:59 AM Page 669


Precalculus—

distance formula, we find that segment is equal to

With no loss of generality, we can rotate sector ACB clockwise, until side coin-cides with the x-axis. This creates new coordinates of (1, 0) for B, and new coordinatesof for A, but the distance remains unchanged! (SeeFigure 7.10.) Recomputing the distance gives

Since both expressions represent the same distance, we can set them equal to eachother and solve for

property of radicals

subtract 2

divide both sides by

The result is called the difference identity for cosine. The sum identity forcosine follows immediately, by substituting for .

difference identity

substitute for

The sum and difference identities can be used to find exact values for the trig func-tions of certain angles (values written in nondecimal form using radicals), simplifyexpressions, and to establish additional identities.

EXAMPLE 1 � Finding Exact Values for Non-Standard AnglesUse the sum and difference identities for cosine to find exact values for

a. b.Check results on a calculator in degree .

Solution � Each involves a direct application of the related identity, and uses special values.

a. difference identity

standard values

combine terms

See Figure 7.11.

cos 15° �16 � 12

4

� a12

2b a13

2b � a12

2b a1

2b

� � 45°, � � 30° cos145° � 30°2 � cos 45° cos 30° � sin 45° sin 30° cos1� � �2 � cos � cos � � sin � sin �

MODE

cos 75° � cos145° � 30°2cos 15° � cos145° � 30°2

cos1��2 � cos �; sin1��2 � �sin � cos1� � �2 � cos � cos � � sin � sin �

�� cos1� � 3�� 4 2 � cos � cos1��2 � sin � sin1��2 cos1� � �2 � cos � cos � � sin � sin �

��

�2 cos1� � �2 � cos � cos � � sin � sin �

�2 cos1� � �2 � �2 cos � cos � � 2 sin � sin �

2 � 2 cos1� � �2 � 2 � 2 cos � cos � � 2 sin � sin �

AB � AB 12 � 2 cos1� � �2 � 12 � 2 cos � cos � � 2 sin � sin �

cos1� � �2. � 12 � 2 cos1� � �2 � 2 3cos21� � �2 � sin21� � �2 4 � 2 cos1� � �2 � 1

� 2cos21� � �2 � 2 cos1� � �2 � 1 � sin21� � �2 AB � 2 3cos1� � �2 � 1 42 � 3sin1� � �2 � 0 42

AB1cos1� � �2, sin1� � �2 2CB

cos2u � sin2u � 1 � 22 � 2 cos � cos � � 2 sin � sin �

� 21cos2� � sin2�2 � 1cos2� � sin2�2 � 2 cos � cos � � 2 sin � sin �

� 2cos2� � 2 cos � cos � � cos2� � sin2� � 2 sin � sin � � sin2�

AB � 21cos � � cos �22 � 1sin � � sin �22AB

binomial squares

regroup

�

A

(cos �, sin �)

(cos �, sin �)B

C

�

� � �1

Figure 7.9

(cos(� � �), sin(� � �))A

B

C � � � (1, 0)

1

Figure 7.10

Figure 7.11

cob19537_ch07_670-680.qxd 1/26/11 5:12 PM Page 670


Precalculus—

b. sum identity

standard values

combine terms

See Figure 7.12.


cos 75° �16 � 12

4

� a12

2b a13

2b � a12

2b a1

2b

� � 45°, � � 30° cos145° � 30°2 � cos 45° cos 30° � sin 45° sin 30° cos1� � �2 � cos � cos � � sin � sin �

CAUTION � Be sure you clearly understand how these identities work. In particular, note that

and in general f 1a � b2� f 1a2� f 1b2.a0 � 12

� 132b30°2 � cos 60° � cos 30°cos160° �

These identities are listed here using the “ ” and “ ” notation to avoid need-less repetition. In their application, use both upper symbols or both lower symbols,with the order depending on whether you’re evaluating the cosine of a sum or differ-ence of two angles. As with the other identities, these can be rewritten to form othermembers of the identity family. One such version is used in Example 2 to consoli-date a larger expression.

��

Figure 7.12

The Sum and Difference Identities for Cosine

cosine family: functions repeat, signs alternate

can be used to expand or contract

EXAMPLE 2 � Using a Sum/Difference Identity to Simplify an ExpressionWrite as a single expression in cosine and evaluate:

Solution � Since the functions repeat and are expressed as a difference, we use the sumidentity for cosine to rewrite the difference as a single expression.

sum identity for cosine

57° � 78° � 135°

The expression is equal to See the figure.


The sum and difference identities can be used to evaluate the cosine of the sum oftwo angles, even when they are not adjacent or expressed in terms of cosine.

cos 135° � �12

2.

� cos 135°� � 57°, � � 78° cos 57° cos 78° � sin 57° sin 78° � cos157° � 78°2

cos � cos � � sin � sin � � cos1� � �2

cos 57° cos 78° � sin 57° sin 78°

cos � cos � � sin � sin � � cos1� � �2cos1� � �2 � cos � cos � � sin � sin �

cob19537_ch07_670-680.qxd 1/26/11 5:20 PM Page 671


Precalculus—

EXAMPLE 3 � Computing the Cosine of a SumGiven with the terminal side of in QI, and with theterminal side of in QII. Compute the value of

Solution � To use the sum formula we need the value of and .Using the given information about the quadrants along with the Pythagoreantheorem, we draw the triangles shown in Figures 7.13 and 7.14, yielding thevalues that follow.

Using gives this result:


To verify the result of Example 3, we can use the inverse trigonometric functionsand our knowledge of reference angles to approximate the values of and . In thefirst line of Figure 7.15, we find the QI value of isapproximately , and store it in memory loca-tion A using (A). Noting the termi-nal side of is in QII, we compute its value

as shown in the second calculation andstore it in memory location B. For the final step of theverification, we evaluate in fractional

form and obtain , as in Example 3.

B. The Sum and Difference Identities for Sine and Tangent

The cofunction identities were introduced in Section 6.6, using the complementary

angles in a right triangle. In this section we’ll verify that and

For the first, we use the difference identity for cosine to obtain

� sin �

� 102 cos � � 112 sin �

cos a�

2� �b � cos

�

2 cos � � sin

�

2 sin �

sin a�

2� �b � cos �.

cos a�

2� �b � sin �

�204

325

cos 1A � B21�106.26°2 �

MATHALPHASTO

22.62°�

��

� �204

325

� �84

325�

120

325

cos 1� � �2 � a12

13b a� 7

25b � a 5

13b a24

25b

cos 1� � �2 � cos � cos � � sin � sin �

cos � �12

13 1QI2, sin � �

5

13 1given2, cos � � �

7

25 1QII2, and sin � �

24

25 1QII2

sin �cos �, sin �, cos �,

cos 1� � �2.�tan � � �24

7�sin � � 513

12

52 � 122 � 132

135

x

y

�

Figure 7.13

�7

2524

x

y

�

(�7)2 � 242 � 252

Figure 7.14


we can develop and use sum

and difference identities for

cosine

Figure 7.15

cob19537_ch07_670-680.qxd 1/26/11 9:04 AM Page 672


Precalculus—

For the second, we use and replace with the real number .

This gives

cofunction identity for cosine

replace with

result, note

This establishes the cofunction relationship for sine: for any

real number t. Both identities can be written in terms of the real number t. See Exercises 19 through 24.

The Cofunction Identities

The sum and difference identities for sine can easily be developed using cofunction

identities. Since we need only rename t as the sum or the

difference and work from there.

cofunction identity

substitute for t

regroup argument

apply difference identity for cosine

result

The difference identity for sine is likewise developed. The sum and differenceidentities for tangent can be derived using ratio identities and their derivation is left asan exercise (see Exercise 78).

The Sum and Difference Identities for Sine and Tangent

sine family: functions alternate, signs repeat

can be used to expand or contract

tangent family:

can be used to expand or contracttan � � tan �

1 � tan � tan �� tan 1� � �2

tan 1� � �2 �tan � � tan �

1 � tan � tan �

sin � cos � � cos � sin � � sin 1� � �2sin 1� � �2 � sin � cos � � cos � sin �

sin 1� � �2 � sin � cos � � cos � sin �

� cos a�

2� �b cos � � sin a�

2� �b sin �

� cos c a�

2� �b � � d

1� � �2 sin 1� � �2 � cos c�2

� 1� � �2 d sin t � cos a�

2� tb

1� � �21� � �2sin t � cos a�

2� tb,

sin a�

2� tb � cos tcos a�

2� tb � sin t

sin a�

2� tb � cos t

c�2

� a�

2� tb d � t cos t � sin a�

2� tb

�

2� t� cos a�

2� c�

2� t d b � sin a�

2� tb

cos a�

2� �b � sin �

�

2� t�cos a�

2� �b � sin �,

signs match original in numerator,signs alternate in denominator

cob19537_ch07_670-680.qxd 1/26/11 9:04 AM Page 673

EXAMPLE 4A � Simplifying Expressions Using Sum/Difference IdentitiesWrite as a single expression in sine:

Solution � Since the functions in each term alternate and the expression is written as a sum,we use the sum identity for sine:

sum identity for sine

substitute 2t for and t for

� sin(3t) simplify

The expression is equal to sin(3t).

With and, the TABLE feature of a calculator set

in radian provides strong support for the result ofExample 4A (see Figure 7.16).

EXAMPLE 4B � Simplifying Expressions Using Sum/Difference Identities

Use the sum or difference identity for tangent to find the exact value of

Solution � Since an exact value is requested, must be the sum or difference of two standard

angles. A casual inspection reveals . This gives

sum identity for tangent

simplify expression


C. Verifying Other Identities

Once the sum and difference identities are established, we can simply add these to thetools we use to verify other identities.

EXAMPLE 5 � Verifying an Identity

Verify that is an identity.tan a� ��

4b �

tan � � 1tan � � 1

�1 � 13

1 � 13

tan a2�

3b � �13, tan a�

4b � 1 �

�13 � 1

1 � 1�132 112

� �2�

3, � �

�

4 tan a2�

3�

�

4b �

tan a2�

3b � tan a�

4b

1 � tan a2�

3b tan a�

4b

tan 1� � �2 �tan � � tan �

1 � tan � tan �

11�

12�

2�

3�

�

4

11�

12

tan 11�

12.

MODE

Y2 � sin 13X2 sin 1X2Y1 � sin 12X2 cos 1X 2 � cos12X2

�� sin 12t2 cos t � cos 12t2 sin t � sin 12t � t2 sin � cos � � cos � sin � � sin 1� � �2

sin 12t2 cos t � cos 12t2 sin t.


Precalculus—


we can use the cofunction

identities to develop the sum

and difference identities for

sine and tangent

Figure 7.16

cob19537_ch07_670-680.qxd 1/26/11 9:04 AM Page 674


Precalculus—

Solution � Using a direct application of the difference formula for tangent we obtain


EXAMPLE 6 � Verifying an IdentityVerify that is an identity.

Solution � Using the sum and difference formulas for sine we obtain


� sin2� � sin2�

� sin2� � sin2� sin2� � sin2� � sin2� sin2�

� sin2� 11 � sin2�2 � 11 � sin2�2 sin2�

� sin2� cos2� � cos2� sin2�

sin 1� � �2 sin 1� � �2 � 1sin � cos � � cos � sin �2 1sin � cos � � cos � sin �2

sin 1� � �2 sin 1� � �2 � sin2� � sin2�

tan a�

4b � 1 �

tan � � 1

1 � tan ��

tan � � 1tan � � 1

� � �, � ��

4 tan a� �

�

4b �

tan � � tan

�

4

1 � tan � tan

�

4

use to write the expression solely in terms of sine

distribute

simplify

cos2x � 1 � sin2x

1A � B2 1A � B2 � A 2 � B 2


we can use the sum and

difference identities to verify

other identities

2. To find an exact value for tan , use the sumidentity for tangent with and .

4. For the sine sum/difference identities, the functionsin each term, with the sign

between them.

6. Discuss/Explain why

is an identity, even though the arguments of cosinehave been reversed. Then verify the identity.

tan1� � �2 �sin1� � �2cos1� � �2

� �� 105°



7.3 EXERCISES

1. Since we know is since in .

3. For the cosine sum/difference identities, thefunctions in each term, with the sign between them.

5. Discuss/Explain how we know the exact value for

will be negative, prior

to applying any identity.

cos 11�

12� cos a2�

3�

�

4b

tan � 6 0tan 60° � tan 105°tan 45° �tan 45° � tan 60° 7 1,

cob19537_ch07_670-680.qxd 1/26/11 9:04 AM Page 675


Precalculus—

Rewrite as a single expression.

25.

26.

27.

28.

Find the exact value of the given expressions.

29.

30.

31.

32.

33. For with terminal side of in QII

and with terminal side of in QIII, find

a. b.

34. For with terminal side of in QI and

with terminal side of in QII, find

a. b.

Find the exact value of the expression given using a sumor difference identity. Some simplifications may involveusing symmetry and the formulas for negatives.

35. 36.

37. 38. sin a11�

12bsin a5�

12b

sin1�75°2sin 105°

tan1� � �2sin1� � �2�cos � � �

12

37

�csc � �29

20

tan1� � �2sin1� � �2�cot � �

15

8

�cos � � �7

25

tan a3�

20b � tan a �

10b

1 � tan a3�

20b tan a �

10b

tan a11�

21b � tan a4�

21b

1 � tan a11�

21b tan a4�

21b

sin a11�

24b cos a5�

24b � cos a11�

24b sin a5�

24b

sin 137° cos 47° � cos 137° sin 47°

tan ax

2b � tan ax

8b

1 � tan ax

2b tan ax

8b

tan15�2 � tan12�21 � tan15�2 tan12�2

sin ax

2b cos ax

3b � cos ax

2b sin ax

3b

sin13x2 cos15x2 � cos13x2 sin15x2


Find the exact value of the expression given using a sumor difference identity. Some simplifications may involveusing symmetry and the formulas for negatives. Checkresults on a calculator.

7. 8.

9. 10.

Use sum/difference identities to verify that bothexpressions give the same result. Check results on acalculator.

11. a. b.

12. a. b.

Rewrite as a single expression in cosine.

13.

14.

Find the exact value of the given expressions. Checkresults on a calculator.

15.

16.

17. For with terminal side of in QIV and


18. For with terminal side of in QII and


Use a cofunction identity to write an equivalentexpression.

19. 20.

21. 22.

23. 24. cos a�

3� �bsin a�

6� �b

sec a �

10btan a5�

12b

sin 18°cos 57°

cos1� � �2.�sec � � �

89

39

�sin � �112

113

cos1� � �2.�tan � � �

5

12

�sin � � �4

5

cos a7�

36b cos a5�

36b � sin a7�

36b sin a5�

36b

cos 183° cos 153° � sin 183° sin 153°

cos a�

3b cos a�

6b � sin a�

3b sin a�

6b

cos17�2 cos12�2 � sin17�2 sin12�2

cos a�

4�

�

3bcos a�

6�

�

4b

cos1120° � 45°2cos145° � 30°2

cosa�5�

12bcos a7�

12b

cos 135°cos 105°

cob19537_ch07_670-680.qxd 1/26/11 9:05 AM Page 676

Precalculus—


52. Use the diagram indicated to compute thefollowing:

a. b. c.

53. For the figure indicated, show that andcompute the following:

a. b. c.

Exercise 53

54. For the figure indicated, show that andcompute the following:

a. b. c.

Exercise 54

Verify each identity.

55. 56.

57.

58.

59. tan ax ��

4b �

1 � tan x

1 � tan x

sin ax ��

4b �

12

2 1sin x � cos x2

cos ax ��

4b �

22

2 1cos x � sin x2

cos1� � �2 � �cos �sin1� � �2 � sin �

� �

�

68 125

tan �cos �sin �

� � � � �

32 24 2845

� ��


� � � � �


39. 40.

41. 42.

Use sum/difference identities to verify that bothexpressions give the same result.

43. a. b.

44. a. b.

45. Find given . SeeExercises 7 and 35.

46. Find given See

Exercises 10 and 37.

47. Given and are acute angles with

and find

a. b. c.

48. Given and are acute angles with

and find

a. b. c.

49. Given and are obtuse angles with

and find

a. b. c.

50. Given and are obtuse angles with

and find

a. b. c.

51. Use the diagram indicated to compute thefollowing:

a. sin A b. cos A c. tan A

tan1� � �2cos1� � �2sin1� � �2sin � �

35

37,

tan � � �60

11��

tan1� � �2cos1� � �2sin1� � �2cos � � �

13

85,

sin � �28

53��

tan1� � �2cos1� � �2sin1� � �2sec � �

25

7,

cos � �8

17��

tan1� � �2cos1� � �2sin1� � �2tan � �

35

12,

sin � �12

13��

2� �5�

12�

19�

12.cos a19�

12b

150° � 105° � 255°sin 255°

sina�

4�

�

6bsina�

3�

�

4b

sin 1135° � 120°2sin 145° � 30°2

tana� �

12btan 75°

tana2�

3btan 150°

12

530��

A

15

8

15

45�

��

Exercise 51 Exercise 52

cob19537_ch07_670-680.qxd 1/26/11 9:05 AM Page 677


Precalculus—

60.

61.

62.

63.

64.

65. sin13t2 � �4 sin3t � 3 sin t

sin12t2 � 2 sin t cos t

cos12t2 � cos2t � sin2t

sin1� � �2 � sin1� � �2 � 2 sin � sin �

cos1� � �2 � cos1� � �2 � 2 cos � cos �

tan ax ��

4b �

tan x � 1

tan x � 1

66.

67. Use a difference identity to show

68. Use sum/difference identities to show

sin ax ��

4b � sin ax �

�

4b � 12 sin x.

cos ax ��

4b �

12

2 1cos x � sin x2.

cos13t2 � 4 cos3t � 3 cos t


69. Force and equilibrium:

The force required to maintain equilibrium when a screw jack is usedcan be modeled by the formula shown, where p is the pitch angle of thescrew, W is the weight of the load, is the angle of friction, with k and cbeing constants related to a particular jack. Simplify the formula using

the difference formula for tangent given and

70. Brewster’s law of reflection:

Brewster’s law of optics states that when unpolarized light strikes a dielectric surface, the transmitted light raysand the reflected light rays are perpendicular to each other. The proof of Brewster’s law involves the expression

. Use the difference identity for sine to verify that this expression leads to

Brewster’s law.

n1 sin �p � n2 sin a�

2� �pb

tan �p �n2

n1

� ��

4.p �

�

6

�

F �Wkc

tan1 p � �2

� APPLICATIONS

71. AC circuits: In a study of AC circuits, the equation

sometimes arises. Use a sum

identity and algebra to show this equation is

equivalent to

72. Fluid mechanics: In studies of fluid mechanics,the equation sometimes arises. Use a difference identity to showthat if the equation is equivalent to

.

73. Art and mathematics: When working in two-pointgeometric perspective, artists must scale their workto fit on the paper or canvas they are using. In

doing so, the equation arises.

Rewrite the expression on the right in terms of sineand cosine, then use the difference identities to

show the equation can be rewritten as .A

B� tan2 �

A

B�

tan �

tan190° � �2

cos � � cot � sin � � 11V1 � 2V2,

1V1sin � � 2V2sin1� � �2

R �1

vC1tan s � tan t2 .

R �cos s cos t

C sin1s � t274. Traveling waves: If two waves of the same

frequency, velocity, and amplitude are travelingalong a string in opposite directions, they can berepresented by the equations and . Use the sum anddifference formulas for sine to show the result

of these waves can be expressed as

75. Pressure on the eardrum: If a frequencygenerator is placed a certain distance from the ear,the pressure on the eardrum can be modeled by thefunction where f is thefrequency and t is the time in seconds. If a secondfrequency generator with identical settings is placedslightly closer to the ear, its pressure on the eardrumcould be represented by ,

where C is a constant. Show that if , the

total pressure on the eardrum is.P1t2 � A 3sin12�ft2 � cos12�ft2 43P11t2 � P21t2 4

C ��

2

P21t2 � A sin12�ft � C2

P11t2 � A sin12�ft2,

YR � 2A sin1kx2cos1vt2.YR � Y1 � Y2

Y2 � A sin1kx � vt2 Y1 � A sin1kx � vt2

cob19537_ch07_670-680.qxd 1/26/11 9:06 AM Page 678


Precalculus—

76. Angle between two cables: Two cables used to steady a radio tower areattached to the tower at heights of 5 ft and 35 ft, with both secured to astake 12 ft from the tower (see figure). Find the value of , where isthe angle between the upper and lower cables.

77. Difference quotient: Given show that the difference quotient

results in the expression

78. Difference identity: Derive the difference identity for tangent using

(Hint: After applying the difference identities,

divide the numerator and denominator by .)cos � cos �

tan1� � �2 � sin1� � �2cos1� � �2 .

sin x acos h � 1

hb � cos x asin h

hb.f 1x � h2 � f 1x2

h

f 1x2 � sin x,

�cos �


A family of identities called the angle reduction formulas will be of use in our study of complex numbers and otherareas. These formulas use the period of a function to reduce large angles to an angle in [0, ) or [0, ) having anequivalent function value: (1) ; (2) . Use the reduction formulas to findvalues for the following functions (note the formulas can also be expressed in degrees).

79. 80. 81. 82. sin 2385°sin a41�

6bcos a91�

6bcos 1665°

sin1t � 2�k2 � sin tcos1t � 2�k2 � cos t2�360°

12 ft

35 ft

� 5 ft

83. An alternative method of proving the differenceformula for cosine uses a unit circle and the factthat equal arcs are subtended by equal chords( in the diagram). Using a combination ofalgebra, the distance formula, and a Pythagoreanidentity, show that

(start by computing D2 and d2). Thendiscuss/explain how the sum identity can be foundusing the fact that

Exercise 83

D

d

(cos �, sin �) (cos �, sin �)

� � �

(1, 0)�

�

(cos(� � �), sin(� � �))

� � �1��2.sin � sin �

cos1� � �2 � cos � cos � �

D � d

84. Verify the Pythagorean theorem for each righttriangle in the diagram, then discuss/explain howthe diagram offers a proof of the sum identities forsine and cosine.

Exercise 84

sin

B c

os A

sin B sin A

sin B

A

BA

cos B

cos B cos A

cos

B s

in A

1

Exercise 76

cob19537_ch07_670-680.qxd 1/26/11 5:20 PM Page 679


Precalculus—


85. (6.5) State the period of the functions given:

a.

b.

86. (2.5) Graph the piecewise-defined function given:

f 1x2 � •3 x 6 �1

x2 �1 � x � 1

x x 7 1

y � 4 tan a2x ��

4b

y � 3 sin a�

8x �

�

3b

87. (6.6) Clarence the Clown is about to be shot from acircus cannon to a safety net on the other side ofthe main tent. If the cannon is 30 ft long and mustbe aimed at for Clarence to hit the net, the endof the cannon must be how high from groundlevel?

88. (1.4) Find the equation of the line parallel to, containing the point (5, ).

Write your answer in standard form.�22x � 5y � �10

40°

7.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities

The derivation of the sum and difference identities in Section 7.3 was a “watershedevent” in the study of identities. By making various substitutions, they lead us verynaturally to many new identity families, giving us a heightened ability to simplifyexpressions, solve equations, find exact values, and model real-world phenomena. Infact, many of the identities are applied in very practical ways, as in a study of projec-tile motion and the conic sections (Chapter 10). In addition, one of the most profoundprinciples discovered in the eighteenth and nineteenth centuries was that electricity,light, and sound could all be studied using sinusoidal waves. These waves often interact with each other, creating the phenomena known as reflection, diffraction,superposition, interference, standing waves, and others. The product-to-sum andsum-to-product identities play a fundamental role in the investigation and study ofthese phenomena.

A. The Double-Angle Identities

The double-angle identities for sine, cosine, and tangent can all be derived using therelated sum identities with two equal angles We’ll illustrate the process herefor the cosine of twice an angle.


assume and substitute for

simplify—double-angle identity for cosine

Using the Pythagorean identity we can easily find two additional members of this family, which are often quite useful. For we have

double-angle identity for cosine

substitute for

double-angle in terms of sine

Using we obtain an additional form:

double-angle identity for cosine

substitute for

double-angle in terms of cosine cos12�2 � 2 cos2� � 1sin2�1 � cos2� � cos2� � 11 � cos2�2

cos12�2 � cos2� � sin2�

sin2� � 1 � cos2�

cos12�2 � 1 � 2 sin2�

cos2�1 � sin2� � 11 � sin2�2 � sin2�

cos12�2 � cos2� � sin2�

cos2� � 1 � sin2�cos2� � sin2� � 1,

cos12�2 � cos2� � sin2�

�� cos1� � �2 � cos � cos � � sin � sin �

cos1� � �2 � cos � cos � � sin � sin �

1� � �2.


how we can:

A. Derive and use the double-angle identities for cosine,tangent, and sine

B. Develop and use thepower reduction and half-angle identities

C. Derive and use theproduct-to-sum and sum-to-product identities

D. Solve applications usingidentities

cob19537_ch07_670-680.qxd 1/26/11 9:06 AM Page 680

7–29 Section 7.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities 681

Precalculus—

The derivations of and are likewise developed and are asked for inExercise 105. The double-angle identities are collected here for your convenience.

The Double-Angle Identities

cosine: sine:

tangent:

EXAMPLE 1 � Using a Double-Angle Identity to F ind Function Values

Given , find the value of

Solution � Using the double-angle identity for cosine interms of sine, we find

double-angle in terms of sine

substitute for

result

If then A calculator check is shown in the figure.


Like the fundamental identities, the double-angle identities can be used to verifyor develop others. In Example 2, we explore one of many multiple-angle identities,verifying that can be rewritten as (in terms of powers of

).

EXAMPLE 2 � Verifying a Multiple Angle IdentityVerify that is an identity.

Solution � Use the sum identity for cosine, with and Note that our goal is anexpression using cosines only, with no multiple angles.


substitute for and for

substitute for and

multiply

substitute for

multiply

combine terms


� 4 cos3� � 3 cos � � 2 cos3 � � cos � � 2 cos � � 2 cos3�

sin2�1 � cos2� � 2 cos3� � cos � � 2 cos �11 � cos2 �2 � 2 cos3� � cos � � 2 cos � sin2�

sin12�2cos 12�2 cos13�2 � 12 cos2� � 12 cos � � 12 sin � cos �2 sin ��2� cos12� � �2 � cos12�2 cos � � sin12�2sin �

cos1� � �2 � cos � cos � � sin � sin �

� � �.� � 2�

cos13�2 � 4 cos3� � 3 cos �

cos �4 cos3� � 3 cos �cos13�2

cos12�2 �7

32.sin � �

5

8,

�7

32

2 a58b2

�2532

� 1 �25

32

sin �58

� 1 � 2 a5

8b2

cos12�2 � 1 � 2 sin2�

cos12�2.sin � �5

8

tan12�2 �2 tan �

1 � tan2�

� 2 cos2� � 1

� 1 � 2 sin2�

sin12�2 � 2 sin � cos � cos12�2 � cos2� � sin2�

tan12�2sin12�2

cob19537_ch07_681-694.qxd 1/26/11 9:08 AM Page 681

EXAMPLE 3 � Using a Double-Angle Formula to Find Exact ValuesFind the exact value of .

Solution � A product of sines and cosines having the same argument hints at the double-angleidentity for sine. Using and dividing by 2 gives

double-angle identity for sine

replace with

multiply


B. The Power Reduction and Half-Angle Identities

Expressions having a trigonometric function raised to a power occur quite frequentlyin various applications. We can rewrite even powers of these trig functions in terms ofan expression containing only cosine to the power 1, using what are called the powerreduction identities. In calculus, this becomes an indispensible tool, making expressionseasier to use and evaluate. It can legitimately be argued that the power reduction identitiesare actually members of the double-angle family, as all three are a direct consequence. Tofind identities for and , we solve the related double-angle identity involving

.

in terms of sine

subtract 1, then divide by

power reduction identity for sine

Using the same approach for gives The identity

for can be derived from (see Exercise 106), but in this

case it’s easier to use the identity The result is

The Power Reduction Identities

EXAMPLE 4 � Using a Power Reduction FormulaWrite in terms of an expression containing only cosines to the power 1 anduse the feature of a calculator to support your result.GRAPH

8 sin4x

cos2� �1 � cos12�2

2 sin2� �

1 � cos12�22

tan2� �1 � cos12�21 � cos12�2

1 � cos12�21 � cos12�2 .tan2� �

sin2�

cos2�.

tan12�2 �2 tan �

1 � tan2�tan2�

cos2� �1 � cos12�2

2.cos2�

sin2� �1 � cos12�2

2

�2 �2 sin2� � cos12�2 � 1

cos 12�2 1 � 2 sin2� � cos12�2cos12�2 sin2�cos2�

sin 45° �122

�

a12

2b

2�12

4

�sin 45°

2

22.5°� sin 22.5° cos 22.5° �sin 32122.5°2 4

2

sin � cos � �sin12�2

2

sin12�2 � 2 sin � cos �

sin 22.5° cos 22.5°


we can derive and use the

double-angle identities for

cosine, tangent, and sine


Precalculus—

cob19537_ch07_681-694.qxd 1/26/11 9:09 AM Page 682

Solution � original expression

substitute for sin2x

multiply

substitute for cos2(2x)

multiply

result

To support our result, we enter 8 sin(X)^4and in our calculator. Recognizing Y1 can onlytake on nonnegative values less than or equal to 8, we set the window as shown in the figure. Anticipating that only one graphwill be seen (since Y1 and Y2 should becoincident), we can vertically shift Y2 down 4 units and graph . The figurethen helps support that .


The half-angle identities follow directly from the power reduction identities, using

algebra and a simple change of variable. For we first take square

roots and obtain Using the substitution gives

and making these substitutions results in the half-angle identity for cosine:

where the radical’s sign depends on the quadrant in which

terminates. Using the same substitution for sine gives

and for the tangent identity, In the case of we can

actually develop identities that are free of radicals by rationalizing the denominator or numerator. We’ll illustrate the former, leaving the latter as an exercise (see Exercise 104).

multiply by the conjugate

rewrite


Since and sin u has the same sign as for all u in its domain,

the relationship can simply be written tan au

2b �

1 � cos u

sin u.

tan au

2b1 � cos u 7 0

2x 2 � �x� � � ` 1 � cos u

sin u`

� �B

11 � cos u22sin2u

� �B

11 � cos u221 � cos2u

tan au

2b � �

B

11 � cos u2 11 � cos u211 � cos u2 11 � cos u2

tan au

2b,tan au

2b � �

A

1 � cos u

1 � cos u.

sin au

2b � �

A

1 � cos u

2,

u

2

cos au

2b � �

A

1 � cos u

2,

� �u

2,u � 2�cos � � �

B

1 � cos12�22

.

cos2� �1 � cos12�2

2,

8 sin4x � 3 � 4 cos12x2 � cos14x2Y3 � Y2 � 4

Y2 � 3� 4 cos12X2�cos14X2Y1 �

� 3 � 4 cos12x2 � cos14x2 � 2 � 4 cos12x2 � 1 � cos14x2

1 � cos 14x22

� 2 c1 � 2 cos12x2 � 1 � cos14x2

2 d

� 2 31 � 2 cos12x2 � cos212x2 41 � cos 12x2

2 � 8 c 1 � cos12x2

2 d 2

8 s in4x � 81sin2x22

�2�

�6

10

2�


Precalculus—

cob19537_ch07_681-694.qxd 1/26/11 9:09 AM Page 683

The Half-Angle Identities

EXAMPLE 5 � Using Half-Angle Formulas to Find Exact ValuesUse the half-angle identities to find exact values for (a) and (b)

Solution � Noting that is one-half the standard angle we can find each value byapplying the respective half-angle identity with in Quadrant I.

a.

Note the verification of part (a) in the figure. Part (b) can be similarly checked witha calculator in degree .


EXAMPLE 6 � Using Half-Angle Formulas to Find Exact Values

For and in QIII, find exact values of and

Solution � With in QIII we know must be in QII

and we choose our signs accordingly: and


�A

16

25�

4

5

�Q

1 � a� 7

25b

2

sina�

2b �A

1 � cos �

2

cos a�

2b 6 0.sin a�

2b 7 0

S�

26

�

26

3�

4

�

2S � 6 � 6

3�

2,�

cos a�

2b.sin a�

2b�cos � � �

7

25

MODE

sin 15° �22 � 13

2

�R

1 �13

2

2

sin a30°

2b �B

1 � cos 30°

2

u � 30°30°,15°

tan 15°.sin 15°

tan au

2b �

sin u

1 � cos utan au

2b �

1 � cos u

sin u

tan au

2b � �

A

1 � cos u

1 � cos usin au

2b � �

A

1 � cos u

2cos au

2b � �

A

1 � cos u

2


Precalculus—

b.

tan 15° �

1 �13

2

1

2

� 2 � 13

tan a30°

2b �

1 � cos 30°

sin 30°

� �A

9

25� �

3

5

� �Q

1 � a� 7

25b

2

cos a�

2b � �

A

1 � cos �

2


we can develop and use the

power reduction and

half-angle identities

cob19537_ch07_681-694.qxd 1/26/11 9:10 AM Page 684


Precalculus—

C. The Product-to-Sum Identities

As mentioned in the introduction, the product-to-sum and sum-to-product identitiesare of immense importance to the study of any phenomenon that travels in waves, likelight and sound. In fact, the tones you hear as you dial a telephone are actually the sumof two sound waves interacting with each other. Each derivation of a product-to-sumidentity is very similar (see Exercise 107), and we illustrate by deriving the identity for

Beginning with the sum and difference identities for cosine, we have

cosine of a difference

cosine of a sum

combine equations

divide by 2

In addition to wave phenomenon, the identities from this family are very useful ina study of calculus and are listed here.

The Product-to-Sum Identities

EXAMPLE 7 � Rewriting a Product as an Equivalent Sum Using IdentitiesWrite the product 2 cos(27t) cos(15t) as the sum of two cosine functions.

Solution � This is a direct application of the product-to-sum identity, with and

product-to-sum identity

substitute

result


There are times we find it necessary to “work in the other direction,” writing a sumof two trig functions as a product. This family of identities can be derived from theproduct-to-sum identities using a change of variable. We’ll illustrate the process for

You are asked for the derivation of in Exercise 108. Tobegin, we use and This creates the sum andthe difference yielding and , respectively.

Dividing the original expressions by 2 gives and which all

together make the derivation a matter of direct substitution. Using these values in any product-to-sum identity gives the related sum-to-product, as shown here.

� �u � v

2,� �

u � v

2

� � � � v� � � � u2� � 2� � 2v,2� � 2� � 2u2� � u � v.2� � u � v

cos u � cos vsin u � sin v.

� cos112t2 � cos142t2 2 c os127t2cos115t2 � 2 a1

2b3cos127t � 15t2 � cos127t � 15t2 4

cos � cos � �1

2 3cos1� � �2 � cos1� � �2 4

� � 15t.� � 27t

cos � sin � �1

2 3sin1� � �2 � sin1� � �2 4 sin � cos � �

1

2 3sin1� � �2 � sin1� � �2 4

sin � sin � �1

2 3cos1� � �2 � cos1� � �2 4 cos � cos � �

1

2 3cos1� � �2 � cos1� � �2 4

cos � cos � �1

2 3cos1� � �2 � cos1� � �2 4

2 c os � cos � � cos1� � �2 � cos1� � �2 � cos � cos � � sin � sin � � cos1� � �2

cos � cos � � sin � sin � � cos1� � �2cos � cos �.

cob19537_ch07_681-694.qxd 1/26/11 9:10 AM Page 685

The sum-to-product identities follow.

The Sum-to-Product Identities

EXAMPLE 8 � Rewriting a Sum as an Equivalent Product Using IdentitiesGiven and express as a product oftrigonometric functions.

Solution � This is a direct application of the sum-to-product identity withand

substitute


For a mixed variety of identities, see Exercises 85 through 102.

D. Applications of Identities

In more advanced mathematics courses, rewriting an expression using identitiesenables the extension or completion of a task that would otherwise be very difficult (oreven impossible). In addition, there are a number of practical applications in the phys-ical sciences.

Projectile Motion

A projectile is any object that is thrown, shot, kicked, dropped, or otherwise given aninitial velocity, but lacking a continuing source of propulsion. If air resistance isignored, the range of the projectile depends only on its initial velocity v and the angle

at which it is propelled. This phenomenon is modeled by the function

r 1�2 �1

16 v2 sin � cos �

�

� 2 sin111�t2 cos1�t2 sin112�t2 � sin110�t2 � 2 sina12�t � 10�t

2b cosa12�t � 10�t

2b

sin u � sin v � 2 sinau � v

2b cos au � v

2b

v � 10�t.u � 12�tsin u � sin v,

y1 � y2y2 � sin110�t2,y1 � sin112�t2

cos u � cos v � �2 sin au � v

2b sin au � v

2b sin u � sin v � 2 cos au � v

2b sin au � v

2b

sin u � sin v � 2 sin au � v

2b cos au � v

2b cos u � cos v � 2 cos au � v

2b cos au � v

2b

2 s in au � v

2b cos au � v

2b � sin u � sin v

sin au � v

2b cos au � v

2b �

1

2 1sin u � sin v2

sin � cos � �1

2 3sin1� � �2 � sin1� � �2 4


Precalculus—

multiply by 2

product-to-sum identity (sum of sines)

substitute for , for ,

substitute u for and v for � � ��

�u � v

2�

u � v2

sum-to-productidentity

substitute for u and for v10�t

12�t


we can derive and use the

product-to-sum and sum-to-

product identities

cob19537_ch07_681-694.qxd 1/26/11 9:10 AM Page 686


Precalculus—

EXAMPLE 9 � Using Identities to Solve an Application

a. Use an identity to show is equivalent to

b. If the projectile is thrown with an initial velocity of ft/sec, how far willit travel if ?

c. From the result of part (a), determine what angle will give the maximumrange for the projectile.

Solution � a. Note that we can use a double-angle identity if we rewrite the coefficient.

Writing as and commuting the factors gives

b. With ft/sec and , the formula gives

Evaluating the result shows the projectile travels a horizontal distance of 144 ft.

c. For any initial velocity v, will be maximized when is a maximum.This occurs when , meaning and The maximumrange is achieved when the projectile is released at an angle of For verification

we’ll assume an initial velocity of 96 ft/sec and

enter the function

as Y1. With an amplitude of 288 and results confined to the first quadrant, we set an appropriate window,graph the function, and use the (CALC) 4:maximum feature. As shown inthe figure, the max occurs at .� � 45°

TRACE2nd

288 sin12�2r 1�2 �

1

32 19622sin12�2 �

45°.� � 45°.2� � 90°sin12�2 � 1sin12�2r 1�2

r 115°2 � a 1

32b19622 sin 30°.� � 15°v � 96

r 1�2 � a 1

32b v212 sin � cos �2 � a 1

32bv2sin12�2.

2 a 1

32b1

16

�

� � 15°v � 96

r 1�2 �1

32 v2 sin12�2.

r1�2 �1

16 v2sin � cos �


Sound Waves

Each tone you hear on a touch-tone phone is actually the combination ofprecisely two sound waves with different frequencies (frequency f is defined as

). This is why the tones you hear sound identical, regardless of what phone

you use. The sum-to-product and product-to-sum formulas help us to understand,study, and use sound in very powerful and practical ways, like sending faxes and usingother electronic media.

EXAMPLE 10 � Using an Identity to Solve an ApplicationOn a touch-tone phone, the sound created by pressing5 is produced by combining a sound wave withfrequency 1336 cycles/sec, with another wave havingfrequency 770 cycles/sec. Their respective equationsare and with the resultant wave being or

Rewrite this sumas a product.

11540�t2.y � cos12672�t2 � cosy � y1 � y2

12� 770t2,y2 � cosy1 � cos12� 1336t2

f �B

2�

1

1209 cps

2 3

4 5 6

7 8 9

* 0 #

1477 cps

697 cps

770 cps

852 cps

941 cps

1336 cps

350

0

900

cob19537_ch07_681-694.qxd 1/27/11 12:12 PM Page 687


Precalculus—

Solution � This is a direct application of a sum-to-product identity, with andComputing one-half the sum/difference of u and v gives

sum-to-product identity


Note we can identify the button pressed when the wave is written as a sum. If wehave only the resulting wave (written as a product), the product-to-sum formula mustbe used to identify which button was pressed.

Additional applications requiring the use of identities can be found in Exercises 117 through 122.

cos12672�t2 � cos11540�t2 � 2 cos12106�t2 cos1566�t2 cos u � cos v � 2 cos au � v

2b cos au � v

2b

2672�t � 1540�t

2� 2106�t and

2672�t � 1540�t

2� 566�t.

v � 1540�t.u � 2672�t

D. You’ve just seen how

we can solve applications

using identities

substitute for uand for v1540�t

2672�t

2. If is in QIII then and must be

in since .

4. For the half-angle identities the sign preceding theradical depends on the in which .

6. In Example 6, we were given and

in QIII. Discuss how the result would differ if westipulate that is in QII instead.�

�cos � � �7

25

u

2

6�

26

�

2180° 6 � 6 270°�



7.4 EXERCISES

1. The double-angle identities can be derived usingthe identities with For weexpand using .

3. Multiple-angle identities can be derived using the sum and difference identities. For sin (3x) usesin ( ).

5. Explain/Discuss how the three different identities

for are related. Verify that

1 � cos u

sin u�

sin u

1 � cos u.

tan au

2b

�

cos1� � �2 cos12�2� � �.

Find exact values for sin(2 ), cos(2 ), and tan(2 ) usingthe information given. Check results on a calculator.

7. ; in QII 8. ; in QII

9. in QII 10. ; in QIII

11. in QIII 12. in QI�sec � �53

28;�tan � �

13

84;

�sin � � �63

65�cos � � �

9

41;

�cos � � �21

29�sin � �

5

13

��13. ;

14. ;

15. ;

16. cos � 7 0cot � � �80

39;

sec � 6 0csc � �5

3

tan � 7 0cos � � �8

17

cos � 6 0sin � �48

73


cob19537_ch07_681-694.qxd 1/26/11 9:11 AM Page 688


Precalculus—

Find exact values for sin , cos , and tan using theinformation given.

17. in QII

18. ; in QIII

19. ; in QII

20. ; in QIV

21. Verify the following identity:

22. Verify the following identity:

Use a double-angle identity to find exact values for thefollowing expressions.

23. 24.

25. 26.

27. 28.

29. Use a double-angle identity to rewrite 9 sin (3x) cos (3x) as a single function. [Hint: .]

30. Use a double-angle identity to rewriteas a single term.

[Hint: Factor out a constant.]

Rewrite in terms of an expression containing onlycosines to the power 1. Verify result with a calculator.

31. sin2x cos2x 32. sin4x cos2x

33. 3 cos4x 34. cos4x sin4x

35. 2 sin6x 36. 4 cos6x

Use a half-angle identity to find exact values for, and for the given value of . Check

results on a calculator.

37. 38.

39. 40.

41. 42.

43. 44. � �11�

12� �

3�

8

� � 112.5°� � 67.5°

� �5�

12� �

�

12

� � 75°� � 22.5°

�tan �sin �, cos �

2.5 � 5 sin2x

9 � 92 122

2 tan1 �12 21 � tan21 �12 2

2 tan 22.5°

1 � tan222.5°

2 cos2 a �

12b � 11 � 2 sin2 a�

8b

cos215° � sin215°cos 75° sin 75°

cos14�2 � 8 cos4� � 8 cos2� � 1

sin13�2 � 3 sin � � 4 sin3�

2�cos12�2 �120

169

2�cos12�2 � �41

841

2�sin12�2 � �240

289

2�sin12�2 �24

25;

�� Use the results of Exercises 37–40 and a half-angleidentity to find the exact value.

45. 46.

47. 48.

Use a half-angle identity to rewrite each expression as asingle, nonradical function.

49. 50.

51. 52.

53. 54.

Find exact values for and using

the information given.

55. ; is obtuse

56. ; is obtuse

57. ; in QII

58. ; in QIII

59. ; in QII

60. ; in QIII

61. ; is acute

62. ; is acute

63.

64. csc � �41

9;

�

26 � 6 �

cot � �21

20; � 6 � 6

3�

2

�cos � �48

73

�sin � �15

113

�sec � � �65

33

�tan � � �35

12

�sin � � �7

25

�cos � � �4

5

�cos � � �8

17

�sin � �12

13

tan a�

2bsin a�

2b, cos a�

2b,

1 � cos 16x2sin 16x2

sin 12x21 � cos 12x2

B

1 � cos 16�21 � cos 16�2B

1 � cos 14�21 � cos 14�2

A

1 � cos 45°

2A

1 � cos 30°

2

cos a5�

24bsin a �

24b

tan 37.5°sin 11.25°

cob19537_ch07_681-694.qxd 1/26/11 9:12 AM Page 689


Precalculus—

Write each product as a sum using the product-to-sumidentities.

65. 66.

67. 68.

69.

70.

Find the exact value using product-to-sum identities.

71. 72.

73. 74.

Write each sum as a product using the sum-to-productidentities.

75. 76.

77. 78.

79.

80.

Find the exact value using sum-to-product identities.

81.

82.

83.

84.

Verify the following identities.

85.

86.

87.

88.

89.

90. sin14x2 � 4 sin x cos x11 � 2 sin2x2cos18�2 � cos214�2 � sin214�21sin2x � 122 � sin4x � cos12x21sin x � cos x22 � 1 � sin12x2

1 � 2 sin2 x

2 sin x cos x� cot12x2

2 sin x cos x

cos2x � sin2x� tan12x2

sin a11�

12b � sin a7�

12b

sin a17�

12b � sin a13�

12b

cos 285° � cos 195°

cos 75° � cos 15°

cos1852�t2 � cos11209�t2cos1697�t2 � cos 11447�t2

cos a7x

6b � cos a5x

6bsin a11x

8b � sin a5x

8b

sin114k2 � sin141k2cos19h2 � cos14h2

sin a7�

12b sin a� �

12bsin a7�

8b cos a�

8b

2 cos 105° cos 165°2 cos 15° sin 135°

2 cos12150�t2 cos1268�t22 cos11979�t2 cos1439�t2

2 sin a5t

2b sin a9t

2b2 cos a7t

2b sin a3t

2b

cos115�2 sin 1�3�2sin1�4�2 sin18�291.

92.

93.

94.

95.

96.

97.

98.

99.

100.

101.

102.

103. Show .

104. Show that is equivalent

to by rationalizing the numerator.

105. Derive the identity for and using

and where .

106. Derive the identity for using

Hint: Solve for and

work in terms of sines and cosines.

107. Derive the product-to-sum identity for .

108. Derive the sum-to-product identity for .cos u � cos v

sin � sin �

tan2�tan12�2 �2 tan1�2

1 � tan21�2 .tan21�2

� � �tan1� � �2,sin1� � �2tan12�2sin12�2

sin u

1 � cos u

tan au

2b � �

A

1 � cos u

1 � cos u

sin2� � 11 � cos �22 � c2 sin a �

2b d 2

sin m � sin n

cos m � cos n� tan am � n

2b

sin1120�t2 � sin180�t2cos1120�t2 � cos180�t2 � �cot120�t2

2 cos2 ax

2b � 1 � cos x

1 � sin212�2 � 1 � 4 sin2� � 4 sin4�

1 � 2 sin2ax

4b � cos ax

2b

cos2 ax

2b � sin2 ax

2b � cos x

csc12x2 �1

2 csc x sec x

tan x � cot x � 2 csc12x2cot � � tan � �

2 cos12�2sin12�2

tan12�2 �2

cot � � tan �

csc2� � 2 �cos12�2sin2�

cos12�2sin2�

� cot2� � 1

cob19537_ch07_681-694.qxd 1/26/11 9:13 AM Page 690


Precalculus—

Range of a projectile: Exercises 111 and 112 refer toExample 9. In Example 9, we noted that the range of aprojectile was maximized at . If or

, the projectile falls short of its maximumpotential distance. In Exercises 111 and 112 assume thatthe projectile has an initial velocity of 96 ft/sec.

111. Compute how many feet short of maximum theprojectile falls if (a) and (b) .Answer in both exact and approximate form.

112. Use a calculator to compute how many feet short ofmaximum the projectile falls if (a) and

and (b) and . Do yousee a pattern? Discuss/explain what you notice andexperiment with other values to confirm yourobservations.

Touch-tone phones: The diagram given in Example 10shows the various frequencies used to create the tonesfor a touch-tone phone. Use a sum-to-product identity towrite the resultant wave when the following numbers arepressed.

113. 3

114. 8

� � 52.5°� � 37.5°� � 50°� � 40°

� � 67.5°� � 22.5°

� 6 45°� 7 45°� � 45°

One button is randomly pressed and the resultant wave ismodeled by y(t) shown. Use a product-to-sum identity towrite the expression as a sum and determine the buttonpressed.

115.

116.

117. Clock angles: KirklandCity has a large clockatop city hall, with aminute hand that is 3 ftlong. Claire and Monicaindependently attempt todevise a function thatwill track the distancebetween the tip of theminute hand at t minutes between the hours, andthe tip of the minute hand when it is in the verticalposition as shown. Claire finds the function

, while Monica devises

. Use the identities

from this section to show the functions areequivalent.

d1t2 �A

18 c1 � cos a�t

30b d

d1t2 � `6 sin a�t

60b `

y1t2 � 2 cos11906�t2 cos1512�t2y1t2 � 2 cos12150�t2 cos1268�t2


109. Supersonic speeds, the sound barrier, and Mach numbers:

The speed of sound varies with temperature and altitude. At , soundtravels about 742 mi/hr at sea level. A jet-plane flying faster than the speedof sound (called supersonic speed) has “broken the sound barrier.” Theplane projects three-dimensional sound waves about the nose of the craftthat form the shape of a cone. The cone intersects the Earth along ahyperbolic path, with a sonic boom being heard by anyone along this path.The ratio of the plane’s speed to the speed of sound is called its Machnumber M, meaning a plane flying at is traveling 3.2 times the speed of sound. This Mach number canbe determined using the formula given here, where is the vertex angle of the cone described. For the followingexercises, use the formula to find or as required. For parts (a) and (b), answer in exact form (using a half-angle identity) and approximate form.

a. b. c.

110. Malus’s law:

When a beam of plane-polarized light with intensity I0 hits an analyzer, the intensity I of the transmitted beam oflight can be found using the formula shown, where is the angle formed between the transmission axes of thepolarizer and the analyzer. Find the intensity of the beam when and candelas (cd). Answer inexact form (using a power reduction identity) and approximate form.

� APPLICATIONS

I0 � 300� � 15°�

I � I0 cos2�

M � 2� � 45°� � 30°

�M

�M � 3.2

32°F

M � csc a�

2b

d

cob19537_ch07_681-694.qxd 1/26/11 9:13 AM Page 691


Precalculus—

118. Origami: TheJapanese art oforigami involves therepeated folding of asingle piece of paperto create various artforms. When theupper right corner of arectangular 21.6-cmby 28-cm piece of paper is folded down until thecorner is flush with the other side, the length L of

the fold is related to the angle by .

(a) Show this is equivalent to ,

(b) find the length of the fold if and (c) find the angle if .

119. Machine gears: A machinepart involves two gears. Thefirst has a radius of 2 cmand the second a radius of 1 cm, so the smaller gearturns twice as fast as thelarger gear. Let representthe angle of rotation in thelarger gear, measured froma vertical and downwardstarting position. Let P be apoint on the circumferenceof the smaller gear, starting at the vertical anddownward position. Four engineers working on animproved design for this component devise

�

L � 28.8 cm�� 30°,

L �21.6 sec �

sin12�2

L �10.8

sin � cos2��

functions that track the height of point P above thehorizontal plane shown, for a rotation of by thelarger gear. The functions they develop are: Engineer A:

Engineer B: Engineer C: andEngineer D: Use any of theidentities you’ve learned so far to show these fourfunctions are equivalent.

120. Working with identities: Compute the value oftwo ways, first using the half-angle identity

for sine, and second using the difference identityfor sine. (a) Find a decimal approximation for eachto show the results are equivalent and (b) verifyalgebraically that they are equivalent. (Hint: Square both sides.)

121. Working with identities: Compute the value oftwo ways, first using the half-angle identity

for cosine, and second using the difference identityfor cosine. (a) Find a decimal approximation foreach to show the results are equivalent and (b) verify algebraically that they are equivalent. (Hint: Square both sides.)

122. Standing waves: A clapotic (or standing) wave isformed when a wave strikes and reflects off a seawallor other immovable object. Against one particularseawall, the standing wave that forms can bemodeled by summing the incoming wave representedby the equation with theoutgoing wave represented by the equation

. Use a sum-to-productidentity to express the resulting wave asa product.

y � yi � yo

yo � 2 sin 11.1x � 0.6t2yi � 2 sin 11.1x � 0.6t2

cos 15°

sin 15°

h1�2 � 1 � cos 12�2.k1�2 � 1 � sin2 � � cos2 �;g 1�2 � 2 sin2�;

f 1�2 � sin12� � 90°2 � 1;

�°

28 cm

L

21.6

cm

�

2 cm

1 cm

P

h2�

�


123. Can you find three distinct, real numbers whose sum is equal to their product? A little known fact fromtrigonometry stipulates that for any triangle, the sum of the tangents of the angles is equal to the products of theirtangents. Use a calculator to test this statement, recalling the three angles must sum to . Our website atwww.mhhe.com/coburn shows a method that enables you to verify the statement using tangents that are allrational values.

124. A proof without words: From elementary geometry we have the following: (a) an angle inscribed in a semicircle is a right angle; and (b) the measure of aninscribed angle (vertex on the circumference) is one-half the measure of itsintercepted arc. Discuss/explain how the unit-circle

diagram offers a proof that . Be detailed and thorough.

125. Using and repeatedly applying the half-angle identity for cosine, show

that is equal to . Verify the result

using a calculator, then use the patterns noted to write the value of inclosed form (also verify this result). As becomes very small, what appears to behappening to the value of ?cos �

�cos 1.875°

22 � 12 � 12 � 132

cos 3.75°

� � 30°

tan ax

2b �

sin x

1 � cos x

180°

1

1 cos �

sin �

s

��2

Exercise 124

cob19537_ch07_681-694.qxd 1/26/11 5:30 PM Page 692

7–41 Reinforcing Basic Concepts 693

Precalculus—

126. (4.2) Use the rational roots theorem to find allzeroes of .

127. (6.1) The hypotenuse of a certain right triangle istwice the shortest side. Solve the triangle.

128. (6.2) Verify that is on the unit circle, thenfind and to verify .1 � tan2� � sec2�sec �tan �

11665, 63

65 2

x4 � x3 � 8x2 � 6x � 12 � 0129. (6.5) Write the equation of

the function graphed interms of a sine function ofthe form

.y � A sin1Bx � C2 � D


�3�2�1

123

y

x2��

MID-CHAPTER CHECK

1. Verify the identity using a multiplication:

2. Verify the identity by factoring:

3. Verify the identity by combining terms:

4. Show the equation given is not an identity.

5. Verify each identity.

a.

b.

6. Verify each identity.

a.

b.cot x � tan x

csc x sec x� cos2x � sin2x

sec2x � tan2x

sec2x� cos2x

1 � sec xcsc x

�1 � cos x

cot x� 0

sin3x � cos3x

sin x � cos x� 1 � sin x cos x

1 � sec2x � tan2x

2 sin xsec x

�cos xcsc x

� cos x sin x

cos2x � cot2x � �cos2x cot2x

sin x 1csc x � sin x2 � cos2x 7. Given and are obtuse angles with

and , find

a.b. cos

c.

8. Use the diagramshown tocompute sin A,cos A, and tan A.

9. Given

and in QII, find

exact values of and .

10. Given with in QIII, find the value

of , and .tan12�2sin12�2, cos12�2�sin � � �

7

25

cos a�

2bsin a�

2b

�

cos � � �15

17

tan1� � �21� � �2

sin1� � �2tan � � �

80

39

sin � �56

65��

48

14

60�

A�

Exercise 8

REINFORCING BASIC CONCEPTS

Identities—Connections and RelationshipsIt is a well-known fact that information is retained longer and used more effectively when it is organized, sequential,and connected. In this feature, we attempt to do just that with our study of identities. In flowchart form we’ll show thatthe entire range of identities has only two tiers, and that the fundamental identities and the sum and difference identitiesare really the keys to the entire range of identities. Beginning with the right triangle definition of sine, cosine, andtangent, the reciprocal identities and ratio identities are more semantic (word related) than mathematical, and thePythagorean identities follow naturally from the properties of right triangles. These form the first tier.

cob19537_ch07_681-694.qxd 1/26/11 9:14 AM Page 693


Precalculus—

Basic Definitions

Fundamental Identities

defined defined derived

Reciprocal Identities Ratio Identities Pythagorean Identities

The reciprocal and ratio identities are actually defined, while the Pythagorean identities are derived from thesetwo families. In addition, the identity is the only Pythagorean identity we actually need tomemorize; the other two follow by division of and as indicated.

In virtually the same way, the sum and difference identities for sine and cosine are the only identities that need to bememorized, as all other identities in the second tier flow from these.

sin2�cos2�cos2� � sin2� � 1

sec �

csc �� tan �

1

tan �� cot �

cot2� � 1 � csc2�1 � tan2� � sec2�

cos �

sin �� cot �

1

cos �� sec �

cos2� � sin2� � 1sin �

cos �� tan �

1

sin �� csc �

tan � �opp

adjcos � �

adj

hypsin � �

opp

hyp

(divide by ) (divide by )sin2�cos2�

Sum/Difference Identities

Double-Angle Identities Power Reduction Identities Half-Angle Identities Product-to-Sum Identitiesuse solve for in solve for combine various

in sum identities related identity and use in the sum/difference identitiespower reduction identities

see Section 7.4

see Section 7.4

(use ) (use )cos2� � 1 � sin2�sin2� � 1 � cos2�

cos 12�2 � 1 � 2 sin2�cos 12�2 � 2 cos2� � 1

sinau

2b � �

A

1 � cos u

2sin2� �

1 � cos 12�22

cos 12�2 � cos2� � sin2�

cosau

2b � �

A

1 � cos u

2cos2� �

1 � cos 12�22

sin 12�2 � 2 sin � cos �

� � u/2cos12�2cos �, sin �cos2�, sin2��

sin 1� � �2 � sin � cos � � cos � sin �

cos 1� � �2 � cos � cos � � sin � sin �

Exercise 1: Starting with the identity derive the other two Pythagorean identities.

cos2� � sin2� � 1, Exercise 2: Starting with the identity derive the double-angle

identities for cosine.cos � cos � � sin � sin �,

cos 1� � �2 �

e

cob19537_ch07_681-694.qxd 1/26/11 9:14 AM Page 694

Precalculus—

We can obtain an implicit equation for the inverse of by interchanging x- and y-values, obtaining By accepted convention, the explicit form of theinverse sine function is written or Since domain and rangevalues have been interchanged, the domain of is and the range is

The graph of can be found by reflecting the portion in red across

the line and using the endpoints of the domain and range (see Figure 7.19).

The Inverse Sine Function

For with domain

and range the inverse sine function is

or

with domain and range

if and only if sin y � xy � sin�1x

c��

2,

�

2d .3�1, 1 4

y � arcsin x,y � sin�1x

3�1, 1 4 ,c��

2,

�

2dy � sin x

y � x

y � sin�1xc��

2,

�

2d .

3�1, 1 4y � sin�1xy � arcsin x.y � sin�1x

x � sin y.y � sin x

x

y

1

��

y � sin x

�2

�2

�2

�

�2

�

�1

2�3�2

x

y

1

�1 1

y � sin x

�2

�2

�2

�

�2

�

�1

, 1��2�

, �1��2��

x

y

1

�1 1

y � sin�1x

y � sin x

y � x

�2

�2

�2

�

�2

�

�1

, 1�� 2�

, �1��2��

Figure 7.17 Figure 7.18 Figure 7.19

x

y

1

1

�1

�1

y � sin�1x

�2

�2

�2

�

�2

�

2��1, �

2��1, � �


how we can:

A. Find and graph the inversesine function and evaluaterelated expressions

B. Find and graph the inversecosine and tangentfunctions and evaluaterelated expressions

C. Apply the definition andnotation of inverse trigfunctions to simplifycompositions

D. Find and graph inversefunctions for sec x, csc x,and cot x

E. Solve applications involvinginverse functions

While we usually associate the number with the features of a circle, it also occurs insome “interesting” places, such as the study of normal (bell) curves, Bessel functions,Stirling’s formula, Fourier series, Laplace transforms, and infinite series. In much thesame way, the trigonometric functions are surprisingly versatile, finding their way intoa study of complex numbers and vectors, the simplification of algebraic expressions,and finding the area under certain curves—applications that are hugely important in acontinuing study of mathematics. As you’ll see, a study of the inverse trig functionshelps support these fascinating applications.

A. The Inverse Sine Function

In Section 5.1 we established that only one-to-one functions have an inverse. All sixtrig functions fail the horizontal line test and are not one-to-one as given. However, bysuitably restricting the domain, a one-to-one function can be defined that makes find-ing an inverse possible. For the sine function, it seems natural to choose the interval

since it is centrally located and the sine function attains all possible range

values in this interval. A graph of is shown in Figure 7.17, with the portion cor-responding to this interval colored in red. Note the range is still (Figure 7.18).3�1, 1 4y � sin x

c��

2,

�

2d

�

7.5 The Inverse Trig Functions and Their Applications

7–43 695

cob19537_ch07_695-710.qxd 1/26/11 5:36 PM Page 695

From the implicit form we learn to interpret the inverse function as, “yis the number or angle whose sine is x.” Learning to read and interpret the explicit formin this way will be helpful.

EXAMPLE 1 � Evaluating y � sin�1x Using Special ValuesEvaluate the inverse sine function for the values given:

a. b. c.

Solution � For x in and y in

a. y is the number or angle whose sine is

so

b. y is the arc or angle whose sine is

so

c. y is the number or angle whose sine is 2 Since 2 is not in is undefined.


In Examples 1(a) and 1(b), note that the equations and

each have an infinite number of solutions, but only one solution in

When x is one of the standard values , and so on , can be

evaluated by reading a standard table “in reverse.” For we locate thenumber in the right-hand column of Table 7.1, and note the “number or angle

whose sine is ” is If x is between and 1 but is not a standard value, we can

use the function on a calculator, which is most often the or function for .

EXAMPLE 2 � Evaluating y � sin�1x Using a CalculatorEvaluate each inverse sine function twice. First in radians rounded to four decimalplaces, then in degrees to the nearest tenth.

a. b.

Solution � For x in we evaluate a. With the calculator in radian , use the keystrokes

0.8492 . We find radians. In degree ,the same sequence of keystrokes gives (note that

.1.0145 rad � 58.1°2 sin�110.84922 � 58.1°MODEsin�110.84922 � 1.0145ENTER)SIN

2ndMODEy � sin�10.8492:y � sin�1x.3�1, 1 4 ,

y � arcsin 1�0.23172y � sin�10.8492

SIN

INV2ndsin�1

�1� �

2.�1,

�1y � arcsin 1�12,

y � sin�1 xba0, 1

2, 13

2, 1

c��

2,

�

2d .

sin y � �1

2sin y �

13

2

3�1, 1 4 , sin�11221 sin y � 2.y � sin�1122:

arcsin a�1

2b � �

�

6.1 sin y � �

1

2,

�1

2y � arcsin a�1

2b:

sin�1a13

2b �

�

3.1 sin y �

13

2,

13

2y � sin�1a13

2b:

c��

2,

�

2d ,3�1, 1 4

y � sin�1122y � arcsin a�1

2by � sin�1a13

2b

x � sin y 3 y � sin�1xy � sin�1x 3 x � sin y

x � sin y,


Precalculus—

Table 7.1

x sin x

0 0

1�2

132

�3

122

�4

12

�6

�12��

6

�122��

4

�132��

3

�1��2

cob19537_ch07_695-710.qxd 1/26/11 12:06 PM Page 696

b. In radian , we find . In degree ,


From our work in Section 5.1, we know that if f and g are inverses, andThis suggests the following properties.

Inverse Function Proper ties for Sine

For and

I. for x in

and

II. for x in

EXAMPLE 3 � Evaluating Expressions Using Inverse Function Proper tiesEvaluate each expression and verify the result on a calculator.

a. b. c.

Solution � a. since is in Property I

b. since is in Property II

c. is not in

This doesn’t mean the expression cannot be evaluated, only that we cannot useProperty II. Since .The calculator verification for each is shown in Figures 7.20 and 7.21. Note �

4� 0.7854.

sin 150° � sin 30°, sin�11sin 150°2 � sin�11sin 30°2 � 30°

3�90°, 90° 4 .sin�11sin 150°2 � 150°, since 150°

c��

2,

�

2d�

4arcsin c sin a�

4b d �

�

4,

3�1, 1 41

2sin c sin�1a1

2b d �

1

2,

sin�1 1sin 150°2arcsin c sina�

4b dsin c sin�1a1

2b d

c��

2,

�

2d1g � f 2 1x2 � sin�11sin x2 � x

3�1, 1 41 f � g2 1x2 � sin1sin�1x2 � x

g1x2 � sin�1x:f 1x2 � sin x

1g � f 2 1x2 � x.1 f � g2 1x2 � x

sin�11�0.23172 � �13.4°.MODE�0.2338 radsin�11�0.23172 �MODEy � arcsin1�0.23172:

7–45 Section 7.5 The Inverse Trig Functions and Their Applications 697

Precalculus—

WORTHY OF NOTEThe notation for the inversesine function is a carryover from the

notation for a general inversefunction, and likewise has nothingto do with the reciprocal of thefunction. The arcsin x notationderives from our work in radians onthe unit circle, where can be interpreted as “y is an arcwhose sine is x.”

y � arcsin x

f �11x2sin�1x


The domain and range concepts at play in Example 3(c) can be further exploredwith the TABLE feature of a calculator. Begin by using the TBLSET screen (

) to set TblStart with . After placing the calculator in degree, go to the screen and input , , and

(the composition ). Then disable Y2 so that only Y1 and Y3 will be displayed(Figure 7.22). Note the inputs are standard angles, the outputs in Y1 are the (expected)

Y2 � Y1

Y3 � Y21Y12Y2 � sin�1XY1 � sin XY=MODE

¢Tbl � �30� 90WINDOW

2nd

Figure 7.20Parts (a) and (b)

Figure 7.21Part (c)

cob19537_ch07_695-710.qxd 1/27/11 12:14 PM Page 697

standard values, and the outputs in Y3 return the original standard angles. Now scrollupward until 180� is at the top of the X column (Figure 7.23), and note that Y3 contin-ues to return standard angles from the interval , including Example 3(c)’sresult: . This is a powerful reminder that the inverse functionproperties cannot always be used when working with inverse trigonometric functions.

sin�11sin 150°2 � 150°3�90°, 90° 4

For the implicit equation of inverse cosine, becomes with thecorresponding explicit forms being or By reflecting thegraph of across the line we obtain the graph of shown inFigure 7.26.

The Inverse Cosine Function

For with domain and range the inverse cosine function is

or

with domain and range

if and only if

EXAMPLE 4 � Evaluating y � cos�1x Using Special ValuesEvaluate the inverse cosine for the values given:

a. b. c. y � cos�11�2y � arccos a�13

2by � cos�1102

cos y � xy � cos�1x

30, � 4 .3�1, 1 4y � arccos xy � cos�1x

3�1, 1 4 , 30, � 4y � cos x

y � cos�1 xy � x,y � cos xy � arccos x.y � cos�1x

x � cos y,y � cos x

�2

�2

� 2�3�2

y � cos(x)

x

2

3

y

�1y � cos(x)

(�, �1)

y

2

3

(0, 1)

�2

� x��2

�1

(�1, �)

y � cos�1(x)

y � cos(x)(�, �1)

�2

� (1, 0) x�

3

y

(0, 1)

2

�1


y � cos�1(x)

x

1

3

� �1��

2

(�1, �)

(1, 0)

y

2

Precalculus—


we can find and graph the

inverse sine function and

evaluate related expressions


B. The Inverse Cosine and Inverse Tangent Functions

Like the sine function, the cosine function is not one-to-one and its domain must alsobe restricted to develop an inverse function. For convenience we choose the interval

since it is again somewhat central and takes on all of its range values in thisinterval. A graph of the cosine function, with the interval corresponding to this intervalshown in red, is given in Figure 7.24. Note the range is still (Figure 7.25).3�1, 1 4x � 30, � 4


cob19537_ch07_695-710.qxd 1/26/11 12:07 PM Page 698

Solution � For x in and y in ,a. y is the number or angle whose cosine is

This shows

b. y is the arc or angle whose cosine is

This shows

c. y is the number or angle whose cosine is Since is undefined. Attempting to evaluate on a calculator will produce the error message shown in the figure.


Knowing that and are inverse functions enables us to stateinverse function properties similar to those for sine.

Inverse Function Proper ties for Cosine

For and

I. for x in

and

II. for x in

EXAMPLE 5 � Evaluating Expressions Using Inverse Function Proper tiesEvaluate each expression.

a. b. c.

Solution � a. since 0.73 is in Property I


c. since is not in

This expression cannot be evaluated using Property II. Since

The results can also be verified using a calculator.


For the tangent function, we likewise restrict the domain to obtain a one-to-one

function, with the most common choice being The corresponding range is �.

The implicit equation for the inverse tangent function is with the explicitforms or With the domain and range interchanged, the domain

of is �, and the range is The graph of for x in a��

2,

�

2by � tan xa��

2,

�

2b.y � tan�1x

y � arctan x.y � tan�1xx � tan y

a��

2,

�

2b.

cos a4�

3b � cos a2�

3b, cos�1 c cos a4�

3b d � cos�1 c cos a2�

3b d �

2�

3.

30, � 4 .4�

3cos�1 c cos a4�

3b d �

4�

3,

30, � 4�

12 arccos c cos a �

12b d �

�

12,

3�1, 1 4cos 3cos�110.732 4 � 0.73,

cos�1 c cos a4�

3b darccos c cos a �

12b dcos 3cos�110.732 4

30, � 41g � f 2 1x2 � cos�1 1cos x2 � x

3�1, 1 41 f � g2 1x2 � cos1cos�1x2 � x

g1x2 � cos�1x:f 1x2 � cos x

y � cos�1xy � cos x

cos�11�2� � 3�1, 1 4 , cos�11�2 � 1 cos y � �.y � cos�11�2:arccos a�13

2b �

5�

6.�

13

21 cos y � �

13

2.

y � arccos a�13

2b:

cos�1102 ��

2.

0 1 cos y � 0.y � cos�1102:30, � 43�1, 1 4


Precalculus—

cob19537_ch07_695-710.qxd 1/26/11 12:07 PM Page 699

The Inverse Tangent Function Inverse Function Proper ties for Tangent

For with domain and

range �, the inverse tangent function is

or

with domain � and range

if and only if tan y � xy � tan�1x

a��

2,

�

2b.

y � arctan x,y � tan�1x

a��

2,

�

2by � tan x For and

I. for x in �

and

II. for x in a��

2,

�

2b.1g � f 2 1x2 � tan�11tan x2 � x

1 f � g2 1x2 � tan1tan�1x2 � x

g1x2 � tan�1x:f 1x2 � tan x

EXAMPLE 6 � Evaluating Expressions Involving Inverse TangentEvaluate each expression.

a. b.

Solution � For x in � and y in

a. since



C. Using the Inverse Trig Functions to Evaluate Compositions

In the context of angle measure, the expression represents an angle—

the angle y whose sine is It seems natural to ask, “What happens if we take the

tangent of this angle?” In other words, what does the expression

mean? Similarly, if represents a real number between and 1, how do

we compute ? Expressions like these occur in many fields of study.sin�1 c cos a�

3b d

�1y � cos a�

3b

tan c sin�1a�1

2b d

�1

2.

y � sin�1a�1

2b

a��

2,

�

2b�0.89arctan 3 tan1�0.892 4 � �0.89,

tan a��

3b � �13tan�11�132 � �

�

3,

a��

2,

�

2b,arctan 3 tan1�0.892 4tan�11�132


Precalculus—


we can find and graph the

inverse cosine and tangent

functions and evaluate related

expressions

is shown in red (Figure 7.27), with the inverse function shown in blue(Figure 7.28).

y � tan�1 x

y � tan x

x

y

��

�3

�1

�2

3

1

2

, 1��4�

, �1��4��

y � tan�1x

��4�1,

��4��1, �

�2�

�2

x

y

��

�3

�1

�2

3

1

2


cob19537_ch07_695-710.qxd 1/26/11 5:37 PM Page 700


Precalculus—

Figure 7.31

EXAMPLE 7 � Simplifying Expressions Involving Inverse Trig FunctionsSimplify each expression:

a. b.

Solution � a. In Example 1 we found Substituting for

gives showing

b. For we begin with the inner function Substituting

for gives With the appropriate checks satisfied we have

showing


If the argument is not a special value and we need theanswer in exact form, we can draw the triangle describedby the inner expression using the definition of the trigono-metric functions as ratios. In other words, for either y or

we draw a triangle with hypotenuse 17

and side 8 opposite to model the statement, “an angle

whose sine is ” (see Figure 7.29). Using the Pythagorean theorem, we find

the adjacent side is 15 and can now name any of the other trig functions.

EXAMPLE 8 � Using a Diagram to Evaluate an Expression In volving Inverse Trig Functions

Evaluate the expression

Solution � The expression is equivalent to where

with in (QIV or QI). For must be in

QIII or QIV. To satisfy both, must be in

QIV. From Figure 7.30 we note

showing

A calculator check is shown in Figure 7.31.


tan c sin�1a� 8

17b d � �

8

15.

tan � � �8

15,

�

sin � � �8

17 1sin � 6 02, �c��

2,

�

2d�

� � sin�1a� 8

17btan �,tan c sin�1a� 8

17b d

tan c sin�1a� 8

17b d .

8

17�

opp

hyp

�

� � sin�1a 8

17b,

sin�1 c cos a�

3b d �

�

6.sin�1a1

2b �

�

6,

sin�1a1

2b.cos a�

3b1

2

cos a�

3b �

1

2.sin�1 c cos a�

3b d ,

tan c arcsin a�1

2b d � �

13

3.tan a��

6b � �

13

3,

arcsin a�1

2b�

�

6arcsin a�1

2b � �

�

6.

sin�1 c cos a�

3b dtan c arcsin a�1

2b d

17

adj

8

�

Figure 7.29

Figure 7.30

15

17

(15, �8)

�8

�

cob19537_ch07_695-710.qxd 1/27/11 12:14 PM Page 701


Precalculus—

x

opp

�

√x2 � 16

These ideas apply even when one side of the triangle is unknown. In other words,

we can still draw a triangle for since “ is an angle whose

cosine is ”

EXAMPLE 9 � Using a Diagram to Evaluate an Expression In volving Inverse Trig Functions

Evaluate the expression Assume and the inverse

function is defined for the expression given.

Solution � Rewrite as where

Draw a triangle with

side x adjacent to and a hypotenuse of (see the figure). The Pythagorean

theorem gives which leads to

giving This shows .


D. The Inverse Functions for Secant, Cosecant, and Cotangent

As with the other functions, we restrict the domains of the secant, cosecant, and cotan-gent functions to obtain one-to-one functions that are invertible (an inverse can befound). Once again the choice is arbitrary, and because some domains are easier towork with than others, these restrictions are not necessarily used uniformly in subse-quent mathematics courses. For we’ve chosen the “most intuitive” restric-tion, one that seems more centrally located (nearer the origin). The graph of is reproduced here, along with its inverse function (see Figures 7.32 and 7.33). Thedomain, range, and graphs of the functions and are given inFigures 7.34 and 7.35.

y � cot�1xy � csc�1x

y � sec xy � sec x,

tan � � tan c cos�1a x

2x2 � 16b d �

4x

opp � 116 � 4.

opp2 � 1x2 � 162 � x2x2 � opp2 � 12x2 � 1622,2x2 � 16

�

� � cos�1a x

2x2 � 16b.

tan �,tan c cos�1a x

2x2 � 16b d

x 7 0tan c cos�1a x

2x2 � 16b d .

x

2x2 � 16�

adj

hyp.

�� cos�1a x

2x2 � 16b,


we can apply the definition

and notation of inverse trig

functions to simplify

compositions

x

y

��

�3

�1

�2

3

1

2x � [0, ) � ( , �]2

�2�

y � sec x

y � (��, �1] � [1, �)x

y

��

�3

�1

�2

3

1

2y � sec�1x

x � (��, �1] � [1, �)

y � [0, ) � ( , �]2�

2�

Figure 7.32y � sec x

Figure 7.33y � sec�1x

cob19537_ch07_695-710.qxd 1/26/11 12:09 PM Page 702

The functions , and can be evaluated by not-ing their relationships to , and , respectively. For

, we have

definition of inverse function

property of reciprocals

reciprocal ratio

rewrite using inverse function notation

substitute sec�1x for y

In other words, to find the value of evaluate .

Similarly, the expression can be evaluated using . The

expression can likewise be evaluated using an inverse tangent function:

EXAMPLE 10 � Evaluating an Inverse Trig FunctionEvaluate using a calculator only if necessary:

a. b.

Solution � a. From our previous discussion, for we evaluate

Since this is a standard value, no calculator is needed and the result is

b. For find on a calculator:


A summary of the highlights from this section follows.

cot�1a �

12b � tan�1a12

�b � 1.3147.

tan�1a12�bcot�1a �

12b,

30°.

cos�1a13

2b.sec�1a 2

13b,

cot�1a �

12bsec�1a 2

13b

cot�1x � tan�1a1xb.

cot�1x

sin�1a1xb, �x� � 1csc�1x

y � cos�1a1xb, �x� � 1y � sec�1x,

sec�1x � cos�1a1xb

y � cos�1a1xb

cos y �1x

1

sec y�

1x

sec y � x

y � sec�1xy � tan�1xy � cos�1x, y � sin�1x

y � cot�1xy � sec�1x, y � csc�1x


Precalculus—

x

y

�1 1

�2

�

�2

y � csc�1x

y � csc�1x

x � (��, �1] � [1, �)

y � [ , 0) � (0, ]�� 2 2

�

x

y

�1 1

y � cot�1x

y � cot�1x

x � (��, �)

y � (0 , �)

�2

�

�2


WORTHY OF NOTEWhile the domains of and both include all realnumbers, evaluating using

involves the restriction

To maintain consistency,

the equation

is often used. The graph of

is that of

reflected across the

x-axis and shifted units up, with

the result identical to the graph of.y � cot�1x

�

2

y � tan�1x

y ��

2� tan�1x

cot�1x ��

2� tan�1x

x � 0.

tan�1a1xb

cot�1xy � tan�1x

y � cot�1x

cob19537_ch07_695-710.qxd 1/26/11 12:09 PM Page 703

Our calculators can lend some insight to the varied domain restrictions the inverse trigfunctions demand, simply by graphing . The result should yield the identityfunction for the domain specified. With the calculator in radian and a 4:ZDecimal window, compare the graphs of and shown in Figures 7.36 and 7.37, respectively. A casual observation verifies property 1:

for any x in the interval and for any x in the

interval . See Exercises 87 through 90 for verifications of properties 2 and 3.c��

2,

�

2d

sin�11sin x2 � x3�1, 1 4sin1sin�1x2 � x

Y2 � sin�11sin X2Y1 � sin1sin�1X2ZOOMMODEy � x

y � f 1g1x2 2


Precalculus—

�4.7

�3.1

3.1

4.7 �4.7

�3.1

3.1

4.7


E. Applications of Inverse Trig Functions

We close this section with one example of the many ways that inverse functions can be applied.

EXAMPLE 11 � Using Inverse Trig Functions to F ind Viewing AnglesBelieve it or not, the drive-inmovie theaters that were sopopular in the 1950s aremaking a comeback! If youarrive early, you can park inone of the coveted “centerspots,” but if you arrive late,you might have to park veryclose and strain your neck towatch the movie. Surprisingly,the maximum viewing angle(not the most comfortableviewing angle in this case) is actually very close to the front. Assume the base of a30-ft screen is 10 ft above eye level (see Figure 7.38).


we can find and graph inverse

functions for sec x, csc x, and

cot x

x

��

�10 ft

30 ft

Figure 7.38

Summary of Inverse Function Proper ties and Compositions

1. For sin x and for any x in the interval ;

for any x in the interval

3. For tan x and for any real number x;

for any x in the interval a��

2,

�

2btan�11tan x2 � x,

tan1tan�1x2 � x,tan�1x,

c��

2,

�

2dsin�11sin x2 � x,

3�1, 1 4sin1sin�1x2 � x,sin�1x, 2. For cos x and

for any x in the interval ;for any x in the interval

4. To evaluate use

use ;

use for all real numbers x�

2� tan�1x,cot�1x,

sin�1a 1 xb, �x� � 1csc�1x,

cos�1a1xb, �x� � 1;sec�1x,

30, � 4cos�11cos x2 � x,3�1, 1 4cos1cos�1x2 � x,

cos�1x,

cob19537_ch07_695-710.qxd 1/26/11 12:09 PM Page 704

a. Use the inverse function concept to find expressions for angle and angle .b. Use the result of part (a) to find an expression for the viewing angle .c. Use a calculator to find the viewing angle (to tenths of a degree) for

distances of 15, 25, 35, and 45 ft, then to determine the distance x (to tenths ofa foot) that maximizes the viewing angle.

Solution � a. The side opposite is 10 ft, and we want to know x — the adjacent side. This

suggests we use giving In the same way, we find

that

b. From the diagram we note that and substituting for and

directly gives

c. After we enter a graphing calculator in degree

gives approximate viewing angles ofand for

25, 35, and 45 ft, respectively. From thisdata, we note the distance x that makes a maximum must be between 15 and 35 ft, and using (CALC)4:maximum shows is a maximum of

at a distance of 20 ft from thescreen (see Figure 7.39).


36.9°�

TRACE2nd

�

x � 15,29.1°,35.8°, 36.2°, 32.9°,MODE

Y1 � tan�1a40

Xb � tan�1a10

Xb,

� � tan�1a40xb � tan�1a10

xb.

�� ,

� � tan�1a40xb.

� � tan�1a10xb.tan � �

10x

,

�

�

�

��


Precalculus—

0 100

50

0

Figure 7.39

E. You’ve just seen how

we can solve applications

involving inverse functions

2. The two most common ways of writing the inversefunction for are and .y � sin x



7.5 EXERCISES

1. All six trigonometric functions fail the test and therefore are not -to- .

3. The domain for the inverse sine function is and the range is .

4. The domain for the inverse cosine function is and the range is .

5. Most calculators do not have a key for evaluatingan expression like Explain how it is doneusing the key.COS

sec�15.6. Discuss/Explain what is meant by the implicit form

of an inverse function and the explicit form. Givealgebraic and trigonometric examples.

cob19537_ch07_695-710.qxd 1/26/11 12:10 PM Page 705

Use the preceding tables to fill in each blank (principalvalues only).

7.

8.

Evaluate without the aid of calculators or tables,keeping the domain and range of each function in mind.Answer in radians.

9. 10.

11. 12. arcsin a�1

2bsin�11

arcsin a13

2bsin�1a12

2b

Evaluate using a calculator. Answer in radians to thenearest ten-thousandth and in degrees to the nearest tenth.

13. arcsin 0.8892 14.

15. 16.

Evaluate each expression, keeping the domain and rangeof each function in mind. Check results using acalculator.

17. 18.

19. 20.

21. 22.

23. 24.

Use the tables given prior to Exercise 7 to fill in eachblank (principal values only).

25.

sin c arcsin a3

5b dsin1sin�1 0.82052

arcsin c sin a�2�

3b dsin�11sin 135°2

sin�11sin 30°2arcsin c sin a�

3b d

sin c arcsin a13

2b dsin c sin�1a12

2b d

sin�1a1 � 15

2bsin�1a 1

17b

arcsin a7

8b


Precalculus—

arcsin 0 � 0°sin 180° �

arcsina�13

2b �sin 1�60°2 � �

13

2

sin�1a13

2b �sin 120° �

13

2

sin�11

2�sin 30° �

1

2

sin�11�12 �sina��

2b � �1

sin�1a�1

2b �sina�5�

6b � �

1

2

arcsin a1

2b �

�

6sina�

6b �

sin�10 �sin 0 � 0

cos�11�12 �cos � � �1

arccosa�1

2b �cos 120° � �

1

2

arccosa13

2b �

�

6cosa�

6b �

cos�11 �cos 0 � 1


The tables here show values of , , and for [ to ]. The restricted domain used to developthe inverse functions is shaded. Use the information from these tables to complete the exercises that follow.

210��180�� tan �cos �sin �

y � sin � y � cos � y � tan �

0

1

0

0 0 �1

2210°

180°�1

2�30°

1

2150°�

13

2�60°

13

2120°�1�90°

90°�13

2�120°

13

260°�

1

2�150°

1

230°�180°

sin ��sin ��

0

0

0 1 �13

2210°

�1180°13

2�30°

�13

2150°

1

2�60°

�1

2120°�90°

90°�1

2�120°

1

260°�

13

2�150°

13

230°�1�180°

cos ��cos ��

0

—

—

0

0 013

3210°

180°�13

3�30°

�13

3150°�13�60°

�13120°�90°

90°13�120°

1360°13

3�150°

13

330°�180°

tan ��tan ��

cob19537_ch07_695-710.qxd 1/26/11 12:11 PM Page 706

26.

Evaluate without the aid of calculators or tables.Answer in radians.

27. 28.

29. 30. arccos (0)

Evaluate using a calculator. Answer in radians to thenearest ten-thousandth, degrees to the nearest tenth.

31. arccos 0.1352 32.

33. 34.

Evaluate each expression, keeping the domain and rangeof each function in mind. Check results using acalculator.

35. 36.

37. 38.

39. 40.

41. 42.

Use the tables presented before Exercise 7 to fill in eachblank.

43.

arccos 1cos 315.8°2cos�1 c cos a5�

4b d

cos c arccos a13

2b dcos c cos�1a�12

2b d

cos c arccos a� 8

17b dcos1cos�1 0.55602

cos�11cos 60°2arccos c cos a�

4b d

cos�1a16 � 1

5bcos�1a15

3b

arccos a4

7b

cos�11�12arccos a�13

2bcos�1a1

2b

44.

Evaluate without the aid of calculators or tables.

45. 46.

47. 48.

Evaluate using a calculator. Answer in radians to thenearest ten-thousandth and in degrees to the nearesttenth.

49. 50.

51. 52.

Simplify each expression without using a calculator.

53. 54.

55. 56.

57. 58.

59. 60.

Explain why the following expressions are not defined.

61. 62. cot(arccos 1)

63. 64.

Use the diagrams below to write the value of: (a) ,(b) , and (c) .

65. 66.

67. 68.

�

10

3x

√100 � 9x2

�

6x

√x2 � 36

�

10

2010√3

�

0.3

0.5 0.4


cos�1 c sec a2�

3b dsin�1 c csc a�

4b d

tan1sin�112

arcsin1cos 135°2arccos 3sin1�30°2 4cot c cos�1a�1

2b dcsc c sin�1a12

2b d

sec c arcsin a1

2b dtan c arccos a13

2b d

cos�1 c sin a��

3b dsin�1 c cos a2�

3b d

arctan1�162arctan a29

21b

tan�110.32672tan�11�2.052

tan�10arctan1132arctan1�12tan�1a�13

3b


Precalculus—

tan�11132 �tana�

3b �

arctan a13

3b �tan 30° �

13

3

arctan1�132 � ��

3tana��

3b �

tan�10 �tan 0 � 0

arctan 1 ��

4tan a�

4b �

arctan 1�132 �tan 120° � �13

tan�10 �tan � � 0

tan�1a13

3b �tan 1�150°2 �

13

3

cos�11 �cos 12�2 � 1

arccosa�1

2b � 120°cos1�120°2 �

cos�1a13

2b �cosa��

6b �

13

2

cos�1a1

2b �cos1�60°2 �

1

2

cob19537_ch07_695-710.qxd 1/26/11 12:12 PM Page 707


Precalculus—

Evaluate each expression by drawing a right triangleand labeling the sides.

69. 70.

71. 72.

73. 74.

75.

76.

Use the tables given prior to Exercise 7 to help fill ineach blank.

77.

tan c sec�1a29 � x2

xb d

cos c sin�1a x

212 � x2b d

tan c arcseca 5

2xb dcot c arcsina3x

5b d

tan c cos�1a123

12b dsin c tan�1a15

2b d

cos c sin�1a�11

61b dsin c cos�1a� 7

25b d

78.

Evaluate using a calculator in degree only asnecessary.

79. arccsc 2 80.

81. 82.

83. arcsec 5.789 84.

85. 86. arccsc 2.9875

Use the graphing feature of a calculator to determinethe interval where the following functions are equivalentto the identity function . If necessary, use the or (CALC) features to determine whether ornot endpoints should be included.

87. 88.

89. 90. Y � tan�11tan x2Y � tan1tan�1x2Y � cos�11cos x2Y � cos1cos�1x2

TRACE2nd

TRACEy � x

sec�117

cot�1a�17

2b

arccot1�12cot�113

csc�1a� 2

13b

MODE


91. The force normal to an object on an inclined plane:

When an object is on an inclined plane, the normal force is the forceacting perpendicular to the plane and away from the force of gravity, and ismeasured in a unit called newtons (N). The magnitude of this forcedepends on the angle of incline of the plane according to the formulaabove, where m is the mass of the object in kilograms and g is the force ofgravity (9.8 m/sec2). Given , find (a) FN for and

and (b) for and .

92. Heat flow on a cylindrical pipe: y � 0

When a circular pipe is exposed to a fan-driven source of heat, the temperature ofthe air reaching the pipe is greatest at the point nearest to the source (see diagram).As you move around the circumference of the pipe away from the source, thetemperature of the air reaching the pipe gradually decreases. One possible model ofthis phenomenon is given by the formula shown, where T is the temperature of theair at a point (x, y) on the circumference of a pipe with outer radius ,T0 is the temperature of the air at the source, and TR is the surrounding roomtemperature. Assuming , and : (a) Find thetemperature of the air at the points (0, 5), (3, 4), (4, 3), (4.58, 2), and (4.9, 1). (b) Why is the temperature decreasing for this sequence of points? (c) Simplify theformula using and use it to find two points on the pipe’s circumference wherethe temperature of the air is .113°

r � 5

r � 5 cmTR � 72°T0 � 220°F

r � 2x2 � y2

T � 1T0 � TR2 sin a y

2x2 � y2b � TR;

FN � 2 NFN � 1 N�� 45°� � 15°m � 225 g

FN � mg cos �

y

x2 � y2 � r2

Pipe

Heat source

Fan

xr

sec�12 � 60°sec160°2 �

arcsec 1 �sec1�360°2 � 1

arcseca� 2

13b �seca7�

6b � �

2

13

arcsec 2 �sec1�60°2 � 2

sec�11�12 � �sec1�2 �

arcseca 2

13b �sec1�30°2 �

2

13

arcsec 2 ��

3seca�

3b �

sec�11 �sec 0 � 1

�g

FN

225 kg

�

225 kg

g

FN

cob19537_ch07_695-710.qxd 1/26/11 12:13 PM Page 708


Precalculus—

� APPLICATIONS

93. Snowconedimensions: Made inthe Shade Snowconessells a colossal sizecone that uses a conicalcup holding 20 oz ofice and liquid. The cupis 20 cm tall and has aradius of 5.35 cm. Findthe angle formed by across-section of thecup.

94. Avalanche conditions:Winter avalanches occurfor many reasons, onebeing the slope of themountain. Avalanches seemto occur most often forslopes between and (snow gradually slides offsteeper slopes). The slopesat a local ski resort have anaverage rise of 2000 ft foreach horizontal run of2559 ft. Is this resort prone to avalanches? Find theangle and respond.

95. Distance to hole: Apopular story on thePGA Tour has GerryYang, Tiger Woods’teammate at Stanfordand occasional caddie,using the Pythagoreantheorem to find thedistance Tiger neededto reach a particularhole. Suppose younotice a marker in theground stating that thestraight-line distancefrom the marker to the hole (H) is 150 yd. If yourball B is 48 yd from the marker (M) and angleBMH is a right angle, determine the angle andyour straight-line distance from the hole.

96. Ski jumps: At awaterskiing contest ona large lake, skiers usea ramp rising out of thewater that is 30 ft longand 10 ft high at thehigh end. What angle does the ramp makewith the lake?

�

�

�

60°35°

�

97. Viewing angles for advertising: A 25-ft-widebillboard is erected perpendicular to a straighthighway, with the closer edge 50 ft away (seefigure). Assume the advertisement on the billboardis most easily read when the viewing angle is or more. (a) Use inverse functions to find anexpression for the viewing angle . (b) Use acalculator to help determine the distance d (totenths of a foot) for which the viewing angle isgreater than . (c) What distance d maximizesthis viewing angle?

98. Viewing angles at an art show: At an art show, apainting 2.5 ft in height is hung on a wall so that itsbase is 1.5 ft above the eye level of an averageviewer (see figure). (a) Use inverse functions tofind expressions for angles and . (b) Use theresult to find an expression for the viewing angle .(c) Use a calculator to help determine the distancex (to tenths of a foot) that maximizes this viewingangle.

�

�

�

x

2.5 ft

1.5 ft

Exercise 98

��

10.5°

�

10.5°20 cm

�

5.35 cm

Exercise 93

�

2000 ft

2559 ft

Exercise 94

150 yd

48 yd�

MarkerM

B

H

Exercise 95

30 ft

10 ft�

Exercise 96

d

� 50 ft

25 ft

Exercise 97

cob19537_ch07_695-710.qxd 1/26/11 12:13 PM Page 709


Precalculus—


Consider a satellite orbiting at an altitude of x mi above Earth. The distance d from thesatellite to the horizon and the length s of the corresponding arc of Earth are shown in thediagram.

100. To find the distance d we use the formula (a) Show how this formula wasdeveloped using the Pythagorean theorem. (b) Find a formula for the angle in terms of rand x, then a formula for the arc length s.

101. If Earth has a radius of 3960 mi and the satellite is orbiting at an altitude of 150 mi, (a) what is the measure of angle ? (b) how much longer is d than s?�

�d � 22rx � x2.


102. (7.4) Use the triangle given with a double-angleidentity to find the exact value of sin12�2.

103. (7.3) Use the triangle given with a sum identity tofind the exact value of sin1� � �2.

104. (4.6) Solve the inequality using zeroesand end-behavior given .

105. (1.4) In 2000, Space Tourists Inc. sold 28 low-orbittravel packages. By 2005, yearly sales of the low-orbit package had grown to 105. Assuming thegrowth is linear, (a) find the equation that modelsthis growth , (b) discuss themeaning of the slope in this context, and (c) use theequation to project the number of packages thatwere sold in 2010.

12000 S t � 02

f 1x2 � x3 � 9xf 1x2 � 0

Earthr

r�

Centerof the Earth

sx d

√85

7

6

�

8939

80

�

�

99. Shooting angles and shots on goal: A soccer player is on a breakaway and isdribbling just inside the right sideline toward the opposing goal (see figure).As the defense closes in, she has just a few seconds to decide when to shoot.(a) Use inverse functions to find an expression for the shooting angle . (b)Use a calculator to help determine the distance d (to tenths of a foot) that willmaximize the shooting angle for the dimensions shown.

�

� d

70 ft

(goal area)

(penalty area)

(sideline)

24 ft

cob19537_ch07_695-710.qxd 1/27/11 5:19 PM Page 710

Precalculus—

EXAMPLE 1 � Visualizing Solutions GraphicallyConsider the equation Using a graph of and

a. state the quadrant of the principal root.b. state the number of roots in and their quadrants.c. comment on the number of real roots.

Solution � We begin by drawing a quick sketch ofand noting that solutions will

occur where the graphs intersect.a. The sketch shows the principal root

occurs between 0 and in QI.

b. For we note the graphs intersecttwice and there will be two solutions in this interval, one in QI and one in QII.

c. Since the graphs of and extend infinitely in both directions,they will intersect an infinite number of times—but at regular intervals! Oncea root is found, adding integer multiples of (the period of sine) to this rootwill give the location of additional roots.


2�

y � 23y � sin x

30, 2�2�

2

y � 23,y � sin x

30, 2�2y � 2

3,y � sin xsin x � 23.

x

y

y � sin x

2��2�

�1

1

�� x

y

y � cos x

2��2�

�1

1

��

x

yy � tan x

2��2��1

�2

�3

3

2

1

��


WORTHY OF NOTE

Note that we refer to as

Quadrant I or QI, regardless ofwhether we’re discussing the unitcircle or the graph of the function.In Example 1(b), the solutionscorrespond to those found in QIand QII on the unit circle, where sin x is also positive.

a0, �

2b

x

y

y � sin x

2��2�

�1

1

��

y � s


how we can:

A. Use a graph to gaininformation aboutprincipal roots, roots in[ ), and roots in �

B. Use inverse functions tosolve trig equations forthe principal root

C. Solve trig equations forroots in [ ) or [ )

D. Solve trig equations forroots in �

0, 360°0, 2�

0, 2�

In this section, we’ll take the elements of basic equation solving and use them to helpsolve trig equations, or equations containing trigonometric functions. All of the alge-braic techniques previously used can be applied to these equations, including the prop-erties of equality and all forms of factoring (common terms, difference of squares,etc.). As with polynomial equations, we continue to be concerned with the number ofsolutions as well as with the solutions themselves, but there is one major difference.There is no “algebra” that can transform a function like into For that we rely on the inverse trig functions from Section 7.5.

A. The Principal Root, Roots in [0, 2 ), and Real Roots

In a study of polynomial equations, making a connection between the degree of anequation, its graph, and its possible roots, helped give insights as to the number, loca-tion, and nature of the roots. Similarly, keeping graphs of basic trig functions in mindhelps you gain information regarding the solution(s) to trig equations. When solving trigequations, we refer to the solution found using and as the principalroot. You will alternatively be asked to find (1) the principal root, (2) solutions in

or and (3) solutions from the set of real numbers �. For convenience,graphs of the basic sine, cosine, and tangent functions are repeated in Figures 7.40through 7.42. Take a mental snapshot of them and keep them close at hand.

30°, 360°2,30, 2�2tan�1cos�1,sin�1,

�

x � solution.sin x � 12

7.6 Solving Basic Trig Equations

7–59 711

cob19537_ch07_711-721.qxd 1/26/11 1:16 PM Page 711

When this process is applied to the equationthe graph shows the principal root occurs

between and 0 in QIV (see Figure 7.43). In the

interval the graphs intersect twice, in QII andQIV where is negative (graphically—below the x-axis). As in Example 1, the graphs continue infinitelyand will intersect an infinite number of times—butagain at regular intervals! Once a root is found, addinginteger multiples of (the period of tangent) to this rootwill give the location of other roots.

B. Inverse Functions and Principal Roots

To solve equations having a single variable term, the basic goal is to isolate the vari-able term and apply the inverse function or operation. This is true for algebraicequations like , or and for trig equationslike . In each case we would add 1 to both sides, divide by 2, thenapply the appropriate inverse. When the inverse trig functions are applied, the resultis only the principal root and other solutions may exist depending on the intervalunder consideration.

EXAMPLE 2 � Finding Principal RootsFind the principal root of .

Solution � We begin by isolating the variable term, then apply the inverse function.

given equation

add 1 and divide by

apply inverse tangent to both sides

result (exact form)


Equations like the one in Example 2 demonstrate the need to be very familiar withthe functions of special angles. They are frequently used in equations and applicationsto ensure results don’t get so messy they obscure the main ideas. For convenience, thevalues of and are repeated in Table 7.2 for . Using symmetry andthe appropriate sign, the table can easily be extended to all values in . Using thereciprocal and ratio relationships, values for the other trig functions can also be found.

C. Solving Trig Equations for Roots in [ ) or [ )

To find multiple solutions to a trig equation, we simply take the reference angle of theprincipal root, and use this angle to find all solutions within a specified range. A men-tal image of the graph still guides us, and the standard table of values (also held inmemory) allows for a quick solution to many equations.

0�, 360�0, 2�

30, 2�2� � 30, � 4cos �sin �

x ��

6

tan�11tan x2 � tan�1a 1

13b

13 tan x �1

13

13 tan x � 1 � 0

13 tan x � 1 � 0

2 sin x � 1 � 02x2 � 1 � 0,2x � 1 � 0, 21x � 1 � 0

�

tan x30, 2�2�

�

2

tan x � �2,


Precalculus—


we can use inverse functions

to solve trig equations for the

principal root

Table 7.2

0 0 1

1 0

0 �1�

�13

2

1

2

5�

6

�12

2

12

2

3�

4

�1

2

13

2

2�

3

�

2

1

2

13

2

�

3

12

2

12

2

�

4

13

2

1

2

�

6

cos �sin ��


we can use a graph to gain

information about principal

roots, roots in [ ), and

roots in �

0, 2�

y � �2

x

yy � tan x

2��2��1

�2

�3

3

2

1

��

Figure 7.43

cob19537_ch07_711-721.qxd 1/26/11 1:17 PM Page 712

EXAMPLE 3 � Finding Solutions in [0, 2 )For find all solutions in

Solution � Isolate the variable term, then apply the inverse function.

given equation

subtract and divide by 2

apply inverse cosine to both sides

result

With as the principal root, we know

Since is negative in QII and QIII, the second

solution is The second solution could also have

been found from memory, recognition, or symmetryon the unit circle. Our (mental) graph verifies theseare the only solutions in .


EXAMPLE 4 � Finding Solutions in [0, 2 )For find all solutions in .

Solution � As with the other equations having a single variable term, we try to isolate thisterm or attempt a solution by factoring.

given equation

add 1 to both sides and take square roots

result

The algebra gives or and we solve each equationindependently.

apply inverse tangent

principal roots

Of the principal roots, only is in the

specified interval. With tan x positive in QI and

QIII, a second solution is While is

not in the interval, we still use it as a referenceto identify the angles in QII and QIV (for

and find the solutions

and The four solutions are

, and which is supported by the graph shown.


7�

4,x �

�

4,

3�

4,

5�

4

7�

4.x �

3�

4

tan x � �12

x � ��

4

5�

4.

x ��

4

x � ��

4 x �

�

4

tan�11tan x2 � tan�11�12 tan�11tan x2 � tan�1112 tan x � �1 tan x � 1

tan x � �1tan x � 1

tan x � �1 2tan 2x � �11

tan2x � 1 � 0

30, 2�2tan2x � 1 � 0,

�

30, 2�2

5�

4.

cos x

�r ��

4.

3�

4

� �3�

4

cos�11cos �2 � cos�1a�12

2b

12 cos � � �12

2

2 c os � � 12 � 0

30, 2�2.2 cos � � 12 � 0,

�

7–61 Section 7.6 Solving Basic Trig Equations 713

Precalculus—

x

yy � cos x

2��2�

�1

1

��

y �� 2√2

WORTHY OF NOTENote how the graph of a trigfunction displays the informationregarding quadrants. From thegraph of we “read” thatcosine is negative in QII and QIII[the lower “hump” of the graph isbelow the x-axis in ] andpositive in QI and QIV [the graph isabove the x-axis in the intervals

and ].13�/2, 2�210, �/22

1�/2, 3�/22

y � cos x

x

y y � tan x

2��

�1

�2

�3

3

2

1

y ��1

y � 1

cob19537_ch07_711-721.qxd 1/26/11 1:17 PM Page 713

For any trig function that is not equal to a standard value, we can use a calculator to ap-proximate the principal root or leave the result in exact form, and apply the same ideasto this root to find all solutions in the interval.

EXAMPLE 5 � Finding Solutions in [ )Find all solutions in for

Solution � Use a u-substitution to simplify the equation and help select an appropriatestrategy. For the equation becomes and factoringseems the best approach. The factored form is with solutions

and Re-substituting for u gives

equations from factored form

apply inverse cosine

principal roots

Both principal roots are in the specified interval. The first is quadrantal, the secondwas found using a calculator and is approximately With positive in QIand QIV, a second solution is The three solutions seen inFigure 7.44 are and although only is exact. With thecalculator still in degree , the solutions and are verifiedin Figure 7.45. While we may believe the second calculation is not exactly zero dueto round-off error, a more satisfactory check can be obtained by storing the result of

as X, and using 3 , as shown in Figure 7.46.cos1X22 � cos1X2 � 2360 � cos�1123 2

� � 311.8°� � 180°MODE

� � 180°311.8°180°,48.2°,1360 � 48.22° � 311.8°.

cos �48.2°.

� � 48.2° � � 180°

cos�11cos �2 � cos�1a2

3b cos�11cos �2 � cos�11�12

cos � �2

3 cos � � �1

cos �u � 23.u � �1

1u � 12 13u � 22 � 0,3u2 � u � 2 � 0u � cos �,

3 cos2� � cos � � 2 � 0.30°, 360°20�, 360�

�

yy � cos �

360�180��360�

�1

1

�180�

y ��1

y � s

Figure 7.44



D. Solving Trig Equations for All Real Roots (�)

As we noted, the intersections of a trig function with a horizontal line occur at regular, pre-dictable intervals. This makes finding solutions from the set of real numbers a simple mat-ter of extending the solutions we found in or To illustrate, consider the

solutions to Example 3. For we found the solutions and

For solutions in , we note the “predictable interval” between roots is identi-

cal to the period of the function. This means all real solutions will be represented by

and (k is an integer). Both are illustrated in

Figures 7.47 and 7.48 with the primary solution indicated with a ✸.

�k � � �5�

4� 2�k,� �

3�

4� 2�k

�� 5�

4.

� �3�

42 cos � � 12 � 0,

30°, 360°2.30, 2�2


we can solve trig equations for

roots in [ ) or [ )0, 360�0, 2�


cob19537_ch07_711-721.qxd 1/26/11 5:42 PM Page 714

EXAMPLE 6 � Finding Solutions in �Find all real solutions to

Solution � In Example 2 we found the

principal root was Since

the tangent function has a periodof , adding integer multiples of

to this root will identify allsolutions:

, as illustrated here.


These fundamental ideas can be extended to many different situations. When asked tofind all real solutions, be sure you find all roots in a stipulated interval before namingsolutions by applying the period of the function. For instance, has two solutions

in and which we can quickly extend to find all real roots.

But using or a calculator limits us to the single (principal) root and

we’d miss all solutions stemming from Note that solutions involving multiples of

an angle (or fractional parts of an angle) should likewise be “handled with care,” as inExample 7.

EXAMPLE 7 � Finding Solutions in �Find all real solutions to .

Solution � Since we have a common factor of cos x, we begin by rewriting the equation asand solve using the zero factor property. The resulting

equations are and

equations from factored formsin12x2 �1

2cos x � 0

2 sin12x2 � 1 � 0 S sin12x2 � 12.cos x � 0

cos x 32 sin12x2 � 1 4 � 0

2 sin12x2 cos x � cos x � 0

3�

2.

x ��

2,x � cos�10

x �3�

2d ,c x �

�

230, 2�2

cos x � 0

�k � x ��

6� �k,

��

x ��

6.

13 tan x � 1 � 0.


Precalculus—

yy � cos �

2�

2� 2� 2�

3� 4��4�

1

��2��3�

� � f � 2�k

etc. etc.

�

s� 4 � h f z4

Figure 7.47y

y � cos x

2�

2� 2� 2�

3� 4��4�

1

��2��3�

� � h � 2�k

etc. etc.

�

z� 4 �f h m4

Figure 7.48

x

yy � tan x

� � � � �

y � 0.58

2� 3��3�

�3

3

��2�

etc. etc.

m'�l�z k

�

cob19537_ch07_711-721.qxd 1/26/11 1:18 PM Page 715

In has solutions and giving and

as solutions in �. Note these can actually be combined and written

as . For we know that sin u is positive in QI and

QII, and the reference angle for This yields the solutions (QI)

and (QII), or in this case and Since we seek all real

roots, we first extend each solution by before dividing by 2, otherwise multiple

solutions would be overlooked.

solutions from sin (2x)

divide by 2


In the process of solving trig equations, we sometimes employ fundamental iden-tities to help simplify an equation, or to make factoring or some other method possible.

EXAMPLE 8 � Solving Trig Equations Using an IdentityFind all real solutions for

Solution � With a mixture of functions, exponents, and arguments, the equation is almostimpossible to solve as it stands. But we can eliminate the sine function using theidentity leaving a quadratic equation in cos x.

given equation

substitute cos2x � sin2x for cos (2x)

combine like terms

subtract 1

Let’s substitute u for to give us a simpler view of the equation. This giveswhich is clearly not factorable over the integers. Using the

quadratic formula with and gives

quadratic formula in u

simplified

To four decimal places we have and To answer in termsof the original variable we re-substitute for u, realizing that has no solution, so solutions in must be provided by and willoccur in QII and QIII. The primary solutions are and

rounded to four decimal places, so all real solutions aregiven by and


x � 4.4048 � 2�k.x � 1.8784 � 2�k2� � 1.8784 � 4.4048

x � cos�11�0.30282 � 1.8784cos x � �0.302830, 2�2 cos x � 3.3028cos x

u � �0.3028.u � 3.3028

�3 � 113

2

u �3 � 21�322 � 4112 1�12

2112

c � �1b � �3, a � 1, u2 � 3u � 1 � 0,

cos x

cos2 x � 3 cos x � 1 � 0 cos2 x � 3 cos x � 1

cos2x � sin2x � sin2x � 3 cos x � 1 cos12x2 � sin2

x � 3 cos x � 1

cos12x2 � cos2x � sin2x,

cos12x2 � sin2x � 3 cos x � 1.

x �5�

12� �k x �

�

12� �k

� 12; k � � 2x �

5�

6� 2�k 2x �

�

6� 2�k

2�k

2x �5�

6.2x �

�

6u �

5�

6

u ��

6sin u �

1

2 is

�

6.

sin12x2 �1

2�k � x �

�

2� �k,

x �3�

2� 2�k

x ��

2� 2�kx �

3�

2,x �

�

2cos x � 030, 2�2,


Precalculus—

WORTHY OF NOTEWhen solving trig equations thatinvolve arguments other than asingle variable, a u-substitution issometimes used. For Example 7,substituting u for 2x gives the

equation , making it “easier

to see” that (since is a

special value), and therefore

and .x ��

122x �

�

6

12

u ��

6

sin u �12

cob19537_ch07_711-721.qxd 1/26/11 1:19 PM Page 716

The TABLE feature of a calculator in radian can partially (yet convincingly)verify the solutions to Example 8. On the screen, enter .This expression is then used in place of X when the left-hand side of the original equa-tion is entered as Y2 (see Figure 7.49). After using the TBLSET screen to setTblStart 0 and , the keystrokes (TABLE) will produce thetable shown in Figure 7.50. Note the values in the Y2 column are all very close to 1 (theright-hand side of the original equation), implying the Y1 values are solutions.

GRAPH2nd¢Tbl � 1�

Y1 � 4.4048 � 2�XY=

MODE


Precalculus—


we can solve trig equations for

roots in �


2. Solving trig equations is similar to solvingalgebraic equations, in that we first thevariable term, then apply the appropriate function.


Fill in each blank with the appropriate word or phrase. Carefully reread the section if necessary.

7.6 EXERCISES

1. For simple equations, a mental graph will tell usthe quadrant of the root, the number ofroots in , and show a pattern for all

roots.

3. For the principal root is ,

solutions in are and ,and an expression for all real roots is and ; .k � �

30, 2�2sin x �

12

24. For , the principal root is ,

solutions in are and ,and an expression for all real roots is .

30, 2�2tan x � �1

5. Discuss/Explain/Illustrate why and

have two solutions in even

though the period of is , while theperiod of is 2�.y � cos x

�y � tan x

30, 2�2,cos x �3

4

tan x �3

46. The equation has four solutions in

. Explain how these solutions can be viewedas the vertices of a square inscribed in the unitcircle.

30, 2�2sin2x �

1

2

cob19537_ch07_711-721.qxd 1/26/11 1:19 PM Page 717


Precalculus—


7. For the equation and the graphs of

and given, state (a) the quadrant

of the principal root and (b) the number of roots in

8. For the equation and the graphs ofand given, state (a) the quadrant of

the principal root and (b) the number of roots in

9. Given the graph shown here, draw thehorizontal line and then for

state (a) the quadrant of theprincipal root and (b) the number of roots in

10. Given the graph of shown, draw thehorizontal line and then for sec state(a) the quadrant of the principal root and (b) thenumber of roots in

11. The table shows in multiples of between 0 and

, with the values for given. Complete the

table without a calculator or references using yourknowledge of the unit circle, the signs of the trigfunctions in each quadrant, memory/recognition,

and so on.tan � �sin �

cos �,

sin �4�

3

�

6�

30, 2�2.x � 5

4,y � 54

y � sec x

x

y y � sec x

2��2�

1

�1

�2

�3

2

3

�� 2

��2

3�2

�3�2

x

y y � tan x

2��2�

�1

�2

�3

1

2

3

��


30, 2�2.tan x � �1.5,

y � �1.5y � tan x

30, 2�2.y � 3

4y � cos xcos x � 3

4

x

yy � cos x

2��2�

�1

1

��

y � E3

x

yy � sin x

2��2�

�1

1

��

y � �34


30, 2�2.y � �

3

4y � sin x

sin x � �3

4

12. The table shows in multiples of between 0 and

, with the values for given. Complete thetable without a calculator or references using yourknowledge of the unit circle, the signs of the trigfunctions in each quadrant, memory/recognition,

, and so on.

Find the principal root of each equation.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

Find all solutions in [ ).

29. 30.

31.

32. 33.2

3 cot x �

5

6� �

3

2

1

2 sec x �

3

4� �

7

4

8 tan x � 713 � �13

6.2 cos x � 4 � 7.19 sin x � 3.5 � 1

0, 2�

413 � 4 tan ��513 � 10 cos �

� tan x � 02 � 4 sin �

�5

3 sin x �

5

6

7

8 cos x �

7

16

4 sec x � �8�6 cos x � 6

�312 csc x � 6213 sin x � �3

�213 tan x � 213 tan x � 1

�4 cos x � 213�4 sin x � 212

2 sin x � �12 cos x � 12

tan � �sin �

cos �

cos �2�

�

4�

Exercise 11

0 0

1

0

�13

2

4�

3

�1

2

7�

6

�

1

2

5�

6

13

2

2�

3

�

2

13

2

�

3

1

2

�

6

tan �cos �sin ��

Exercise 12

0 1

0

0

12�

12

2

7�

4

3�

2

�12

2

5�

4

�1�

�12

2

3�

4

�

2

12

2

�

4

tan �cos �sin ��

cob19537_ch07_711-721.qxd 1/26/11 1:20 PM Page 718

34. 35.

36. 37.

38. 39.

40. 41.

42.

Solve the following equations by factoring. State allsolutions in [0°, 360°). Round to one decimal place if theresult is not a standard value.

43.

44.

45.

46. 47.

48. 49.

50.

Find all real solutions. Note that identities are notrequired to solve these exercises.

51. 52.

53. 54.

55. 56.

57. 58.

59. 60.

61.

62. �8 sin a1

2xb � �413

�213 cos a1

3xb � 213

213 tan13x2 � 613 tan12x2 � �13

2 sin13x2 � �126 cos12x2 � �3

213 tan x � 213 tan x � �13

4 sin x � 213�4 cos x � 212

2 cos x � 1�2 sin x � 12

4 cos2x � 3 � 0

4 sin2x � 1 � 02 cos3x � cos2x � 0

sec2x � 6 sec x � 162 sin2x � 7 sin x � 4

2 cos x sin x � cos x � 0

6 tan2� � 213 tan � � 0

3 cos2� � 14 cos � � 5 � 0

2

3 cos2x �

5

6�

4

3

412 sin2x � 412613 cos2x � 313

�4 csc2x � �83 sec2x � 6

�7 tan2x � �214 sin2x � 1

4 cos2x � 3�110 sin x � �5513 63.

64.

65.

66.

Solve each equation using a calculator and inverse trigfunctions to determine the principal root (not bygraphing). Clearly state (a) the principal root and (b) all real roots.

67. 68.

69. 70.

71. 72.

73. 74.

Solve the following equations using an identity. State allreal solutions in radians using the exact form wherepossible and rounded to four decimal places if the resultis not a standard value.

75.

76.

77.

78.

79.

80.

81.

82. 3 sin12�2 � cos212�2 � 1 � 0

cos12�2 � 2 sin2� � 3 sin � � 0

1cos � � sin �22 � 2

1cos � � sin �22 � 1

12 sin12x2cos13x2 � 12 sin13x2cos12x2 � 1

2 cosa1

2xbcos x � 2 sina1

2xbsin x � 1

4 sin2x � 4 cos2x � 213

cos2x � sin2x �1

2

6 sin12�2 � 3 � 2�5 cos12�2 � 1 � 0

2

5 cos12�2 �

1

4

1

2 sin12�2 �

1

3

13 csc x � 2 � 1112 sec x � 3 � 7

5 sin x � �23 cos x � 1

13 sin12x2 sec12x2 � 2 sin12x2 � 0

cos13x2 csc 12x2 � 2 cos13x2 � 0

13 sin x tan12x2 � sin x � 0

12 cos x sin12x2 � 3 cos x � 0


Precalculus—


83. Range of a projectile:

The distance a projectile travels is called its range and is modeled by the formula shown,where R is the range in meters, v is the initial velocity in meters per second, and is theangle of release. Two friends are standing 16 m apart playing catch. If the first throw hasan initial velocity of 15 m/sec, what two angles will ensure the ball travels the 16 mbetween the friends?

84. Fine-tuning a golf swing:

A golf pro is taking specific measurements on a client’s swing to help improve hergame. If the angle is too small, the ball is hit late and “too thin” (you top the ball). If is too large, the ball is hit early and “too fat” (you scoop the ball). Approximate theangle formed by the club and the extended (left) arm using the given measurementsand formula shown.

�

��

(arm length)2 � 2 (club length)(arm length)cos �(club head to shoulder)2 � (club length)2 �

�

R �549

v2 sin12�2

37 in.

27 in.

39 in. �

Exercise 84

cob19537_ch07_711-721.qxd 1/26/11 1:22 PM Page 719


Precalculus—

� APPLICATIONS

Acceleration due to gravity: When a steel ball isreleased down an inclined plane, the rate of the ball’sacceleration depends on the angle of incline. Theacceleration can be approximated by the formula

where is in degrees and theacceleration is measured in meters per second/persecond. To the nearest tenth of a degree,

85. What angle produces an acceleration of 0 m/sec2

when the ball is released? Explain why this isreasonable.

86. What angle produces an acceleration of9.8 m/sec2? What does this tell you about theacceleration due to gravity?

87. What angle produces an acceleration of 5 m/sec2?Will the angle be larger or smaller for anacceleration of 4.5 m/sec2?

88. Will an angle producing an acceleration of2.5 m/sec2 be one-half the angle required for anacceleration of 5 m/sec2? Explore and discuss.

Snell’s law states thatwhen a ray of lightpasses from onemedium into another,the sine of the angle ofincidence variesdirectly with the sine ofthe angle of refraction (see the figure). Thisphenomenon is modeledby the formula where k is called the indexof refraction. Note the angle is the angle at which thelight strikes the surface, so that Use thisinformation to work Exercises 89 to 92.

89. A ray of light passes from air into water, strikingthe water at an angle of Find the angle ofincidence and the angle of refraction , if theindex of refraction for water is

90. A ray of light passes from air into a diamond,striking the surface at an angle of Find theangle of incidence and the angle of refraction ,if the index of refraction for a diamond is

91. Find the index of refraction for ethyl alcohol if abeam of light strikes the surface of this medium atan angle of and produces an angle of refraction

Use this index to find the angle ofincidence if a second beam of light created anangle of refraction measuring 15°.

� � 34.3°.40°

k � 2.42.��

75°.

k � 1.33.��

55°.

� � 90° � �.�

sin � � k sin �,

�

�

�A1�2 � 9.8 sin �,

92. Find the index of refraction for rutile (a type ofmineral) if a beam of light strikes the surface ofthis medium at an angle of and produces anangle of refraction Use this index tofind the angle of incidence if a second beam oflight created an angle of refraction measuring

93. Roller coasterdesign: As part ofa science fairproject, Hadrabuilds a scalemodel of a roller

coaster using the equation

where y is the height of the model in inches andx is the distance from the “loading platform” ininches. (a) How high is the platform? (b) Whatdistances from the platform does the model attain aheight of 9.5 in.?

94. Company logo: Part of the logo for an engineeringfirm was modeled by a cosine function. The logowas then manufactured in steel and installed on theentrance marquee of the home office. The positionand size of the logo is modeled by the function

where y is the height of thegraph above the base of the marquee in inches andx represents the distance from the edge of themarquee. Assume the graph begins flush with theedge. (a) How far above the base is the beginningof the cosine graph? (b) What distances from theedge does the graph attain a height of 19.5 in.?

y � 9 cos x � 15,

y � 5 sin a1

2xb � 7,

10°.

� � 18.7°.30°

new medium

light

sin(�) � k sin(�)

� �

�

�

refraction

incidence

��

reflection

Exercises 89 to 92

Loading platform

International Engineering

cob19537_ch07_711-721.qxd 1/26/11 1:22 PM Page 720

7–69 Section 7.7 General Trig Equations and Applications 721

Precalculus—


95. Find all real solutions to in twoways. First use a calculator with and to determine the regular intervalsbetween points of intersection. Second, simplify byadding x to both sides, and draw a quick sketch ofthe result to locate x-intercepts. Explain why bothmethods give the same result, even though the firstpresents you with a very different graph.

Y2 � �XY1 � 5 cos X � X

5 cos x � x � �x 96. Once the fundamental ideas of solving a given familyof equations are understood and practiced, a studentusually begins to generalize them—making thenumbers or symbols used in the equation irrelevant.(a) Use the inverse sine function to find the principalroot of by solving for x interms of y, A, B, C, and D. (b) Solve the followingequation using the techniques addressed in thissection, and then using the “formula” from part (a):

Do the results agree?5 � 2 sin a1

2x �

�

4b � 3.

y � A sin1Bx � C2 � D,


97. (3.1) Use a substitution to show that is azero of .

98. (3.3) Currently, tickets to productions of theShakespeare Community Theater cost $10.00, withan average attendance of 250 people. Due tomarket research, the theater director believes thatfor each $0.50 reduction in price, 25 more peoplewill attend. What ticket price will maximize thetheater’s revenue? What will the averageattendance projected to become at that price?

f 1x2 � x2 � 4x � 5x � 2 � i 99. (7.5) Evaluate without using a calculator:

a. b.

100. (6.1) The largest Ferris wheel in the world, locatedin Yokohama, Japan, has a radius of 50 m. To thenearest hundredth of a meter, how far does a seaton the rim travel as the wheel turns through

?� � 292.5°

sin 3 tan�11�12 4tan c sin�1a�1

2b d

7.7 General Trig Equations and Applications

At this point you’re likely beginning to understand the true value of trigonometry to thescientific world. Essentially, any phenomenon that is cyclic or periodic is beyond thereach of polynomial (and other) functions, and may require trig for an accurate under-standing. And while there is an abundance of trig applications in oceanography, astron-omy, meteorology, geology, zoology, and engineering, their value is not limited to thehard sciences. There are also rich applications in business and economics, and a grow-ing number of modern artists are creating works based on attributes of the trig func-tions. In this section, we try to place some of these applications within your reach, withthe Exercise Set offering an appealing variety from many of these fields.

A. Trig Equations and Algebraic Methods

We begin this section with a follow-up to Section 7.6, by introducing trig equationsthat require slightly more sophisticated methods to work out a solution.


how we can:

A. Use additional algebraictechniques to solve trigequations

B. Solve trig equations usingmultiple angle, sum anddifference, and sum-to-product identities

C. Solve trig equations usinggraphing technology

D. Solve trig equations of theform A sin(Bx � C ) � D � k

E. Use a combination ofskills to model and solvea variety of applications

cob19537_ch07_711-721.qxd 1/26/11 1:22 PM Page 721

EXAMPLE 1 � Solving a Trig Equation by Squaring Both SidesFind all solutions in

Solution � Our first instinct might be to rewrite the equation in terms of sine and cosine, butthat simply leads to a similar equation that still has two different functions

Instead, we square both sides and see if the Pythagoreanidentity will be of use. Prior to squaring, we separate thefunctions on opposite sides to avoid the mixed term 2 tan x sec x.

given equation

subtract tan x and square

result

Since we substitute directly and obtain an equation in tangentalone.

substitute 1 � tan2x for sec2x

simplify

solve for tan x

in QI and QIII

The proposed solutions are [QI] and [QIII]. Since squaring an equation

sometimes introduces extraneous roots, both should be checked in the original equation.

The check shows only is a solution.


Here is one additional example that uses a factoring strategy commonly employedwhen an equation has more than three terms.

EXAMPLE 2 � Solving a Trig Equation by FactoringFind all solutions in

Solution � The four terms in the equation share no common factors, so we attempt to factor bygrouping. We could factor from the first two terms but instead elect to groupthe terms and begin there.

given equation

rearrange and group terms

remove common factors

remove common binomial factors

Using the zero factor property, we write two equations and solve eachindependently.

resulting equations

isolate variable term

solve

in QI and QIV in QI and QIIin QIII and QIV

solutions� � 30°, 150°, 210°, 330°sin � 6 0� � 60°, 300°sin � 7 0cos � 7 0

sin � � �1

2 cos � �

1

2

sin2� �1

4 2 c os � � 1

4 s in2 � � 1 � 0 2 c os � � 1 � 0

12 cos � � 12 14 sin2� � 12 � 0 4 s in2 �12 cos � � 12 � 112 cos � � 12 � 0

18 sin2� cos � � 4 sin2 �2 � 12 cos � � 12 � 0 8 s in2� cos � � 2 cos � � 4 sin2� � 1 � 0

sin2�2 cos �

30°, 360°2: 8 sin2� cos � � 2 cos � � 4 sin2� � 1 � 0.

x � �

6

7�

6x �

�

6

tan x 7 0

1

13� tan x

�2 � �213 tan x 1 � tan2x � 3 � 213 tan x � tan2x

sec2x � 1 � tan2x,

sec2x � 3 � 213 tan x � tan2x 1sec x22 � 113 � tan x22

sec x � tan x � 13

1 � tan2x � sec2x313 cos x � sin x � 1 4 .

30, 2�2: sec x � tan x � 13.


Precalculus—

cob19537_ch07_722-732.qxd 1/26/11 5:54 PM Page 722

Initially factoring from the first two terms and proceeding from there wouldhave produced the same result.


B. Solving Trig Equations Using Various Identities

To solve equations effectively, a student should strive to develop all of the necessary“tools.” Certainly the underlying concepts and graphical connections are of primaryimportance, as are the related algebraic skills. But to solve trig equations effectivelywe must also have a ready command of commonly used identities. Observe howExample 3 combines a double-angle identity with factoring by grouping.

EXAMPLE 3 � Using Identities and Algebra to Solve a T rig EquationFind all solutions in Round solutions tofour decimal places as necessary.

Solution � Noting that one of the terms involves a double angle, we attempt to replace thatterm to make factoring a possibility. Using the double identity for sine, we have

substitute 2 sin x cos x for sin (2x)

set equal zero and group terms

factor using 3 cos x � 1

common binomial factor

Use the zero factor property to solve each equation independently.

resulting equations

isolate variable term

in QII and QIII in QI and QII

4.3726 solutions

Should you prefer the exact form, the solutions from the cosine equation could be

written as and


C. Trig Equations and Graphing Technology

A majority of the trig equations you’ll encounter in your studies can be solved usingthe ideas and methods presented both here and in the previous section. But there aresome equations that cannot be solved using standard methods because they mix poly-nomial functions (linear, quadratic, and so on) that can be solved using algebraicmethods, with what are called transcendental functions (trigonometric, logarithmic,and so on). By definition, transcendental functions are those that transcend the reachof standard algebraic methods. These kinds of equations serve to highlight the value ofgraphing and calculating technology to today’s problem solvers.

x � 2� � cos�1a�1

3b.x � cos�1a�1

3b

x ��

6,

5�

6 x � 1.9106,

sin x 7 0cos x 6 0

sin x �1

2 cos x � �

1

3

2 s in x � 1 � 0 3 c os x � 1 � 0

13 cos x � 12 12 sin x � 12 � 0 2 s in x13 cos x � 12 � 113 cos x � 12 � 0

16 sin x cos x � 2 sin x2 � 13 cos x � 12 � 0 312 sin x cos x2 � 2 sin x � 3 cos x � 1

30, 2�2: 3 sin12x2 � 2 sin x � 3 cos x � 1.

2 cos �


Precalculus—


we can use additional

algebraic techniques to solve

trig equations


we can solve trig equations

using various identities

cob19537_ch07_722-732.qxd 1/26/11 5:54 PM Page 723

EXAMPLE 4 � Solving Trig Equations Using TechnologyUse a graphing calculator in radian mode to find all real roots of

Round solutions to four decimal places.

Solution � When using graphing technology our initial concern is the size of the viewingwindow. After carefully entering the equation on the screen, we note the term2 sin x will never be larger than 2 or less than for any real number x. On the other

hand, the term becomes larger for larger values of x, which would seem to cause

to “grow” as x gets larger. We conclude the standard window is a good

place to start, and the resulting graph is shown in Figure 7.51.

2 sin x �3x

5

3x

5

�2Y=

2 sin x �3x

5� 2 � 0.

From this screen it appears there are three real roots, but to be sure none are hiddento the right, we extend the Xmax value to 20 (Figure 7.52). Using (CALC) 2:zero, we follow the prompts and enter a left bound of 0 (a number tothe left of the zero) and a right bound of 2 (a number to the right of the zero—seeFigure 7.52). The calculator then prompts you for a GUESS, which you can bypassby pressing . The smallest root is approximately . Repeating thissequence we find the other roots are and


Some equations are very difficult to solve analytically, and even with the use of agraphing calculator, a strong combination of analytical skills with technical skills is

required to state the solution set. Consider the equation and

solutions in . There appears to be no quick analytical solution, and the first

attempt at a graphical solution holds some hidden surprises. Enter

and on the screen. Pressing 7:ZTrig gives the screen in Fig-

ure 7.53, where we note there are at least twoand possibly three solutions, depending on howthe sine graph intersects the cotangent graph.We are also uncertain as to whether the graphs

intersect again between and Increasing

the maximum Y-value to Ymax � 8 shows theydo indeed. But once again, are there now threeor four solutions? In situations like this it may

�

2.�

�

2

ZOOMY=Y2 �1

tan112X2

Y1 � 5 sin a12

Xb� 5

3�2�, 2�25 sin a1

2xb � 5 � cot a1

2xb

x � 5.5541.x � 3.0593x � 0.8435ENTER

TRACE2nd

�10

�10

10

10Figure 7.51

�10

10

0 20

Figure 7.52

�2� 2�

�4

4Figure 7.53


cob19537_ch07_722-732.qxd 1/27/11 12:18 PM 4

be helpful to use the Zeroes Method forsolving graphically. On the screen, dis-able Y1 and Y2 and enter Y3 as Pressing 7:ZTrig at this point clearlyshows that there are four solutions in thisinterval (Figure 7.54), which can easily be found using (CALC) 2:zero:

, and 0.3390.See Exercises 33 and 34 for more practicewith these ideas.

D. Solving Equations of the Form A sin (Bx C) D � k

You may remember equations of this form from Section 6.5. They actually occur quitefrequently in the investigation of many natural phenomena and in the modeling of datafrom a periodic or seasonal context. Solving these equations requires a good combina-tion of algebra skills with the fundamentals of trig.

EXAMPLE 5 � Solving Equations That Involve Transformations

Given and , for what real numbers x

is f(x) less than 240?

Algebraic Solution � We reason that to find values where we should begin by finding valueswhere . The result is

equation

subtract 320 and divide by 160; isolate variable term

At this point we elect to use a u-substitution for to obtaina “clearer view.”

substitute u for

in QIII and QIV

solutions in u

To complete the solution we re-substitute for u and solve.

re-substitute for u

multiply both sides by

solutions

We now know when and but when will f(x) be less than

240? By analyzing the equation, we find the function has period of

and is shifted to the left 1 units. This would indicate the graph peaks early in the interval with a “valley” in the interior. We conclude in theinterval (2.5, 4.5).

f 1x2 6 24030, 2�2

P �2��3

� 6

x � 4.5x � 2.5f 1x2 � 240

x � 4.5 x � 2.5

3�

x � 1 �11

2 x � 1 �

7

2

�

31x � 12

�

31x � 12 �

11�

6 �

31x � 12 �

7�

6

�

31x � 12

u �11�

6 u �

7�

6

sin u 6 0

�

31x � 12sin u � �0.5

a�

3x �

�

3b �

�

31x � 12

sin a�

3x �

�

3b � �0.5

160 s in a�

3x �

�

3b � 320 � 240

f 1x2 � 240f 1x2 6 240,

x � 30, 2�2f 1x2 � 160 sin a�

3x �

�

3b � 320

��

x � �5.7543, �4.0094, �3.1416TRACE2nd

ZOOM

Y1 � Y2.Y=


Precalculus—


we can solve trig equations

using graphing technology �4

4

�2� 2�

Figure 7.54

cob19537_ch07_722-732.qxd 1/27/11 12:18 PM Page 725

Graphical Solution � With the calculator in radian , enter and

on the screen. To set an appropriate window, note the amplitude ofY1 is 160 and that the graph has been vertically shifted up 320 units. With the x-valuesranging from 0 to , the (CALC) 5:intersect feature determines (2.5,240) is a point of intersection of the two graphs (see Figure 7.55). In Figure 7.56,we find a second point of intersection is (4.5, 240). Observing that the graph of Y1

falls below 240 between these two points, we conclude that in the interval [0, ),for (2.5, 4.5).


There is a mixed variety of equation types in Exercises 39 through 48.

E. Applications Using Trigonometric Equations

Using characteristics of the trig functions, we can often generalize and extend many ofthe formulas that are familiar to you. For example, the formulas for the volume of aright circular cylinder and a right circular cone are well known, but what about the vol-ume of a nonright figure (see Figure 7.57)? Here, trigonometry provides the answer, asthe most general volume formula is where V0 is a “standard” volume for-mula and is the complement of angle of deflection (see Exercises 51 and 52).

As for other applications, consider the following from the environmental sciences.Natural scientists are very interested in the discharge rate of major rivers, as this givesan indication of rainfall over the inland area served by the river. In addition, the dis-charge rate has a large impact on the freshwater and saltwater zones found at the river’sestuary (where it empties into the sea).

EXAMPLE 6 � Solving an Equation Modeling the Discharge Rate of a RiverFor May through November, the discharge rate of the Ganges River (Bangladesh)

can be modeled by where represents

May 1, and D(t) is the discharge rate in m3/sec.Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

a. What is the discharge rate in mid-October?b. For what months (within this interval) is the discharge rate over

26,050 m3/sec?

t � 1D1t2 � 16,580 sin a�

3t �

2�

3b � 17,760

�V � V0 sin �,

0

600

0 2�

Figure 7.55

0

600

0 2�

Figure 7.56

x �f 1x2 6 2402�

TRACE2nd2�

Y=Y2 � 240

sin a�

3X �

�

3b � 320Y1 � 160MODE

Figure 7.57


Precalculus—

�

�

h


we can solve trig equations of

the form A sin(Bx � C) � D � k

cob19537_ch07_722-732.qxd 1/26/11 5:55 PM Page 726

Solution � a. To find the discharge rate in mid-October we simply evaluate the function at

given function

substitute 6.5 for t

compute result on a calculator

In mid-October the discharge rate is 1180 m3/sec.b. We first find when the rate is equal to 26,050 m3/sec:

substitute 26,050 for D(t)

subtract 17,760; divide by 16,580

Using a u-substitution for we obtain the equation

in QI and QII

solutions in u

To complete the solution we re-substitute for u and solve.

re-substitute for u

multiply both sides by

solutions

The Ganges River will have a flow rate of over 26,050 m3/sec between mid-June (2.5) and mid-August (4.5).


To obtain a graphical solution to Example 6(b),

enter on

the screen, then . Set the windowas shown in Figure 7.58 and locate the points ofintersection. The graphs verify that in theinterval [1, 8], for (2.5, 4.5).

There is a variety of additional exercises in theExercise Set. See Exercises 57 through 64.

t � D1t2 7 26,050

Y2 � 26,050Y=

sin a�

3X �

2�

3b � 17,760Y1 � 16,580

t � 4.5 t � 2.5

3

� t � 2 � 2.5 t � 2 � 0.5

�

31t � 22�

3 1t � 22 �

5�

6

�

3 1t � 22 �

�

6

�

3t �

2�

3�

�

31t � 22

u �5�

6u �

�

6

sin u 7 00.5 � sin u

a�

3t �

2�

3b

0.5 � sin a�

3t �

2�

3b

26,050 � 16,580 sin a�

3t �

2�

3b � 17,760

D1t2 � 26,050.

� 1180

D16.52 � 16,580 sin c�316.52 �

2�

3d � 17,760

D1t2 � 16,580 sin a�

3t �

2�

3b � 17,760

t � 6.5:


Precalculus—

E. You’ve just seen how

we can use a combination of

skills to model and solve a

variety of applications

0 8

0

40,000

Figure 7.58

cob19537_ch07_722-732.qxd 1/27/11 12:19 PM Page 727


Precalculus—

2. One strategy to solve equations with four terms andno common factors is by .



7.7 EXERCISES

1. The three Pythagorean identities are ,, and .

4. To combine two sine or cosine terms with differentarguments, we can use the to formulas.

3. When an equation contains two functions from aPythagorean identity, sometimes bothsides will lead to a solution.

6. Regarding Example 6, discuss/explain how todetermine the months of the year the discharge rateis under 26,050 m3/sec, using the solution setgiven.

5. Regarding Example 4, discuss/explain the

relationship between the line and the

graph shown in Figure 7.52.

y �3

5x � 2

Solve each equation in [0, 2 ) using the methodindicated. Round nonstandard values to four decimalplaces.

• Squaring both sides

7. 8.

9. 10.

11. 12.

• Factor by grouping

13.

14.

15.

16.

• Using identities

17. 18.

19.

20. 3 cos12x2 � cos x � 1 � 0

3 cos12x2 � 7 sin x � 5 � 0

1 � tan2x

tan2x�

4

3

1 � cot2x

cot2x� 2

413 sin2x sec x � 13 sec x � 2 � 8 sin2x

3 tan2x cos x � 3 cos x � 2 � 2 tan2x

4 sin x cos x � 213 sin x � 2 cos x � 13 � 0

cot x csc x � 2 cot x � csc x � 2 � 0

sec x � tan x � 2cos x � sin x �4

3

sin x � cos x � 12tan x � sec x � �1

cot x � csc x � 13sin x � cos x �16

2

�21.

22.

23.

24.

25.

26.

Find all roots in [ ) using a graphing calculator.State answers in radians rounded to four decimalplaces.

27. 28.

29. 30.

31. 32.

33.

34. 4 sin x � 2 cos2 ax

2b

11 � sin x22 � cos 12x2 � 4 cos x11 � sin x2cos12x2 � x2 � �5x2 � sin12x2 � 1

sin212x2 � 2x � 1cos212x2 � x � 3

3 sin x � x � 45 cos x � x � 3

0, 2�

tan4x � 2 sec2x tan2x � sec4x � cot2x

sec4x � 2 sec2x tan2x � tan4x � tan2x

cos17x2sin14x2 � cos x � 0sin17x2cos14x2 � sin15x2 �

sin15x2sin12x2 � 1 � 0cos13x2 � cos15x2cos12x2 �

2 cos2 ax

3b � 3 sin ax

3b � 3 � 0

2 sin2 ax

2b � 3 cos ax

2b � 0


cob19537_ch07_722-732.qxd 1/26/11 5:56 PM Page 728

State the period P of each function and find all solutionsin [0, P). Round to four decimal places as needed.

35.

36.

37.

38.

Solve each equation in [ ) using any appropriatemethod. Round nonstandard values to four decimalplaces.

39.

40. 5 sec2x � 2 tan x � 8 � 0

cos x � sin x �12

2

0, 2�

�0.075 sin a�

2x �

�

3b � 0.023 � �0.068

1235 cos a �

12x �

�

4b � 772 � 1750

�7512 sec a�

4x �

�

6b � 150 � 0

250 sin a�

6x �

�

3b � 125 � 0

41.

42.

43.

44.

45.

46.

47.

48. 2 tan a�

2� xb sin2x �

13

2

sec2x tan a�

2� xb � 4

sin a�

2� xb csc x � 13

sec x cos a�

2� xb � �1

1 � sin2x

cot2x�12

2

csc x � cot x � 1

5 csc2x � 5 cot x � 5 � 0

1 � cos2x

tan2x�13

2


Precalculus—

49. The equation of a line in trigonometric form:

The trigonometric form of a linear equation is given by the formula shown, where D is the perpendicular distance from the origin to the line and is the angle between theperpendicular segment and the x-axis. For each pair of perpendicular lines given, (a) find

the point (a, b) of their intersection; (b) compute the distance and the

angle , and give the equation of the line L1 in trigonometric form; and (c)

use the or the TABLE feature of a graphing calculator to verify that both equationsname the same line.

I. II. III.

50. Rewriting as a single function:

Linear terms of sine and cosine can be rewritten as a single function using the formula shown, where

and . Rewrite the equations given using these relationships and verify they are

equivalent using the or the TABLE feature of a graphing calculator:

a.b.

The ability to rewrite a trigonometric equation in simpler form has a tremendous number of applications ingraphing, equation solving, working with identities, and solving applications.

y � 4 cos x � 3 sin x

y � 2 cos x � 213 sin x

GRAPH

� � sin�1aa

kbk � 2a2 � b2

y � k sin(x � �)y � a cos x � b sin x

L2: y � 13xL2: y � 2xL2: y � x

L1: y � �13

3x �

413

3L1: y � �

1

2x � 5L1: y � �x � 5

GRAPH

� � tan�1abab

D � 2a2 � b2

�

y �D � x cos �

sin �


D

�

y

x

Exercise 49

cob19537_ch07_722-732.qxd 1/26/11 5:56 PM Page 729


Precalculus—

51. Volume of a cylinder: Thevolume of a cylinder is given bythe formula wherer is the radius and h is the heightof the cylinder, and is theindicated complement of the angleof deflection . Note that when

the formula becomes that

of a right circular cylinder (if then h is called

the slant height or lateral height of the cylinder).An old farm silo is built in the form of a rightcircular cylinder with a radius of 10 ft and a heightof 25 ft. After an earthquake, the silo became tiltedwith an angle of deflection (a) Find thevolume of the silo before the earthquake. (b) Findthe volume of the silo after the earthquake. (c) Whatangle is required to bring the original volume ofthe silo down 2%?

52. Volume of a cone: Thevolume of a cone is given by

the formula ,

where r is the radius and h isthe height of the cone, and is the indicated complementof the angle of deflection .

Note that when the

formula becomes that of a

right circular cone (if

then h is called the slant height or lateral height ofthe cone). As part of a sculpture exhibit, an artist isconstructing three such structures each with aradius of 2 m and a slant height of 3 m. (a) Find thevolume of the sculptures if the angle of deflectionis (b) What angle was used if thevolume of each sculpture is 12 m3?

53. River discharge rate: For June through February,the discharge rate of the La Corcovada River(Venezuela) can be modeled by the function

where t represents

the months of the year with corresponding toJune, and D(t) is the discharge rate in cubic metersper second. (a) What is the discharge rate in mid-September ? (b) For what months of theyear is the discharge rate over 50 m3/sec?Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

1t � 4.52

t � 1

D1t2 � 36 sin a�

4t �

9

4b � 44,

�� 15°.

� ��

2,

� ��

2,

�

�

V �1

3�r2h sin �

�

� � 5°.

� ��

2,

� ��

2,

�

�

V � �r2h sin �,

54. River discharge rate: For February through June,the average monthly discharge of the Point WolfeRiver (Canada) can be modeled by the function

where t represents

the months of the year with corresponding toFebruary, and D(t) is the discharge rate in cubicmeters/second. (a) What is the discharge rate inmid-March ? (b) For what months of theyear is the discharge rate less than 7.5 m3/sec?Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

55. Seasonal sales: Hank’s Heating Oil is a veryseasonal enterprise, with sales in the winter farexceeding sales in the summer. Monthly sales forthe company can be modeled by

where S(x) is

the average sales in month x(a) What is the average sales amount for July? (b) Forwhat months of the year are sales less than $4000?

56. Seasonal income: As a roofing company employee,Mark’s income fluctuates with the seasons and theavailability of work. For the past several years hisaverage monthly income could be approximated by

the function

where I(m) represents income in month(a) What is Mark’s average

monthly income in October? (b) For what months ofthe year is his average monthly income over $4500?

57. Seasonal ice thickness: The average thickness ofthe ice covering an arctic lake can be modeled by

the function where

T(x) is the average thickness in month(a) How thick is the ice in

mid-March? (b) For what months of the year is theice at most 10.5 in. thick?

58. Seasonal temperatures: The function

models the average monthlytemperature of the water in amountain stream, where T(x)is the temperature of the water in month x

. (a) Whatis the temperature of the waterin October? (b) What two months are most likely togive a temperature reading of (c) For whatmonths of the year is the temperature below 50°F?

62°F?

1x � 1 S January21°F2

T1x2 � 19 sin a�

6x �

�

2b � 53

x 1x � 1 S January2.T1x2 � 9 cos a�

6xb � 15,

m 1m � 1 S January2.I1m2 � 2100 sin a�

6m �

�

2b � 3520,

1x � 1 S January2.S1x2 � 1600 cos a�

6x �

�

12b � 5100,

1t � 2.52

t � 1

D1t2 � 4.6 sin a�

2t � 3b � 7.4,

� APPLICATIONS

�

�

h

Exercise 51

�

h�

Exercise 52

cob19537_ch07_722-732.qxd 1/26/11 5:57 PM Page 730

59. Coffee sales: Coffee sales fluctuate with theweather, with a great deal more coffee sold in thewinter than in the summer. For Joe’s Diner, assume

the function

models daily coffee sales (for non-leap years),where G(x) is the number of gallons sold and xrepresents the days of the year(a) How many gallons are projected to be sold onMarch 21? (b) For what days of the year are morethan 40 gal of coffee sold?

60. Park attendance: Attendance at a popular state park varies with the weather, with a great deal more visitors coming in during thesummer months. Assume daily attendance at the park can be modeled by the function

(for non-leap

years), where V(x) gives the number of visitors onday . (a) Approximately howmany people visited the park on November 1

(b) For what days of theyear are there more than 900 visitors?

61. Exercise routine: As part of his yearly physical,Manu Tuiosamoa’s heart rate is closely monitoredduring a 12-min, cardiovascular exercise routine.His heart rate in beats per minute (bpm) is modeled

by the function

where x represents the duration of the workout inminutes. (a) What was his resting heart rate?(b) What was his heart rate 5 min into the workout?(c) At what times during the workout was his heartrate over 170 bpm?

B1x2 � 58 cos a�

6x � �b � 126

111 � 30.5 � 335.52?x 1x � 1 S January 12

V1x2 � 437 cos a 2�

365 x � �b � 545

1x � 1 S January 12.

G1x2 � 21 cos a 2�

365x �

�

2b � 29

62. Exercise routine: As part ofher workout routine, Sara Leeprograms her treadmill tobegin at a slight initial grade(angle of incline), graduallyincrease to a maximum grade,then gradually decrease backto the original grade. For theduration of her workout, the grade is modeled by

the function where

G(x) is the percent grade x minutes after the workouthas begun. (a) What is the initial grade for herworkout? (b) What is the grade at min? (c) At

how long has she been working out?(d) What is the duration of the treadmill workout?

Geometry applications: Solve Exercises 63 and 64graphically using a calculator. For Exercise 63, give inradians rounded to four decimal places. For Exercise 64,answer in degrees to the nearest tenth of a degree.

63. The area of a circularsegment (the shaded portionshown) is given by the

formula

where is in radians. If thecircle has a radius of 10 cm,find the angle that gives an area of 12 cm2.

64. The perimeter of a trapezoidwith parallel sides B and b,altitude h, and base angles and is given by the formula

If , and find the angle that gives a perimeter of 105 m.�

� � 45°,b � 30 m, B � 40 m, h � 10 mP � B � b � h1csc � � csc �2.

��

�

�

A �1

2 r21� � sin �2,

�

G1x2 � 4.9%,x � 4

G1x2 � 3 cos a�

5x � �b � 4,


Precalculus—

65. As we saw in Chapter 6, cosine is the cofunction ofsine and each can be expressed in terms of the other:

and .

This implies that either function can be used tomodel the phenomenon described in this section byadjusting the phase shift. By experimentation,(a) find a model using cosine that will produceresults identical to the sine function in Exercise 58and (b) find a model using sine that will produceresults identical to the cosine function in Exercise 59.

66. Use multiple identities to find all real solutions forthe equation given: cos12x2sin x � 0.

sin15x2 � sin12x2cos x �

sin a�

2� �b � cos �cos a�

2� �b � sin �

67. A rectangularparallelepiped with squareends has 12 edges and sixsurfaces. If the sum of alledges is 176 cm and thetotal surface area is 1288 cm2, find (a) thelength of the diagonal ofthe parallelepiped (shownin bold) and (b) the anglethe diagonal makes withthe base (two answers are possible).


Exercise 67

�

r r�

b

B

h

� �

Exercise 64

cob19537_ch07_722-732.qxd 1/26/11 5:57 PM Page 731


Precalculus—

68. (6.7) Find the values of all six trig functions of anangle, given is on its terminal side.

69. (4.3) Sketch the graph of f by locating its zeroesand using end-behavior:

70. (5.3) Use a calculator and the change-of-baseformula to find the value of log5279.

71. (6.6) The Willis Tower (formerly known as theSears Tower) in Chicago, Illinois, remains one ofthe tallest structures in the world. The top of theroof reaches 1450 ft above the street below and the

f 1x2 � x4 � 3x3 � 4x.

P1�51, 682 antenna extends anadditional 280 ftinto the air. Find theviewing angle forthe antenna from adistance of 1000 ft(the angle formedfrom the base of theantenna to its top).

�


280 ft

1450 ft

1000 ft

�

Exercise 71

MAKING CONNECTIONS

Making Connections: Graphically , Symbolically , Numerically, and V erballyEight graphs A through H are given. Match the characteristics shown in 1 through 16 to one of the eight graphs.

�4

4

2��2�

y

�4

4

4�4

y

�2

2

2��2�

y

�4

4

2��2�

y

1. ____

2. ____

3. ____

4. ____

5. ____

6. ____ sin(2t)

7. ____

8. ____ is on the grapha�

6, 13

2b

x � cos y

sec t � 2

y � arcsin t

f 1t2 � sin t, t � c��

2,

�

2d

tan t � �1

f 1t2 � cos t, t � 30, � 4

(a) (b) (c) (d)

�4

4

4�4

y

�4

4

2��2�

y

�4

4

4�4

y

�4

4

4�4

y(e) (f) (g) (h)

9. ____

10. ____

11. ____ 2 sin t cos t

12. ____

13. ____

14. ____

15. ____

16. ____ y � cos2t � sin2t

x � sin y

f a1

2b �

�

3

t � �5�

3, �

�

3,

�

3,

5�

3

t � �5�

4, �

�

4,

3�

4,

7�

4

f 1t2 � cos�1t

f a1

2b �

�

6

cob19537_ch07_722-732.qxd 2/7/11 4:39 PM Page 732

7–81 Summary and Concept Review 733

Precalculus—

SECTION 7.1 Fundamental Identities and Families of Identities

KEY CONCEPTS• The fundamental identities include the reciprocal, ratio, and Pythagorean identities.

• A given identity can algebraically be rewritten to obtain other identities in an identity “family.”

• Standard algebraic skills like distribution, factoring, combining terms, and special products play an important rolein working with identities.

• The pattern gives an efficient method for combining rational terms.

• Using fundamental identities, a given trig function can be expressed in terms of any other trig function.

• Once the value of a given trig function is known, the value of the other five can be uniquely determined usingfundamental identities, if the quadrant of the terminal side is known.

EXERCISESVerify using the method specified and fundamental identities.

1. multiplication 4. combine terms using

2. factoring

3. special product

Find the value of all six trigonometric functions using the information given.

5. in QIII 6. in QIV

SECTION 7.2 More on Verifying Identities

KEY CONCEPTS• The sine and tangent functions are odd functions, while cosine is even.

• The steps used to verify an identity must be reversible.

• If two expressions are equal, one may be substituted for the other and the result will be equivalent.

• To verify an identity we mold, change, substitute, and rewrite one side until we “match” the other side.

• Verifying identities often involves a combination of algebraic skills with the fundamental trig identities. A collection and summary of the Guidelines for Verifying Identities can be found on page 662.

• To show an equation is not an identity, find any one value where the expressions are defined but the equation isfalse, or graph both functions on a calculator to see if the graphs are identical.

EXERCISESVerify that each equation is an identity.

7. 8.

9. 10.1sin x � cos x22

sin x cos x� csc x sec x � 2

sin4x � cos4x

sin x cos x� tan x � cot x

cot xsec x

�csc x

tan x� cot x1cos x � csc x2csc2x11 � cos2x2

tan2x� cot2x

sec � �25

23; �cos � � �

12

37; �

1sec x � tan x2 1sec x � tan x2csc x

� sin x

sec2xcsc x

� sin x �tan2xcsc x

tan2x csc x � csc x

sec2x� csc x

AB

�CD

�AD � BC

BDsin x1csc x � sin x2 � cos2 x

A

B�

C

D�

AD � BC

BD

SUMMAR Y AND CONCEPT REVIEW

cob19537_ch07_733-742.qxd 1/26/11 4:07 PM Page 733

Precalculus—


SECTION 7.3 The Sum and Dif ference Identities

KEY CONCEPTSThe sum and difference identities can be used to

• Find exact values for nonstandard angles that are a sum or difference of two standard angles.

• Verify the cofunction identities and to rewrite a given function in terms of its cofunction.

• Find coterminal angles in for very large angles (the angle reduction formulas).

• Evaluate the difference quotient for sin x, cos x, and tan x.

• Rewrite a sum as a single expression: .

The sum and difference identities for sine and cosine can be remembered by noting

• For , the function repeats and the signs alternate:

• For the signs repeat and the functions alternate:

EXERCISESFind exact values for the following expressions using sum and difference formulas.

11. a. b. 12. a. b.

Evaluate exactly using sum and difference formulas.13. a. b.

Rewrite as a single expression using sum and difference formulas.

14. a. b.

Evaluate exactly using sum and difference formulas, by reducing the angle to an angle in or

15. a. b.

Use a cofunction identity to write an equivalent expression for the one given.

16. a. b.

17. Verify that both expressions yield the same result using sum and difference formulas: and .

18. Use sum and difference formulas to verify the following identity.

SECTION 7.4 The Double-Angle, Half-Angle, and Pr oduct-to-Sum Identities

KEY CONCEPTS• When multiple angle identities (identities involving ) are used to find exact values, the terminal side of must

be determined so the appropriate sign can be used.

• The power reduction identities for cos2x and sin2x are closely related to the double-angle identities, and can bederived directly from and .

• The half-angle identities can be developed from the power reduction identities by using a change of variable andtaking square roots. The sign is then chosen based on the quadrant of the half angle.

• The product-to-sum and sum-to-product identities can be derived using the sum and difference formulas, and haveimportant applications in many areas of science.

cos12x2 � 1 � 2 sin2xcos12x2 � 2 cos2x � 1

�n�

cos ax ��

6b � cos ax �

�

6b � 13 cos x

tan 15° � tan1135° � 120°2 tan 15° � tan145° � 30°2sin ax �

�

12bcos ax

8b

sin a57�

4bcos 1170°

30, 2�2.30, 360°2sin ax

4b cos a3x

8b � cos ax

4b sin a3x

8bcos 13x2 cos 1�2x2 � sin 13x2 sin 1�2x2

sin 139° cos 19° � cos 139° sin 19°cos 109° cos 71° � sin 109° sin 71°

sin a� �

12btan 15°tana �

12bcos 75°

sin1� � �2 � sin � cos � � cos � sin �sin1� � �2cos1� � �2 � cos � cos � � sin � sin �cos1� � �2

cos � cos � � sin � sin � � cos1� � �230°, 360°2

cob19537_ch07_733-742.qxd 1/26/11 4:07 PM Page 734

7–83 Summary and Concept Review 735

Precalculus—

EXERCISESFind exact values for , and using the information given.

19. a. in QIV b. in QIII

Find exact values for , and using the information given.

20. a. in QII b. in QII

Find exact values using the appropriate double-angle identity.

21. a. b.

Find exact values for and using the appropriate half-angle identity.

22. a. b.

Find exact values for and using the given information.

23. a. in QIV b. in QIV

24. Verify the equation is an identity. 25. Solve using a sum-to-product formula.

26. The area of an isosceles triangle (two equal sides) is given by the formula where the equal

sides have length x and the vertex angle measures (a) Use this formula and the half-angle identities to find thearea of an isosceles triangle with vertex angle and equal sides of 12 cm. (b) Use substitution and a

double-angle identity to verify that then recompute the triangle’s area. Do the

results match?

SECTION 7.5 The Inverse T rig Functions and Their Applications

KEY CONCEPTS• In order to create one-to-one functions, the domains of , and are restricted as follows:

(a) ; (b) ; and (c) .

• For the inverse function is given implicitly as and explicitly as or

• The expression is read, “y is the angle or real number whose sine is x.” The other inverse functions aresimilarly read/understood.



• The domains of and are likewise restricted to create one-to-one functions:

(a) (b) and (c)

• In some applications, inverse functions occur in a composition with other trig functions, with the expression bestevaluated by drawing a diagram using the ratio definition of the trig functions.

y � cot t, t � 10, �2.y � csc t, t � c��

2, 0b ´ a0,

�

2d ;y � sec t; t � c0,

�

2b ´ a�

2, � d ;

y � cot ty � sec t, y � csc t,y � arctan x.y � tan�1xx � tan yy � tan x,y � arccos x.y � cos�1xx � cos yy � cos x,

y � sin�1xy � arcsin x.y � sin�1xx � sin yy � sin x,

y � tan t; t � a��

2,

�

2by � cos t, t � 30, � 4y � sin t, t � c��

2,

�

2d

y � tan ty � sin t, y � cos t

x2sin a�

2b cos a�

2b �

1

2x2sin �,

� � 30°�°.

A � x2sin a�

2b cos a�

2b,

cos13x2 � cos x � 0cos13�2 � cos �

cos13�2 � cos ��

2 tan2�

sec2� � 2

csc � � �65

33; �90° 6 � 6 0; �cos � �

24

25; 0° 6 � 6 360°; �

cos a�

2bsin a�

2b

� �5�

8� � 67.5°

cos �sin �

1 � 2 sin2a �

12bcos222.5° � sin222.5°

sin12�2 � �336

625; �cos12�2 � �

41

841; �

tan �sin �, cos �

csc � � �29

20; �cos � �

13

85; �

tan12�2sin12�2, cos12�2

cob19537_ch07_733-742.qxd 1/26/11 4:08 PM Page 735

• To evaluate we use for use and so on.

• Trigonometric substitutions can be used to simplify certain algebraic expressions.

EXERCISESEvaluate without the aid of calculators or tables. State answers in both radians and degrees in exact form.

27. 28. 29.

Evaluate the following using a calculator. Answer in radians to the nearest ten-thousandth and in degrees to the nearesttenth. Some may be undefined.

30. 31. 32.

Evaluate the following without the aid of a calculator, keeping the domain and range of each function in mind. Somemay be undefined.

33. 34. 35.

Evaluate the following using a calculator. Some may be undefined.

36. 37. 38.

Evaluate each expression by drawing a right triangle and labeling the sides.

39. 40. 41.

Use an inverse function to solve the following equations for in terms of x.

42. 43. 44.

SECTION 7.6 Solving Basic T rig Equations

KEY CONCEPTS• When solving trig equations, we often consider either the principal root, roots in or all real roots.

• Keeping the graph of each function in mind helps to determine the desired solution set.

• After isolating the trigonometric term containing the variable, we solve by applying the appropriate inversefunction, realizing the result is only the principal root.

• Once the principal root is found, roots in or all real roots can be found using reference angles and theperiod of the function under consideration.

• Trig identities can be used to obtain an equation that can be solved by factoring or other solution methods.

EXERCISESSolve each equation without the aid of a calculator (all solutions are standard values). Clearly state (a) the principalroot; (b) all solutions in the interval ; and (c) all real roots.45. 46. 47.

Solve using a calculator and the inverse trig functions (not by graphing). Clearly state (a) the principal root;(b) solutions in and (c) all real roots. Answer in radians to the nearest ten-thousandth as needed.

48. 49. 50. 12 csc x � 3 � 72

5 sin12�2 �

1

49 cos x � 4

30, 2�2;

8 tan x � 713 � �133 sec x � �62 sin x � 1230, 2�2

30, 2�2

30, 2�2,

x � 4 sin a� ��

6b713 sec � � xx � 5 cos �

�

cot c sin�1a x

281 � x2b dtan c arcseca 7

3xb dsin c cos�1a12

37b d

cot�1 c cot a11�

4b darccos 3cos1�60°2 4sin�11sin 1.02452

cos1cos�122arcsec c sec a�

4b dc sin c sin�1a1

2b d

f 1x2 � arccos a7

8by � sin�10.8892y � tan�14.3165

y � arccos a�13

2by � csc�12y � sin�1a12

2b

tan�1a1

tb;y � cot�1t,y � cos�1a1

tb;y � sec�1t,


Precalculus—

cob19537_ch07_733-742.qxd 1/26/11 4:08 PM Page 736

SECTION 7.7 General Trig Equations and Applications

KEY CONCEPTS• In addition to the basic solution methods from Section 7.6, additional strategies include squaring both sides,

factoring by grouping, and using the full range of identities to simplify an equation.

• Many applications result in equations of the form . To solve, isolate the factor (subtract D and divide by A), then apply the inverse function.

• Once the principal root is found, roots in or all real roots can be found using reference angles and theperiod of the function under consideration.

EXERCISESFind solutions in using the method indicated. Round nonstandard values to four decimal places.51. squaring both sides 52. using identities

53. factor by grouping 54. using any appropriate method

State the period P of each function and find all solutions in [0, P). Round to four decimal places as needed.

55. 56.

57. The revenue earned by Waipahu Joe’s Tanning Lotions fluctuates with the seasons, with a great deal more lotion

sold in the summer than in the winter. The function models the monthly sales of

lotion nationwide, where R(x) is the revenue in thousands of dollars and x represents the months of the year( ). (a) How much revenue is projected for July? (b) For what months of the year does revenue exceed $37,000?

58. The area of a circular segment (the shaded portion shown in the diagram) is given by the

formula ,where is in radians. If the circle has a radius of 10 cm, find

the angle that gives an area of 12 cm2.�

�A �1

2r21� � sin �2

x � 1 S Jan

R1x2 � 15 sin a�

6x �

�

2b � 30

80 cos a�

3x �

�

4b � 4012 � 0�750 sin a�

6x �

�

2b � 120 � 0

csc x � cot x � 14 sin x cos x � 213 sin x � 2 cos x � 13 � 0

3 cos12x2 � 7 sin x � 5 � 0sin x � cos x �16

2

30, 2�2

30, 2�2sin1Bx � C2Asin1Bx � C2 � D � k

7–85 Practice Test 737

Precalculus—

r r�

PRACTICE TEST

Verify each identity using fundamental identities andthe method specified.

1. special products

2. factoring

3. Find the value of all six trigonometric functions

given in QIVcos � �48

73; �

sin3x � cos3x

1 � cos x sin x� sin x � cos x

1csc x � cot x2 1csc x � cot x2sec x

� cos x

4. Find the exact value of using a sum ordifference formula.

5. Rewrite as a single expression and evaluate:

6. Evaluate exactly using an angle reductionformula.

7. Use sum and difference formulas to verify

sin ax ��

4b � sin ax �

�

4b � 12 cos x.

cos 1935°

cos 81° cos 36° � sin 81° sin 36°

tan 15°

cob19537_ch07_733-742.qxd 1/26/11 4:08 PM Page 737

8. Find exact values for , and given

in QI.

9. Use a double-angle identity to evaluate.

10. Find exact values for and given

in QI.

11. The area of a triangle isgiven geometrically as

. The

trigonometric formula for

the triangle’s area is where is the

angle formed by the sides b and c. In a certaintriangle, and Use theformula for A given here and a half-angle identity tofind the area of the triangle in exact form.

12. The equation can be writtenin an alternative form that makes it easier to graph.This is done by eliminating the mixed xy-term using

the relation to find We can then

find values for and , which are used in aconversion formula. Find and for

assuming in QI.

13. Evaluate without the aid of calculators or tables.

a. b.

c.

14. Evaluate the following. Use a calculator for part(a), give exact answers for part (b), and find thevalue of the expression in part (c) without using acalculator. Some may be undefined.a. b.

c.

Evaluate the expressions by drawing a right triangleand labeling the sides.

15.

16.

17. Solve without the aid of a calculator (all solutionsare standard values). Clearly state (a) the principalroot, (b) all solutions in the interval and(c) all real roots.I. II. 13 sec x � 2 � 48 cos x � �412

30, 2�2,

cot c cos�1a x

225 � x2b d

cos c tan�1a56

33b d

y � sec�1 c seca7�

24b d

y � arctan1tan 78.5°2y � sin�10.7528

y � arccos1cos 30°2y � sin c sin�1a1

2b dy � tan�1a 1

13b

2�17x2 � 513xy � 2y2 � 0,cos �sin �

cos �sin �

�.tan12�2 �B

A � C

Ax2 � Bxy � Cy2 � 0

� � 22.5°.b � 8, c � 10,

�A �1

2bc sin �,

A �1

2 base # height

tan � �12

35; �

cos a�

2bsin a�

2b

2 cos275° � 1

cos12�2 � �161

289; �

tan �sin �, cos � 18. Solve each equation using a calculator and inversetrig functions to find the principal root (not bygraphing). Then state (a) the principal root, (b) allsolutions in the interval , and (c) all real roots.

I. II.

19. Use a graphing calculator to solve the equations inthe indicated interval. State answers in radiansrounded to the nearest ten-thousandth.a.b.

20. Solve the following equations for usinga combination of identities and/or factoring. Statesolutions in radians using the exact form wherepossible.a.

b.

Solve each equation in [ ) by squaring both sides,factoring, using identities, or by using any appropriatemethod. Round nonstandard values to four decimalplaces.

21.

22.

23. The revenue for Otake’s Mower Repair is veryseasonal, with business in the summer months farexceeding business in the winter months. Monthlyrevenue for the company can be modeled by the

function where

R(x) is the average revenue (in thousands of dollars)for month (a) What is the averagerevenue for September? (b) For what months of theyear is revenue at least $12,500?

24. The lowest temperature onrecord for the even monthsof the year are given in thetable for the city of Denver,Colorado. The equation

is a fairlyaccurate model for thisdata. Use the equation toestimate the record lowtemperatures for the oddnumbered months.Source: 2004 Statistical Abstract of the United States,Table 379.

25. Write the product as a sum using a product-to-sumidentity: 2 cos11979�t2cos1439�t2.

2.5892 � 6sin10.576x �y � 35.223

x 1x � 1 S Jan2.R1x2 � 7.5 cos a�

6x �

4�

3b � 12.5,

2

3 sin a2x �

�

6b �

3

2�

5

6

3 sin 12x2 � cos x � 0

0, 2�

1cos x � sin x22 �1

2

2 sin x sin12x2 � sin12x2 � 0

x � 30, 2�221x � 1 � 3 cos2 x; x � 30, 2�23 cos12x � 12 � sin x; x � 3��, � 4

�3 cos12x2 � 0.8 � 02

3 sin12x2 �

1

4

30, 2�2


Precalculus—

c

ab

�

Month Low ( ) Temp. ( )

2

4

6 30

8 41

10 3

12 �25

�2

�30

�FJan S 1

cob19537_ch07_733-742.qxd 1/26/11 4:08 PM Page 738

Precalculus—

7–87 Strengthening Core Skills 739

Seeing the Beats as the Beats Go OnWhen two sound waves of slightly different frequencies are combined, the resultant wave varies periodically in amplitudeover time. These amplitude pulsations are called beats. In this Exploration and Discovery, we’ll look at ways to “see” thebeats more clearly on a graphing calculator, by representing sound waves very simplistically as and

and noting a relationship between m, n, and the number of beats in Using a sum-to-productformula, we can represent the resultant wave as a single term. For and the result is

The window used and resulting graph areshown in Figures 7.59 and 7.60, and it appearsthat “silence” occurs four times in this interval—where the graph of the combined waves is tangentto (bounces off of) the x-axis. This indicates atotal of four beats. Note the number of beats isequal to the difference Further experimentation will show this is not acoincidence, and this enables us to construct twoadditional functions that will frame thesepulsations and make them easier to see. Since the maximum amplitude of the resulting wave is 2, we use functions of the form

to construct the frame, where k is

the number of beats in the interval For and we have

and the functions we use will be

and as shown in Figure 7.61. The result is shown inFigure 7.62, where the frame clearly shows the four beats or more precisely, the four moments of silence.

For each exercise, (a) express the sum as a product, (b) graph YR on a graphing calculator for

and identify the number of beats in this interval, and (c) determine what value of k in would be used to

frame the resultant YR, then enter these as Y4 and Y5 to check the result.

Exercise 1: Exercise 2:

Exercise 3: Exercise 4: Y1 � cos111X2; Y2 � cos110X2Y1 � cos114X2; Y2 � cos16X2Y1 � cos112X2; Y2 � cos19X2Y1 � cos114X2; Y2 � cos18X2

�2 cos ak

2xb

x � 30, 2� 4Y1 � Y2

Y5 � �2 cos12X2Y4 � 2 cos12X2k

2�

12 � 8

2� 2

Y2 � cos18X2,Y1 � cos112X2 1m � n � k2.�2 cos ak

2xb

m � n: 12 � 8 � 4.

� 2 cos110X2cos12X2Y3 � cos112X2 � cos18X2 � 2 cos a12X � 8X

2b cos a12X � 8X

2b

Y2 � cos18X2Y1 � cos112X2 30, 2� 4 .Y2 � cos1nX2 Y1 � cos1mX2

CALCULATOR EXPLORA TION AND DISCOVER Y

Figure 7.59

�3

3

2�0

Figure 7.60

Figure 7.61

�3

3

2�0

Figure 7.62

Trigonometric Equations and InequalitiesThe ability to draw a quick graph of the trigonometric functions is a tremendous help in understanding equations andinequalities. A basic sketch can help reveal the number of solutions in and the quadrant of each solution. Fornonstandard angles, the value given by the inverse function can then be used as a basis for stating the solution set forall real numbers. We’ll illustrate the process using a few simple examples, then generalize our observations to solve

30, 2�2

STRENGTHENING CORE SKILLS

cob19537_ch07_733-742.qxd 1/27/11 12:21 PM Page 739

more realistic applications. Consider the function a sine wave withamplitude 2, and a vertical translation of To find intervals in where wereason that f has a maximum of and a minimum of since

With no phase shift and a standard period of we can easily draw a quicksketch of f by vertically translating x-intercepts and max/min points 1 unit up. After drawing theline (see Figure 7.63), it appears there are two intersections in the interval, one in QI andone in QII. More importantly, it is clear that between these two solutions. Substituting2.5 for in we solve for sin x to obtain which we use to statethe solution in exact form: for In approximate form the solution interval is

If the function involves a horizontal shift, the graphical analysis will reveal which intervals should bechosen to satisfy the given inequality.

The basic ideas remain the same regardless of the complexity of the equation, and we illustrate by studying the

function and the inequality Remember—our current goal is not a

supremely accurate graph, just a sketch that will guide us to the solution using the inverse functions and the correct

quadrants. Perhaps that greatest challenge is recalling that when the horizontal shift is but other than this a

fairly accurate sketch can quickly be obtained.

Illustration 1 � Given find intervals in where

Solution � This is a sine wave with a period of 365 (days), an amplitude of 750, shifted units to the right

and 950 units up. The maximum value will be 1700 and the minimum value will be 200. Forconvenience, scale the axes from 0 to 360 (as though the period were 360 days), and plot the x-intercepts and maximum/minimum values for a standard sine wave with amplitude 750 (byscaling the axes). Then shift these points about 90 units in the positive direction (to the right), and950 units up, again using a scale that makes this convenient (see Figure 7.64). This sketch alongwith the graph of is sufficient to show that solutions to occur early in thesecond quarter and late in the third quarter, with solutions to occurring between

these solutions. For we substitute 1250 for and isolate the sine function,

obtaining which leads to exact form solutions of and

. In approximate form the solution interval is

Practice with these ideas by finding solutions to the following inequalities in the intervals specified.

Exercise 1:

Exercise 2:

Exercise 3:

Exercise 4: f 1x2 � 15,750 sin a 2�

360x �

�

4b � 19,250; f 1x2 7 25,250; x � 30, 3602

h 1x2 � 125 sin a�

6x �

�

2b � 175; h 1x2 � 150; x � 30, 122

x � 30, 2�2g1x2 � 4 sin ax ��

3b � 1; g1x2 � �2;

x � 30, 2�2f 1x2 � 3 sin x � 2; f 1x2 7 3.7;

x � 3115, 250 4 .d � a� � sin�10.4 ��

2b a365

2�b

d � asin�10.4 ��

2b a365

2�bsin a 2�

365 d �

�

2b � 0.4,

R 1d2R 1d2 � 750 sina 2�

365 d �

�

2b � 950,

R 1d2 7 1250R 1d2 � 1250y � 1250

�C

B� 91.25

R 1d2 7 1250.30, 365 4R 1d2 � 750 sin a 2�

365 d �

�

2b � 950,

�C

B,B � 1,

R 1d2 7 1250.R 1d2 � 750 sin a 2�

365 d �

�

2b � 950

x � 10.85, 2.292. x � 1sin�1 0.75, � � sin�1 0.752.f 1x2 7 2.5sin x � 0.75,f 1x2 � 2 sin x � 1,f 1x2 f 1x2 7 2.5

y � 2.5

2�,�1 � sin x � 1.21�12 � 1 � �1,2112 � 1 � 3

f 1x2 7 2.5,30, 2�2�1.f 1x2 � 2 sin x � 1,


Precalculus—

y � 2.5

f (x)

2��

�3

3

y

x

Figure 7.63

R(d)

360180�500

2000

1000

y

d

Figure 7.64

cob19537_ch07_733-742.qxd 1/26/11 4:08 PM Page 740

7–89 Cumulative Review Chapters 1–7 741

Precalculus—

1. Find for all six trigfunctions, given

is on theterminal side with inQII.

2. Find the lengths of themissing sides.

3. Verify that is a zero of

4. Determine the domain of Answerin interval notation.

5. Standing 5 mi (26,400 ft) from the base of MountLogan (Yukon) the angle of elevation to the summitis . How much taller is Mount McKinley(Alaska) which stands at 20,320 ft high?

6. Use the Guidelines for Graphing PolynomialFunctions to sketch the graph of

7. Use the Guidelines for Graphing Rational Functions

to sketch the graph of

8. The Petronas Towers in Malaysia are two of the talleststructures in the world. The top of the roof reaches1483 ft above the street below and the stainless steelpinnacles extend an additional 241 ft into the air (seefigure). Find the viewing angle for the pinnaclesfrom a distance of 1000 ft (the angle formed from thebase of the antennae to its top).

9. A wheel with radius 45 cm is turning at 5revolutions per second. Find the linear velocity of apoint on the rim in kilometers per hour, rounded tothe nearest hundredth.

�

241 ft

1483 ft

1000 ft

�

h1x2 �x � 1

x2 � 4

f 1x2 � x3 � 3x2 � 4.

36° 56¿

r1x2 � 29 � x2.

g1x2 � x2 � 4x � 1.x � 2 � 13

�P1�13, 842

f 1�2 10. Solve for

11. Solve for x:

12. Earth has a radius of 3960 mi. Tokyo, Japan, islocated at N latitude, very near the Elatitude line. Adelaide, Australia, is at Slatitude, and also very near E latitude. Howmany miles separate the two cities?

13. Since 1970, sulphur dioxide emissions in the UnitedStates have been decreasing at a nearly linear rate. In1970, about 31 million tons were emitted into theatmosphere. In 2000, the amount had decreased toapproximately 16 million tons. (a) Find a linearequation that models sulphur dioxide emissions.(b) Discuss the meaning of the slope ratio in thiscontext. (c) Use the equation model to estimate theemissions in 1985 and 2010.Source: 2004 Statistical Abstract of the United States, Table 360.

14. List the three Pythagorean identities and threeidentities equivalent to

15. For what values

of x in satisfy ?

16. Write as a single logarithmic expression in simplestform:

17. After doing some market research, the manager of asporting goods store finds that when a four-pack ofpremium tennis balls are priced at $9 per pack,20 packs per day are sold. For each decrease of $0.25,1 additional pack per day will be sold. Find the priceat which four-packs of tennis balls should be sold inorder to maximize the store’s revenue on this item.

18. Write the equation of thefunction whose graph isgiven, in terms of a sinefunction.

19. Verify that the following is an

identity:

20. Given the zeroes ofare and

estimate the following:a. the domain of the functionb. intervals where and f 1x2 � 0f 1x2 7 0

x � �12,x � �4f 1x2 � x4 � 18x2 � 32

cos x � 1

tan2x�

cos x

sec x � 1

log1x2 � 92 � log1x � 12 � log1x2 � 2x � 32.

f 1x2 7 330.530, 2�2f 1x2 � 325 cos a�

6x �

�

2b � 168,

cos12�2.

139°34.6°

139°35.4°

�3 ` x �1

2` � 5 �10

x: 21x � 3234 � 1 � 55.

CUMULATIVE REVIEW CHAPTERS 1–7

60

5√3

Exercise 2

y

8�4��8�

�6

6

�4� x

Exercise 18

cob19537_ch07_733-742.qxd 1/26/11 4:08 PM Page 741

21. Use the triangle shown to find the exact value of

22. Use the triangle shown to find the exact value of

23. The amount of waste product released by amanufacturing company varies according to itsproduction schedule, which is much heavier duringthe summer months and lighter in the winter. Wasteproduct amount reaches a maximum of 32.5 tons inthe month of July, and falls to a minimum of21.7 tons in January . (a) Use thisinformation to build a sinusoidal equation thatmodels the amount of waste produced each month.(b) During what months of the year does outputexceed 30 tons?

24. At what interest rate will $2500 grow to $3500 if it’sleft on deposit for 6 yr and interest is compoundedcontinuously?

25. Identify each geometric formula:a. b.

c. d. y �1

2bhy � 2�r

y � LWHy � �r2h

1t � 12

68

51

�

�

sin1� � �2.

√202

11

9

�

sin12�2. Exercises 26 through 30 require the use of a graphingcalculator.

26. Use the and TABLE features of a calculator tosupport the validity of the identity

. Thenconvert all functions to sines and cosines, rewrite theidentity, and verify this new identity using a sum-to-product identity.

27. On a graphing calculator, the graph of

appears to be

continuous. However, Descartes’ rule of signs tellsus the denominator must have at least one real zero.Since the graph does not appear to have any verticalasymptotes, (a) what does this tell you about r(x)?(b) Use the TABLE feature of a calculator with

to help determine the x-coordinate ofthis discontinuity.

28. Solve the following equation graphically. Roundyour answer to two decimal places.

29. Use the intersection-of-graphs method to solve theinequality , for . Answerusing exact values.

30. The range of a certain projectile is modeled by the

function , where is the

angle at launch and is in feet. (a) Rewrite thefunction using a double angle formula, (b) determinethe maximum range of the projectile, and(c) determine the launch angle that results in thismaximum range.

r1�2�cos �r1�2 �

625

4 sin �

x � 30, 2�2sin x cos x

cos�1x � tan x

¢Tbl � 0.1

r1x2 �1.5x2 � 0.9x � 2.4

0.5x3 � 0.8x2 � x � 1.6

sec10.9x2 cos10.8x2 � sec10.9x2 cos12.6x22 cos11.7x2 �

GRAPH


Precalculus—

cob19537_ch07_733-742.qxd 1/26/11 4:09 PM Page 742

Precalculus—

7–91 743

CONNECTIONS TO CALCULUS

In calculus, as in college algebra, we often encounter expressions that are difficult touse in their given form, and so attempt to write the expression in an alternative formmore suitable to the task at hand. Often, we see algebra and trigonometry workingtogether to achieve this goal.

Simplifying Expressions Using a Trigonometric Substitution

For instance, it is difficult to apply certain concepts from calculus to the equation

, and we attempt to rewrite the expression using a trig substitution and a

Pythagorean identity. When doing so, we’re careful to ensure the substitution used rep-resents a one-to-one function, and that it maintains the integrity of the domain.

y �x

29 � x2

EXAMPLE 1 � Simplifying Algebraic Expressions Using Trigonometry

Simplify using the substitution for , then

verify that the result is equivalent to the original function.

Solution � substitute 3 for x

; factor

substitute for

since

result

Check � Using the notation for inverse functions, we can rewrite as a function of x and use a calculator to compare it with the original function. For we

obtain or . Substituting for in gives

. With the calculator in radian

, enter and

on the screen. Using (due to

the domain), the resulting table seems to indicatethat the functions are indeed equivalent (see the figure).


Trigonometric Identities and Equations

While the tools of calculus are very powerful, it is the application of rudimentary con-cepts that makes them work. Earlier, we saw how basic algebra skills were needed to sim-plify expressions that resulted from applications of calculus. Here we illustrate the use ofbasic trigonometric skills combined with basic algebra skills to achieve the same end.

TblStart � �3Y=

Y2 � tan csin�1aX

3bdY1 �

X

29 �X2MODE

c sin�1ax

3b dy � tan

y � tan ��sin�1ax

3b� � sin�1ax

3bx

3� sin �

x � 3 sin �y � tan �

� tan �

��

2� � �

�

229 cos2� � 3 cos � �

3 sin �

3 cos �

1 � sin2�cos2� �3 sin �

29 cos2�

13 sin �22 � 9 sin2��3 sin �

2911 � sin2�2 �3 sin �

29 � 9 sin2�

sin � y �3 sin �

29 � 13 sin �22

��

26 � 6

�

2x � 3 sin �y �

x

29 � x2

cob19537_ch07_743-744.qxd 1/26/11 4:43 PM Page 743

EXAMPLE 2 � Finding Maximum and Minimum ValuesUsing the tools of calculus, it can be shown that the maximum and/or minimum

values of will occur at the zero(s) of the function

Simplify the right-hand side and use the result to find the location of any maximumor minimum values that occur in the interval

Solution � Begin by simplifying the numerator.

distribute

simplify

factor, commute terms

This shows that when or when In the

interval this gives and . The function f has a maximum value

of at , with a minimum value of at .


5�

4�12

�

412

5�

4� �

�

430, 2�2,

� ��

4� �k, k � �.cot � � 1,f 1�2 � 0

�cot � � 1

csc � �1csc2� � 12 � cot � � csc2�

csc �

�csc �1cot2� � cot � � csc2�2

csc2�

��csc3� � csc � cot � � csc � cot2�

csc2�

f 1�2 �csc �1�csc2�2 � 1�csc � cot � � csc � cot2�2

csc2�

30, 2�2.

f 1�2 �csc �1�csc2�2 � 11 � cot�2 1�csc � cot �2

csc2�.

f 1�2 �1 � cot �

csc �

744 Connections to Calculus 7–92

Precalculus—

simplify, substitute for ; resultcot2�csc2� � 1

Connections to Calculus Exercises

For the functions given, (a) use the substitution indicated to find an equivalent function of , (b) rewrite theresulting function in terms of x using an inverse trig function, and (c) use the TABLE feature of a graphingcalculator to verify the two functions are equivalent for .

1. 2.

Rewrite the following expressions using the substitution indicated.

3. 5. 4. 6.

Using the tools of calculus, it can be shown that for each function f(x) given, the zeroes of f (x) give the locationof any maximum and/or minimum values. Find the location of these values in the interval [0, 2 ), using trigidentities as needed to solve f (x) � 0. Verify solutions using a graphing calculator.

7. 9.

8. 10.

f 1x2 �12 � sin x2 1�sin x2 � cos x cos x

12 � sin x22f 1x2 � sin x sec2x � tan x cos x

f 1x2 �cos x

2 � sin x;f 1x2 � sin x tan x;

f 1x2 � 2 sin x1�sin x2 � 2 cos x cos xf 1x2 �

sec x1�sin x2 � 11 � cos x2sec x tan x

sec2x

f 1x2 � 2 sin x cos x;f 1x2 �1 � cos x

sec x;

�

8

1x2 � 4232; x � 2 sec �281 � x2

x; x � 9 sin �

x

29 � x2; x � 3 tan �

x2

216 � x2; x � 4 sin �

x � 12 sin �, � � a��

2,

�

2by �

2144 � x2

x;x � 13 tan �, � � a��

2,

�

2by �

2169 � x2

x;

x � 0

�

cob19537_ch07_743-744.qxd 1/26/11 4:43 PM Page 744

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