Trigonometric Identities and Equationsavijitmaths.weebly.com/.../5/19956057/book1-_trigonometric_identities.pdf · 606 CHAPTER 7 Trigonometric Identities and Equations 7.1 Fundamental

605

7Trigonometric Identitiesand Equations

In 1831 Michael Faraday discovered thatwhen a wire passes by a magnet, a smallelectric current is produced in the wire.Now we generate massive amounts of elec-tricity by simultaneously rotating thousandsof wires near large electromagnets. Becauseelectric current alternates its direction onelectrical wires, it is modeled accurately byeither the sine or the cosine function.

We give many examples of applicationsof the trigonometric functions to electricityand other phenomena in the examples andexercises in this chapter, including a modelof the wattage consumption of a toaster inSection 7.4, Example 5.

7.1 Fundamental Identities

7.2 Verifying Trigonometric Identities

7.3 Sum and Difference Identities

7.4 Double-Angle Identities and Half-Angle Identities

Summary Exercises on VerifyingTrigonometric Identities

7.5 Inverse Circular Functions

7.6 Trigonometric Equations

7.7 Equations Involving InverseTrigonometric Functions

LIALMC07_0321227638.QXP 2/26/04 10:47 AM Page 605

606 CHAPTER 7 Trigonometric Identities and Equations

7.1 Fundamental IdentitiesNegative-Angle Identities ■ Fundamental Identities ■ Using the Fundamental Identities

Figure 1

TEACHING TIP Point out that intrigonometric identities, � can bean angle in degrees, a real num-ber, or a variable.

Negative-Angle Identities As suggested by the circle shown in Figure 1,an angle � having the point on its terminal side has a corresponding angle

with the point on its terminal side. From the definition of sine,

(Section 5.2)

so and are negatives of each other, or

Figure 1 shows an angle � in quadrant II, but the same result holds for � in anyquadrant. Also, by definition,

(Section 5.2)

so

We can use these identities for and to find in terms of

Similar reasoning gives the following identities.

This group of identities is known as the negative-angle or negative-numberidentities.

Fundamental Identities In Chapter 5 we used the definitions of thetrigonometric functions to derive the reciprocal, quotient, and Pythagorean identities. Together with the negative-angle identities, these are called the fundamental identities.

Fundamental Identities

Reciprocal Identities

Quotient Identities

(continued)

tan � �sin �cos �

cot � �cos �sin �

cot � �1

tan �sec � �

1cos �

csc � �1

sin �

csc�� csc �, sec�� sec �, cot�� cot �

tan�� tan �.

tan�� sin��cos��

��sin �

cos ��

sin �

cos �

tan �:tan��cos��sin��

cos�� cos �.

cos�� x

rand cos � �

x

r,

sin�� sin �.

sin �sin��

sin�� y

rand sin � �

y

r,

�x, �y��x, y�

sin(–�) = – = –sin �yr

Ox

y

yx

(x, y)

–y

r

(x, –y)

–�

�

r


7.1 Fundamental Identities 607

TEACHING TIP Encourage studentsto memorize the identities pre-sented in this section as well assubsequent sections. Point outthat numerical values can be usedto help check whether or not anidentity was recalled correctly.

Pythagorean Identities

Negative-Angle Identities

N O T E The most commonly recognized forms of the fundamental identitiesare given above. Throughout this chapter you must also recognize alternativeforms of these identities. For example, two other forms of are

Using the Fundamental Identities One way we use these identities is tofind the values of other trigonometric functions from the value of a given trigono-metric function. Although we could find such values using a right triangle, thisis a good way to practice using the fundamental identities.

EXAMPLE 1 Finding Trigonometric Function Values Given One Value and theQuadrant

If and � is in quadrant II, find each function value.

(a) (b) (c)

Solution

(a) Look for an identity that relates tangent and secant.

Pythagorean identity

Combine terms.

Take the negative square root. (Section 1.4)

Simplify the radical. (Section R.7)

We chose the negative square root since sec � is negative in quadrant II.

��34

3� sec �

��34

9� sec �

34

9� sec2 �

25

9� 1 � sec2 �

tan � � �53 ��

5

3�2

� 1 � sec2 �

tan2 � � 1 � sec2 �

cot��sin �sec �

tan � � �53

sin2 � � 1 � cos2 � and cos2 � � 1 � sin2 �.

sin2 � � cos2 � � 1

csc(��) � �csc � sec(��) � sec � cot(��) � �cot �

sin(��) � �sin � cos(��) � cos � tan(��) � �tan �

sin2 � � cos2 � � 1 tan2 � � 1 � sec2 � 1 � cot2 � � csc2 �

TEACHING TIP Warn students thatthe given information in

Example 1,

does not mean that and Ask them whythese values cannot be correct.

cos � � 3.sin � � �5

tan � � �53

��53

,



(b) Quotient identity

Multiply by cos �.

Reciprocal identity

(c) Reciprocal identity

Negative-angle identity

Simplify. (Section R.5)

Now try Exercises 5, 7, and 9.

C A U T I O N To avoid a common error, when taking the square root, be sure tochoose the sign based on the quadrant of � and the function being evaluated.

Any trigonometric function of a number or angle can be expressed in termsof any other function.

EXAMPLE 2 Expressing One Function in Terms of Another

Express cos x in terms of tan x.

Solution Since sec x is related to both cos x and tan x by identities, start with

Take reciprocals.

Reciprocal identity

Take square roots.

Quotient rule (Section R.7); rewrite.

Rationalize the denominator.(Section R.7)

Choose the � sign or the � sign, depending on the quadrant of x.

Now try Exercise 43.

cos x ��1 � tan2 x

1 � tan2 x

cos x ��1

�1 � tan2 x

�� 1

1 � tan2 x� cos x

1

1 � tan2 x� cos2 x

1

1 � tan2 x�

1

sec2 x

1 � tan2 x � sec2 x.

tan � � �53 ; cot��

1

��53� �

3

5

cot�� 1

�tan �

cot�� 1

tan��

sin � �5�34

34

��3�34

34 ��5

3� � sin �

� 1

sec ��tan � � sin �

cos � tan � � sin �

tan � �sin �

cos �

From part (a),

tan � � �53

1sec � � �

3�34 � �

3�3434 ;



Figure 2Write each fraction with theLCD. (Section R.5)

Concept Check Fill in the blanks.

1. If , then .

2. If , then .

3. If , then .

4. If and , then .

Find sin s. See Example 1.

5. , s in quadrant I 6. , s in quadrant IV

7. , 8. ,

9. , 10.

11. Why is it unnecessary to give the quadrant of s in Exercise 10?

csc s � �8

5tan s � 0sec s �

11

4

sec s � 0tan s � ��7

2tan s � 0cos��s� �

�5

5

cot s � �1

3cos s �

3

4

tan��x� �sin x � .6cos x � .8

cot x �tan x � 1.6

cos��x� �cos x � �.65

tan��x� �tan x � 2.6

1. �2.6 2. �.65 3. .625

4. �.75 5. 6.

7. 8.

9. 10. �5

8�

�105

11

��77

11�

2�5

5

�3�10

10�7

4

7.1 Exercises

9

4

–4

–2� 2�

Y1 = Y2

4

–4

–2� 2�

The graph supports the resultin Example 3. The graphs ofy1 and y2 appear to be iden-tical.

y1 = tan x + cot x

y2 =cos x sin x

1

We can use a graphing calculator to decide whether two functions areidentical. See Figure 2, which supports the identity With anidentity, you should see no difference in the two graphs. ■

All other trigonometric functions can easily be expressed in terms of and/or We often make such substitutions in an expression to simplify it.

EXAMPLE 3 Rewriting an Expression in Terms of Sine and Cosine

Write in terms of and and then simplify theexpression.

Solution

Quotient identities

Add fractions.

Pythagorean identity


C A U T I O N When working with trigonometric expressions and identities, besure to write the argument of the function. For example, we would not write

an argument such as � is necessary in this identity.sin2 � cos2 � 1;

tan � � cot � �1

cos � sin �

�sin2 � � cos2 �

cos � sin �

�sin2 �

cos � sin ��

cos2 �

cos � sin �

tan � � cot � �sin �

cos ��

cos �

sin �

cos �,sin �tan � � cot �

cos �.sin �

sin2 x � cos2 x � 1.



12. �sin x 13. odd 14. cos x15. even 16. �tan x 17. odd18.19.20.

21. ;

; ;

;

22. ; ;

; ;

23. ;

; ;

;

24. ;

; ;

;

25. ; ;

; ;

26. ; ;

; ;

27. ; ;

; ;

csc � � �4�7

7

cot � � �3�7

7tan � � �

�7

3

cos � �3

4sin � � �

�7

4

csc � � �5

4

sec � � �5

3cot � �

3

4

tan � �4

3cos � � �

3

5

csc � �5

3sec � �

5

4tan � �

3

4

cos � �4

5sin � �

3

5

sec � � �5�21

21cot � �

�21

2

tan � �2�21

21cos � � �

�21

5

sin � � �2

5

csc � � ��17sec � ��17

4

cot � � �4cos � �4�17

17

sin � � ��17

17

csc � �5�6

12

sec � � 5cot � ��6

12

tan � � 2�6sin � �2�6

5

csc � �3

2sec � � �

3�5

5

cot � � ��5

2tan � � �

2�5

5

cos � � ��5

3

f ��x� � �f �x�f ��x� � �f �x�f ��x� � f �x�

Relating ConceptsFor individual or collaborative investigation

(Exercises 12–17)

A function is called an even function if for all x in the domain of f. A func-tion is called an odd function if for all x in the domain of f. WorkExercises 12–17 in order, to see the connection between the negative-angle identitiesand even and odd functions.

12. Complete the statement: .

13. Is the function defined by even or odd?





Concept Check For each graph, determine whether or is true.

18. 19.

20.

Find the remaining five trigonometric functions of . See Example 1.

21. , in quadrant II 22. , in quadrant I

23. , in quadrant IV 24. , in quadrant III

25. , 26. ,

27. , 28. , sin � � 0cos � � �1

4sin � � 0sec � �

4

3

cos � � 0sin � � �4

5sin � � 0cot � �

4

3

�csc � � �5

2�tan � � �

1

4

�cos � �1

5�sin � �

2

3

�

4

–4

–2� 2�

4

–4

–2� 2�

4

–4

–2� 2�

f ��x� � �f �x�f ��x� � f �x�

f �x� � tan x

tan��x� �

f �x� � cos x

cos��x� �

f �x� � sin x

sin��x� �

f ��x� � �f �x�f ��x� � f �x�



28. ;

; ;

;

29. B 30. D 31. E 32. C33. A 34. C 35. A 36. E37. D 38. B

41.

42.

43.

44.

45.

46.

47.

48.

49. 50. 1 51.52. 53. 54.55.56. 57.

58. 59.

60. 61.62. tan2 �

cos2 �cot � � tan �

sin2 �

cos �sin2 � cos2 �

cot � � tan �1 � cot �sec � � cos �

tan2 �cos2 �csc2 �cot �cos �

sec x ��1 � sin2 x

1 � sin2 x

csc x ��1 � cos2 x

1 � cos2 x

cot x � ��csc2 x � 1

tan x � ��sec2 x � 1

cot x ��1 � sin2 x

sin x

sin x � ��1 � cos2 x

tan � ��2�2p � 4

p

sin � ��2x � 1

x � 1

csc � �4�15

15sec � � �4

cot � � ��15

15tan � � ��15

sin � ��15

4

99

Concept Check For each expression in Column I, choose the expression fromColumn II that completes an identity.

I II

29. A.

30. B. cot x

31. C.

32. D.

33. E. cos x

Concept Check For each expression in Column I, choose the expression fromColumn II that completes an identity. You may have to rewrite one or both expressions.

I II

34. A.

35. B.

36. C.

37. D.

38. E. tan x

39. A student writes “ .” Comment on this student’s work.

40. Another student makes the following claim: “Since , I should beable to also say that if I take the square root of both sides.”Comment on this student’s statement.

41. Concept Check Suppose that . Find .

42. Concept Check Find if .

Write the first trigonometric function in terms of the second trigonometric function. SeeExample 2.

43. sin x; cos x 44. cot x; sin x 45. tan x; sec x

46. cot x; csc x 47. csc x; cos x 48. sec x; sin x

Write each expression in terms of sine and cosine, and simplify it. See Example 3.

49. 50. 51.

52. 53. 54.

55. 56.

57. 58.

59. 60.

61. 62.1 � tan2 �

1 � cot2 �sin � �csc � � sin ��

�sec � � csc �� cos � � sin ��sec � � cos �

1 � sin2 �

1 � cot2 �

cos2 � � sin2 �

sin � cos �

cos � � sin �

sin ��1 � cos �� 1 � sec ��

�sec � � 1� �sec � � 1�sin2 � �csc2 � � 1�cot2 � �1 � tan2 ��cos � csc �sec � cot � sin �cot � sin �

sec � �p � 4

ptan �

sin �cos � � xx � 1

sin � � cos � � 1sin2 � � cos2 � � 1

1 � cot2 � csc2

cos2 x �

csc2 x � cot2 x � sin2 x1 � sin2 x �

sin��x�sec x

csc x�

1

sec2 xsec2 x � 1 �

sin2 x

cos2 x�tan x cos x �

1 �

sin x

cos xtan2 x � 1 �

sec2 xcos��x� �

tan x �

sin2 x � cos2 xcos x

sin x�



63. 64.

65. ;

66. ;

67.68. It is the negative of .69.70. It is the same function.71. (a) (b)(c)72. identity 73. not an identity74. not an identity 75. identity76. not an identity

5 sin�3x�cos�2x��sin�4x�

cos�4x�sin�2x�

�sin�2x�

�2�2 � 8

9

2�2 � 8

9

�25�6 � 60

12

25�6 � 60

12

�sin �sec2 �

7.2 Verifying Trigonometric IdentitiesVerifying Identities by Working with One Side ■ Verifying Identities by Working with Both Sides

Recall that an identity is an equation that is satisfied for all meaningful replace-ments of the variable. One of the skills required for more advanced work inmathematics, especially in calculus, is the ability to use identities to write ex-pressions in alternative forms. We develop this skill by using the fundamentalidentities to verify that a trigonometric equation is an identity (for those valuesof the variable for which it is defined). Here are some hints to help you getstarted.

63. 64.

65. Concept Check Let . Find all possible values for .

66. Concept Check Let . Find all possible values for .


(Exercises 67–71)

In Chapter 6 we graphed functions defined by

with the assumption that . To see what happens when , work Ex-ercises 67–71 in order.

67. Use a negative-angle identity to write as a function of 2x.

68. How does your answer to Exercise 67 relate to ?

69. Use a negative-angle identity to write as a function of 4x.

70. How does your answer to Exercise 69 relate to ?

71. Use your results from Exercises 67–70 to rewrite the following with a positive valueof b.

(a) (b) (c)

Use a graphing calculator to decide whether each equation is an identity. (Hint: InExercise 76, graph the function of x for a few different values of y (in radians).)

72. 73.

74. 75.

76. cos�x � y� � cos x � cos y

cos 2x � cos2 x � sin2 xsin x � �1 � cos2 x

2 sin s � sin 2scos 2x � 1 � 2 sin2 x

�5 sin��3x�cos��2x�sin��4x�

y � cos�4x�y � cos��4x�

y � sin�2x�y � sin��2x�

b � 0b � 0

y � c � a f �b�x � d�

sin x � cos xsec xcsc x � �3

sec x � tan xsin xcos x � 1

5

tan��sec �

sin2 � � tan2 � � cos2 �


7.2 Verifying Trigonometric Identities 613

TEACHING TIP There is no substi-tute for experience when it comesto verifying identities. Guide stu-dents through several examples,giving hints such as “Apply a recip-rocal identity,” or “Use a differentform of the Pythagorean identity

”sin2 � � cos2 � � 1.

Looking Ahead to CalculusTrigonometric identities are used in

calculus to simplify trigonometric ex-

pressions, determine derivatives of

trigonometric functions, and change

the form of some integrals.

Hints for Verifying Identities

1. Learn the fundamental identities given in the last section. Whenever yousee either side of a fundamental identity, the other side should come tomind. Also, be aware of equivalent forms of the fundamental identities.For example, is an alternative form of the identity

2. Try to rewrite the more complicated side of the equation so that it is iden-tical to the simpler side.

3. It is sometimes helpful to express all trigonometric functions in the equa-tion in terms of sine and cosine and then simplify the result.

4. Usually, any factoring or indicated algebraic operations should be per-formed. For example, the expression can be factoredas The sum or difference of two trigonometric expressions,such as can be added or subtracted in the same way as anyother rational expression.

5. As you select substitutions, keep in mind the side you are not changing,because it represents your goal. For example, to verify the identity

try to think of an identity that relates tan x to cos x. In this case, since and the secant function is the best link

between the two sides.

6. If an expression contains multiplying both numerator and de-nominator by would give which could be replacedwith Similar results for and maybe useful.

C A U T I O N Verifying identities is not the same as solving equations.Techniques used in solving equations, such as adding the same terms to bothsides, or multiplying both sides by the same term, should not be used whenworking with identities since you are starting with a statement (to be verified)that may not be true.

Verifying Identities by Working with One Side To avoid the temptationto use algebraic properties of equations to verify identities, work with only oneside and rewrite it to match the other side, as shown in Examples 1–4.

1 � cos x1 � cos x,1 � sin x,cos2 x.1 � sin2 x,1 � sin x

1 � sin x,

sec2 x � tan2 x � 1,1cos xsec x �

tan2 x � 1 �1

cos2 x,

�cos � � sin �

sin � cos �

1

sin ��

1

cos ��

cos �

sin � cos ��

sin �

sin � cos �

1sin � �

1cos � ,

�sin x � 1�2.sin2 x � 2 sin x � 1

cos2 � � 1.sin2 � �sin2 � � 1 � cos2 �



4

–4

–2� 2�

The screen supports the con-clusion in Example 2.

tan2 x (1 + cot2 x) = 1 – sin2 x

1

4

–4

–2� 2�

The graphs coincide, support-ing the conclusion in Exam-ple 1.

For s = x,cot x + 1 = csc x (cos x + sin x)

EXAMPLE 1 Verifying an Identity (Working with One Side)

Verify that the following equation is an identity.

Solution We use the fundamental identities from Section 7.1 to rewrite oneside of the equation so that it is identical to the other side. Since the right side ismore complicated, we work with it, using the third hint to change all functionsto sine or cosine.

Steps ReasonsRight side of

given equation

Distributive property (Section R.1)

Left side ofgiven equation

The given equation is an identity since the right side equals the left side.




Solution We work with the more complicated left side, as suggested in the sec-ond hint. Again, we use the fundamental identities from Section 7.1.

Distributive property

Since the left side is identical to the right side, the given equation is an identity.


cos2 x � 1 � sin2 x �1

1 � sin2 x

sec2 x � 1cos2 x �

1

cos2 x

tan2 x � 1 � sec2 x � sec2 x

tan2 x 1

tan2 x � 1 � tan2 x � 1

cot2 x � 1tan2 x � tan2 x � tan2 x

1

tan2 x

tan2 x�1 � cot2 x� � tan2 x � tan2 x cot2 x

tan2 x�1 � cot2 x� �1

1 � sin2 x

cos ssin s � cot s; sin s

sin s � 1 � cot s � 1

�cos s

sin s�

sin s

sin s

csc s � 1sin s csc s�cos s � sin s� �

1

sin s�cos s � sin s�

cot s � 1 � csc s�cos s � sin s�



TEACHING TIP Show that the iden-tity in Example 4 can also be veri-fied by multiplying the numeratorand denominator of the left sideby 1 � sin x.



Solution We transform the more complicated left side to match the right side.

(Section R.5)

The third hint about writing all trigonometric functions in terms of sine and cosine was used in the third line of the solution.




Solution We work on the right side, using the last hint in the list given earlierto multiply numerator and denominator on the right by

Multiply by 1. (Section R.1)

(Section R.3)

Lowest terms (Section R.5)


Verifying Identities by Working with Both Sides If both sides of anidentity appear to be equally complex, the identity can be verified by workingindependently on the left side and on the right side, until each side is changedinto some common third result. Each step, on each side, must be reversible.

�cos x

1 � sin x

1 � sin2 x � cos2 x �cos2 x

cos x�1 � sin x�

�x � y� �x � y� � x2 � y2 �1 � sin2 x

cos x�1 � sin x�

1 � sin x

cos x�

�1 � sin x� �1 � sin x�cos x�1 � sin x�

1 � sin x.

cos x

1 � sin x�

1 � sin x

cos x

1cos2 t � sec2 t; 1

sin2 t � csc2 t � sec2 t � csc2 t

�1

cos2 t�

1

sin2 t

tan t � sin tcos t ; cot t � cos t

sin t �sin t

cos t

1

sin t cos t�

cos t

sin t

1

sin t cos t

ab � a

1b � tan t

1

sin t cos t� cot t

1

sin t cos t

a � bc � a

c �bc

tan t � cot t

sin t cos t�

tan t

sin t cos t�

cot t

sin t cos t

tan t � cot t

sin t cos t� sec2 t � csc2 t



(Section R.4)x2 � 2xy � y2 � �x � y�2

With all steps reversible, the procedure is as shown in the margin. The left sideleads to a common third expression, which leads back to the right side. This pro-cedure is just a shortcut for the procedure used in Examples 1–4: one side ischanged into the other side, but by going through an intermediate step.

EXAMPLE 5 Verifying an Identity (Working with Both Sides)


Solution Both sides appear equally complex, so we verify the identity bychanging each side into a common third expression. We work first on the left,multiplying numerator and denominator by cos �.

Multiply by 1.

Distributive property

On the right side of the original equation, begin by factoring.

Factor

Lowest terms

We have shown that

Left side of Common third Right side ofgiven equation expression given equation

verifying that the given equation is an identity.


sec � � tan �

sec � � tan ��

1 � sin �

1 � sin ��

1 � 2 sin � � sin2 �

cos2 �,

�1 � sin �

1 � sin �

1 � sin2 �. ��1 � sin ��2

�1 � sin �� 1 � sin ��

cos2 � � 1 � sin2 � ��1 � sin ��2

1 � sin2 �

1 � 2 sin � � sin2 �

cos2 ��

�1 � sin ��2

cos2 �

�1 � sin �

1 � sin �

tan � � sin �cos � �

1 �sin �

cos � cos �

1 �sin �

cos � cos �

sec � cos � � 1 �1 � tan � cos �

1 � tan � cos �

�sec � cos � � tan � cos �

sec � cos � � tan � cos �

sec � � tan �


�sec � � tan �� cos �

�sec � � tan �� cos �

sec � � tan �


1 � 2 sin � � sin2 �

cos2 �

left � right

common thirdexpression



CL

An Inductor and a Capacitor

Figure 3

C A U T I O N Use the method of Example 5 only if the steps are reversible.

There are usually several ways to verify a given identity. For instance, an-other way to begin verifying the identity in Example 5 is to work on the left asfollows.

Fundamental identities (Section 7.1)

Compare this with the result shown in Example 5 for the right side to see thatthe two sides indeed agree.

EXAMPLE 6 Applying a Pythagorean Identity to Radios

Tuners in radios select a radio station by adjusting the frequency. A tuner maycontain an inductor L and a capacitor C, as illustrated in Figure 3. The energystored in the inductor at time t is given by

and the energy stored in the capacitor is given by

where F is the frequency of the radio station and k is a constant. The total energyE in the circuit is given by

Show that E is a constant function. (Source: Weidner, R. and R. Sells, Ele-mentary Classical Physics, Vol. 2, Allyn & Bacon, 1973.)

Solution

Given equation

Substitute.

Factor. (Section R.4)

Since k is constant, is a constant function.


E�t�

� k

sin2 � � cos2 � � 1 (Here � � 2Ft). � k�1� � k�sin2�2Ft� � cos2�2Ft� � k sin2�2Ft� � k cos2�2Ft�

E�t� � L�t� � C�t�

E�t� � L�t� � C�t�.

C�t� � k cos2�2Ft�,

L�t� � k sin2�2Ft�

�1 � sin �

1 � sin �

�

1 � sin �

cos �

1 � sin �

cos �

sec � � tan �


1

cos ��

sin �

cos �

1

cos ��

sin �

cos �

Add and subtract fractions.(Section R.5)

Simplify the complex fraction.(Section R.5)



Perform each indicated operation and simplify the result.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

Factor each trigonometric expression.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

Each expression simplifies to a constant, a single function, or a power of a function. Use fundamental identities to simplify each expression.

23. 24. 25. sec r cos r

26. cot t tan t 27. 28.

29. 30. 31.

32.

In Exercises 33–68, verify that each trigonometric equation is an identity. SeeExamples 1–5.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

45.

46. tan2 � sin2 � � tan2 � � cos2 � � 1

�1 � cos2 �� 1 � cos2 �� 2 sin2 � � sin4 �

cos �

sin � cot �� 1sin4 � � cos4 � � 2 sin2 � � 1

sin2 �

cos �� sec � � cos �

cos �

sec ��

sin �

csc �� sec2 � � tan2 �

sin2 � � tan2 � � cos2 � � sec2 �cot s � tan s � sec s csc s

sin2 � �1 � cot2 �� 1cos2 � �tan2 � � 1� � 1

tan2 � � 1

sec �� sec �

1 � sin2 �

cos �� cos �

tan �

sec �� sin �

cot �

csc �� cos �

1

tan2 �� cot � tan �

sin2 x

cos2 x� sin x csc xcsc2 t � 1sec2 x � 1

csc � sec �

cot �

sin � tan �

cos �

cot � sin �tan � cos �

sin3 � � cos3 �sin3 x � cos3 x

cot4 x � 3 cot2 x � 2cos4 x � 2 cos2 x � 1

4 tan2 � � tan � � 32 sin2 x � 3 sin x � 1

�tan x � cot x�2 � �tan x � cot x�2�sin x � 1�2 � �sin x � 1�2

sec2 � � 1sin2 � � 1

�sin � � cos ��21

1 � cos x�

1

1 � cos x�1 � tan s�2 � 2 tan s

�1 � sin t�2 � cos2 tcos �

sin ��

sin �

1 � cos �

cos x

sec x�

sin x

csc x

1

sin � � 1�

1

sin � � 1

1

csc2 ��

1

sec2 �cos � �sec � � csc ��

tan s�cot s � csc s�sec x

csc x�

csc x

sec xcot � �

1

cot �

1. or

2. csc x sec x or

3. 4.

5. 1 6. or

7. 1 8. or

9. 10.

11. or

12.13.14.15. 4 sin x 16. 4 tan x cot x or 417.18.19.20. or

21.22.23. 24. 25. 126. 1 27. 28.29. 30. 31.32. csc2 �

sec2 xcot2 ttan2 xsec2 �tan2 �

cos �sin ��sin � � cos �� 1 � sin � cos ��sin x � cos x� �1 � sin x cos x�

csc2 x�cot2 x � 2��cot2 x � 2� �cot2 x � 1��cos2 x � 1�2�4 tan � � 3� �tan � � 1��2 sin x � 1� �sin x � 1�

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

�2 cot x csc x�2 cos x

sin2 x

sec2 s2 � 2 sin t

csc �1

sin �

�2 sec2 ��2

cos2 �

1 � cot �1 � sec s

1

sin x cos x

1

sin � cos �csc � sec �

7.2 Exercises



47. 48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

63.

64.

65.

66.

67.

68.

69. A student claims that the equation is an identity, since by lettingor radians we get , a true statement. Comment on this stu-

dent’s reasoning.

70. An equation that is an identity has an infinite number of solutions. If an equation hasan infinite number of solutions, is it necessarily an identity? Explain.

0 � 1 � 1�2�� 90�

cos � � sin � � 1

sin4 � � cos4 �

sin2 � � cos2 �� 1

�sec � � csc �� cos � � sin �� cot � � tan �

�sec � � tan ��2 �1 � sin �

1 � sin �

1 � cos x

1 � cos x�

1 � cos x

1 � cos x� 4 cot x csc x

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

tan2 t � 1

sec2 t�

tan t � cot t

tan t � cot t

cot2 t � 1

1 � cot2 t� 1 � 2 sin2 t

sec4 s � tan4 s

sec2 s � tan2 s� sec2 s � tan2 s

sin �

1 � cos ��

sin � cos �

1 � cos �� csc � �1 � cos2 ��

sin � � cos � �sin �

1 �cos �

sin �

�cos �

1 �sin �

cos �

1 � sin �

1 � sin �� sec2 � � 2 sec � tan � � tan2 �

sec4 x � sec2 x � tan4 x � tan2 x

csc � � cot �

tan � � sin �� cot � csc �

sin2 � sec2 � � sin2 � csc2 � � sec2 �

1

tan � � sec ��

1

tan � � sec �� 2 tan �

cot � � 1

cot � � 1�

1 � tan �

1 � tan �

1 � cos x

1 � cos x� �cot x � csc x�2

tan s

1 � cos s�

sin s

1 � cos s� cot s � sec s csc s

1

sec � � tan �� sec � � tan �

1

1 � sin ��

1

1 � sin �� 2 sec2 �

�sec � � tan ��2 � 1

sec � csc � � tan � csc �� 2 tan �

cos � � 1

tan2 ��

cos �

sec � � 1

9

9



Concept Check Graph each expression and conjecture an identity. Then verify yourconjecture algebraically.

71. 72.

73. 74.

Graph the expressions on each side of the equals sign to determine whether the equationmight be an identity. (Note: Use a domain whose length is at least 2 .) If the equationlooks like an identity, prove it algebraically. See Example 1.

75. 76.

77. 78.

By substituting a number for s or t, show that the equation is not an identity.

79. 80.

81. 82.

83. When is a true statement?

(Modeling) Work each problem.

84. Intensity of a Lamp According to Lambert’s law, theintensity of light from a single source on a flat surface atpoint P is given by

,

where k is a constant. (Source: Winter, C., Solar PowerPlants, Springer-Verlag, 1991.)

(a) Write I in terms of the sine function.(b) Explain why the maximum value of I occurs when

.

85. Oscillating Spring The distance or displacement y of aweight attached to an oscillating spring from its natural posi-tion is modeled by

,

where t is time in seconds. Potential energy is the energy ofposition and is given by

,

where k is a constant. The weight has the greatest potential energy when the springis stretched the most. (Source: Weidner, R. and R. Sells, Elementary ClassicalPhysics, Vol. 1, Allyn & Bacon, 1973.)

(a) Write an expression for P that involves the cosine function.(b) Use a fundamental identity to write P in terms of .

86. Radio Tuners Refer to Example 6. Let the energy stored in the inductor be given by

and the energy in the capacitor be given by

,C�t� � 3 sin2�6,000,000t�

L�t� � 3 cos2�6,000,000t�

sin�2t�

P � ky2

y � 4 cos�2t�

� � 0

I � k cos2 �

sin x � �1 � cos2 x

cos t � �1 � sin2 tcsc t � �1 � cot2 t

�cos2 s � cos ssin�csc s� � 1

1

1 � sin s�

1

1 � sin s� sec2 s

tan s � cot s

tan s � cot s� 2 sin2 s

1 � cot2 s �sec2 s

sec2 s � 1

2 � 5 cos s

sin s� 2 csc s � 5 cot s

tan � sin � � cos �cos � � 1

sin � � tan �

�csc � � cot �� sec � � 1��sec � � tan �� 1 � sin ��

71.

72.

73.

74.75. identity 76. identity77. not an identity78. not an identity83. It is true when 84. (a)(b) For for all integers n,

, its maximum value,and I attains a maximum value of k.85. (a)(b) P � 16k�1 � sin2�2t�

P � 16k cos2�2t�

cos2 � � 1� � 2nI � k�1 � sin2 ��

sin x 0.

tan � sin � � cos � � sec �

cos � � 1

sin � � tan �� cot �

� tan ��csc � � cot �� sec � � 1�

� cos ��sec � � tan �� 1 � sin ��

9 P

�

x

y


where t is time in seconds. The total energy E in the circuit is given by.

(a) Graph L, C, and E in the window by , with and. Interpret the graph.

(b) Make a table of values for L, C, and E starting at , incrementing by .Interpret your results.

(c) Use a fundamental identity to derive a simplified expression for .E�t�

10�7t � 0Yscl � 1

Xscl � 10�7��1, 4�0, 10�6E�t� � L�t� � C�t�

7.3 Sum and Difference Identities 621

86. (a) The sum of L and Cequals 3.

4

0 10–6

–1

7.3 Sum and Difference IdentitiesCosine Sum and Difference Identities ■ Cofunction Identities ■ Sine and Tangent Sum and Difference Identities

Cosine Sum and Difference Identities Several examples presented ear-lier should have convinced you by now that does not equal

For example, if and then

while

We can now derive a formula for We start by locating angles Aand B in standard position on a unit circle, with Let S and Q be thepoints where the terminal sides of angles A and B, respectively, intersect thecircle. Locate point R on the unit circle so that angle POR equals the difference

See Figure 4.

Figure 4

Point Q is on the unit circle, so by the work with circular functions inChapter 6, the x-coordinate of Q is the cosine of angle B, while the y-coordinateof Q is the sine of angle B.

Q has coordinates

In the same way,

S has coordinates

and R has coordinates �cos�A � B�, sin�A � B��.

�cos A, sin A�,

�cos B, sin B�.

Ox

y

(cos(A – B), sin(A – B)) R

BA A – B

(cos A, sin A) S

P (1, 0)

Q (cos B, sin B

A � B.

B � A.cos�A � B�.

cos A � cos B � cos

2� cos 0 � 0 � 1 � �1.

cos�A � B� � cos�

2� 0� � cos

2� 0,

B � 0,A � 2cos A � cos B.

cos�A � B�

(b) Let , , and . for all inputs.(c) E�t� � 3

Y3 � 3Y3 � E�t�Y2 � C�t�Y1 � L�t�



Angle SOQ also equals Since the central angles SOQ and POR areequal, chords PR and SQ are equal. By the distance formula, since

(Section 2.1)

Squaring both sides and clearing parentheses gives

Since for any value of x, we can rewrite the equation as

Subtract 2; divide by �2.

Although Figure 4 shows angles A and B in the second and first quadrants,respectively, this result is the same for any values of these angles.

To find a similar expression for rewrite as and use the identity for

Cosine difference identity

Negative-angle identities (Section 7.1)

Cosine of a Sum or Difference

These identities are important in calculus and useful in certain applications.Although a calculator can be used to find an approximation for cos 15°, for example, the method shown below can be applied to get an exact value, as wellas to practice using the sum and difference identities.

EXAMPLE 1 Finding Exact Cosine Function Values

Find the exact value of each expression.

(a) cos 15° (b) (c)

Solution

(a) To find cos 15°, we write 15° as the sum or difference of two angles withknown function values. Since we know the exact trigonometric functionvalues of 45° and 30°, we write 15° as (We could also use

) Then we use the identity for the cosine of the difference of two angles.60� � 45�.

45� � 30�.

cos 87� cos 93� � sin 87� sin 93�cos 5

12

cos(A � B) � cos A cos B � sin A sin B

cos(A � B) � cos A cos B � sin A sin B

cos�A � B� � cos A cos B � sin A sin B

� cos A cos B � sin A��sin B� � cos A cos��B� � sin A sin��B�

cos�A � B� � cos�A � ��B�

cos�A � B�.A � ��B�A � Bcos�A � B�,

cos�A � B� � cos A cos B � sin A sin B.

2 � 2 cos�A � B� � 2 � 2 cos A cos B � 2 sin A sin B

sin2 x � cos2 x � 1

� cos2 A � 2 cos A cos B � cos2 B � sin2 A � 2 sin A sin B � sin2 B.

cos2�A � B� � 2 cos�A � B� � 1 � sin2�A � B�

� ��cos A � cos B�2 � �sin A � sin B�2.

��cos�A � B� � 12 � �sin�A � B� � 02

PR � SQ,A � B.



TEACHING TIP In Example 1(b),students may benefit from con-

verting radians to 75° in

order to realize that

can be used in place of 5

12.

6�

4� 30° � 45°

5

12

TEACHING TIP Mention that theseidentities state that the trigono-metric function of an acute angleis the same as the cofunction of itscomplement. Verify the cofunctionidentities for acute angles usingcomplementary angles in a righttriangle along with the right-triangle-based definitions of thetrigonometric functions. Emphasizethat these identities apply to anyangle �, not just acute angles.

The screen supports the solu-tion in Example 1(b) byshowing that

cos = .5�12

√6 – √24


(Section 5.3)

(b)

Cosine sum identity

(Section 6.2)

(c)Cosine sum identity

(Section 5.2)


Cofunction Identities We can use the identity for the cosine of the differ-ence of two angles and the fundamental identities to derive cofunctionidentities.

Cofunction Identities

Similar identities can be obtained for a real number domain by replacing90° with

Substituting 90° for A and � for B in the identity for gives

This result is true for any value of � since the identity for is true forany values of A and B.

cos�A � B�

� sin �.

� 0 cos � � 1 sin �

cos�90� � �� cos 90� cos � � sin 90� sin �

cos�A � B�

2 .

tan(90� � �) � cot � csc(90� � �) � sec �

sin(90� � �) � cos � sec(90� � �) � csc �

cos(90� � �) � sin � cot(90� � �) � tan �

� �1

� cos 180�

cos 87� cos 93� � sin 87� sin 93� � cos�87� � 93��

��6 � �2

4

��3

2

�2

2�

1

2

�2

2

� cos

6 cos

4� sin

6 sin

4

6 � 2

12 ; 4 � 312 cos

5

12� cos�

6�

4 � �

�6 � �2

4

��2

2

�3

2�

�2

2

1

2

� cos 45� cos 30� � sin 45� sin 30�

cos 15� � cos�45� � 30��



TEACHING TIP Verify results fromExample 2 using a calculator.

EXAMPLE 2 Using Cofunction Identities to Find �

Find an angle � that satisfies each of the following.

(a) (b) (c)

Solution

(a) Since tangent and cotangent are cofunctions,

Cofunction identity

Set angle measures equal.

(b)

Cofunction identity

(c)

Cofunction identity

Combine terms.

Now try Exercises 25 and 27.

N O T E Because trigonometric (circular) functions are periodic, the solutionsin Example 2 are not unique. We give only one of infinitely many possibilities.

If one of the angles A or B in the identities for and is a quadrantal angle, then the identity allows us to write the expression in termsof a single function of A or B.

EXAMPLE 3 Reducing to a Function of a Single Variable

Write as a trigonometric function of �.

Solution Use the difference identity. Replace A with 180° and B with �.

(Section 5.2)


� �cos �

� ��1� cos � � �0� sin �

cos�180� � �� cos 180� cos � � sin 180� sin �

cos�180� � ��

cos�A � B�

cos�A � B�cos�A � B�

�

4� �

sec��

4 � � sec �

sec�

2�

3

4 � � sec �

csc 3

4� sec �

� � 120�

90� � � � �30�

cos�90� � �� cos��30�� sin � � cos��30��

� � 65�

90� � � � 25�

tan�90� � �� tan 25�

cot � � tan 25�

tan�90� � �� cot �.

csc 3

4� sec �sin � � cos��30��cot � � tan 25�


Sine and Tangent Sum and Difference Identities We can use the cosine sum and difference identities to derive similar identities for sine and tangent. Since we replace � with to get

Cofunction identity


Cofunction identities

Now we write as and use the identity for

Sine sum identity

Negative-angle identities

Sine of a Sum or Difference

To derive the identity for we start with

We express this result in terms of the tangent function by multiplying both nu-merator and denominator by

Replacing B with and using the fact that gives theidentity for the tangent of the difference of two angles.

tan��B� � �tan B�B

tan � � sin �cos � tan�A � B� �

tan A � tan B

1 � tan A tan B

�

sin A

cos A�

sin B

cos B

1 �sin A

cos A

sin B

cos B

�

sin A cos B

cos A cos B�

cos A sin B

cos A cos B

cos A cos B

cos A cos B�

sin A sin B

cos A cos B

tan�A � B� �

sin A cos B � cos A sin B

1

cos A cos B � sin A sin B

1

1

cos A cos B

1

cos A cos B

1cos A cos B .

�sin A cos B � cos A sin B

cos A cos B � sin A sin B.

tan�A � B� �sin�A � B�cos�A � B�

tan�A � B�,

sin(A � B) � sin A cos B � cos A sin B

sin(A � B) � sin A cos B � cos A sin B

sin�A � B� � sin A cos B � cos A sin B

� sin A cos��B� � cos A sin��B� sin�A � B� � sin�A � ��B�

sin�A � B�.sin�A � ��B�sin�A � B�

sin�A � B� � sin A cos B � cos A sin B.

� cos�90� � A� cos B � sin�90� � A� sin B

� cos��90� � A� � B sin�A � B� � cos�90� � �A � B�

A � Bsin � � cos�90� � ��,


Fundamental identity(Section 7.1)

Sum identities

Simplify thecomplex fraction.(Section R.5)

Multiply numerators;multiply denominators.



Tangent of a Sum or Difference

EXAMPLE 4 Finding Exact Sine and Tangent Function Values

Find the exact value of each expression.

(a) sin 75° (b) (c)

Solution

(a)

Sine sum identity

(Section 5.3)

(b)

Tangent sum identity

(Section 6.2)

Rationalize the denominator. (Section R.7)

Multiply. (Section R.7)

Combine terms.

Factor out 2. (Section R.5)

Lowest terms

(c)Sine difference identity

Negative-angle identity

(Section 5.3)


� ��3

2

� �sin 120�

� sin��120��

sin 40� cos 160� � cos 40� sin 160� � sin�40� � 160��

� �2 � �3

�2�2 � �3�

�2

�4 � 2�3

�2

��3 � 3 � 1 � �3

1 � 3

��3 � 1

1 � �3

1 � �3

1 � �3

��3 � 1

1 � �3 1

�tan 3 � tan 4

1 � tan 3 tan 4

tan 7

12� tan�

3�

4 � �

�6 � �2

4

��2

2

�3

2�

�2

2

1

2

� sin 45� cos 30� � cos 45� sin 30�

sin 75� � sin�45� � 30��

sin 40� cos 160� � cos 40� sin 160�tan 7

12

tan(A � B) �tan A � tan B

1 � tan A tan Btan(A � B) �

tan A � tan B1 � tan A tan B



EXAMPLE 5 Writing Functions as Expressions Involving Functions of �

Write each function as an expression involving functions of �.

(a) (b) (c)

Solution

(a) Using the identity for

(b)

(c)


EXAMPLE 6 Finding Function Values and the Quadrant of

Suppose that A and B are angles in standard position, with and Find each of the following.

(a) (b) (c) the quadrant of

Solution

(a) The identity for requires sin A, cos A, sin B, and cos B. We aregiven values of sin A and cos B. We must find values of cos A and sin B.

Fundamental identity

Subtract

Since A is in quadrant II,

In the same way, Now use the formula for

(b) To find first use the values of sine and cosine from part (a) toget and

tan�A � B� ��

43 �

125

1 � ��43� �12

5 ��

1615

1 �4815

�16156315

�16

63

tan B � 125 .tan A � �

43

tan�A � B�,

� �20

65�

36

65�

16

65

sin�A � B� �4

5 ��5

13� � ��3

5��12

13�sin�A � B�.sin B � �

1213 .

cos A � 0. cos A � �3

5

1625 . cos2 A �

9

25

sin A � 45

16

25� cos2 A � 1

sin2 A � cos2 A � 1

sin�A � B�

A � Btan�A � B�sin�A � B�

� B �32 .cos B � �

513 ,

2 � A � ,sin A � 4

5 ,

A � B

� �sin �

� 0 cos � � ��1� sin �

sin�180� � �� sin 180� cos � � cos 180� sin �

tan�45� � �� tan 45� � tan �

1 � tan 45� tan ��

1 � tan �

1 � tan �

�1

2 cos � �

�3

2 sin �.

sin�30� � �� sin 30� cos � � cos 30� sin �

sin�A � B�,

sin�180� � ��tan�45� � ��sin�30� � ��



(c) From parts (a) and (b), and both posi-tive. Therefore, must be in quadrant I, since it is the only quadrant inwhich both sine and tangent are positive.


EXAMPLE 7 Applying the Cosine Difference Identity to Voltage

Common household electric current is called alternating current because thecurrent alternates direction within the wires. The voltage V in a typical 115-voltoutlet can be expressed by the function where � is the angu-lar speed (in radians per second) of the rotating generator at the electrical plantand t is time measured in seconds. (Source: Bell, D., Fundamentals of ElectricCircuits, Fourth Edition, Prentice-Hall, 1988.)

(a) It is essential for electric generators to rotate at precisely 60 cycles per secso household appliances and computers will function properly. Determine �for these electric generators.

(b) Graph V in the window by

(c) Determine a value of so that the graph of is thesame as the graph of

Solution

(a) Each cycle is 2 radians at 60 cycles per sec, so the angular speed isper sec.

(b) Because the amplitude of the functionis 163 (from Section 6.3), is an appropriate interval for therange, as shown in Figure 5.

Figure 5

(c) Using the negative-angle identity for cosine and a cofunction identity,

Therefore, if then


V�t� � 163 cos��t �

2 � � 163 sin �t.

� � 2 ,

cos�x �

2 � � cos��

2� x�� cos�

2� x� � sin x.

200

0 .05

–200

For x = t,V(t) = 163 sin 120�t

��200, 200V�t� � 163 sin �t � 163 sin 120 t.

120 radians� � 60�2� �

V�t� � 163 sin �t.V�t� � 163 cos��t � ��

��200, 200.�0, .05

V�t� � 163 sin �t,

A � Btan�A � B� � 16

63,sin�A � B� � 1665


Concept Check Match each expression in Column I with the correct expression inColumn II to form an identity.

I II

1. A.

2. B.

3. C.

4. D.

E.

F.

Use identities to find each exact value. (Do not use a calculator.) See Example 1.

5. cos 75� 6.

7. cos 105� 8.(Hint: ) (Hint: )

9. 10.

11. 12.

Write each function value in terms of the cofunction of a complementary angle. SeeExample 2.

13. tan 87� 14. sin 15� 15. 16.

17. 18. 19. 20.

Use the cofunction identities to fill in each blank with the appropriate trigonometricfunction name. See Example 2.

21. 22.

23. 24.

Find an angle that makes each statement true. See Example 2.

25. 26.

27. 28.

Use identities to find the exact value of each of the following. See Example 4.

29. 30. 31.

32. 33. 34.

35. 36.

37. 38.tan 80� � tan��55��

1 � tan 80� tan��55��tan 80� � tan 55�

1 � tan 80� tan 55�

sin 40� cos 50� � cos 40� sin 50�sin 76� cos 31� � cos 76� sin 31�

tan��7

12�sin��7

12�sin

12

tan

12tan

5

12sin

5

12

cot�� 10�� tan�2� � 20��sin�3� � 15�� cos�� 25��sin � � cos�2� � 10��tan � � cot�45� � 2��

�

72� � cot 18�33� � sin 57�

��

6 �sin 2

3�

6cot

3�

tan 174� 3�sec 146� 42�cot 9

10sin

5

8

sin 2

5cos

12

cos 7

9 cos

2

9� sin

7

9 sin

2

9cos 40� cos 50� � sin 40� sin 50�

cos��

12�cos 7

12

�105� � �60� � ��45��105� � 60� � 45�cos��105��cos��15��

cos x cos y � sin x sin y

cos x sin y � sin x cos y

sin x cos y � cos x sin ysin�x � y� �

sin x cos y � cos x sin ysin�x � y� �

sin x sin y � cos x cos ycos�x � y� �

cos x cos y � sin x sin ycos�x � y� �


1. F 2. A 3. C 4. D

5. 6.

7. 8.

9. 10.

11. 0 12. �1 13. cot 3�

14. cos 75� 15.

16. 17.

18. 19.

20. 21. tan22. cos 23. cos 24. tan

25. 15� 26. 27. 20�

28. 29.

30. 31.

32. 33.

34. 35. 36. 1

37. �1 38. �1

�2

22 � �3

��6 � �2

4�6 � �2

4

2 � �32 � �3

�6 � �2

4

80

3

°

100

3

°

cot��84� 3��

csc��56� 42��tan��2

5 �cos��

8 �cos

10

sin 5

12

�6 � �2

4�2 � �6

4

�2 � �6

4�2 � �6

4

�6 � �2

4�6 � �2

4

7.3 Exercises



39. 40.41. �sin x 42. �sin x

43.

44.

45.

46.

47.

48.

49. 50.

51. (a) (b) (c) (d) I

52. (a) (b)

(c) (d) IV

53. (a)

(b)

(c) (d) II

54. (a) (b) (c) (d) IV

55. (a) (b) (c)

(d) II

56. (a) (b) (c) (d) III

58.

59.

60. (a)

(b)

61. 62.

63. 64.

65. 66.�6 � �2

4�2 � �3

�2 � �3��6 � �2

4

�2 � �3�6 � �2

4

��6 � �2

4

�2 � �6

4

��6 � �2

4

��6 � �2

4

36

77

84

85�

77

85

�77

36

13

85�

36

85

�63

16

33

65

16

65

�8�5 � 5�2

20 � 2�10

4�2 � �5

9

�2�10 � 2

9

�8�6 � 3

4 � 6�6

�8�6 � 3

25

4 � 6�6

25

63

16

33

65

16

65

1 � tan x

1 � tan x�3 tan x � 1

�3 � tan x

�3 tan � � 1

�3 � tan �

�3 � tan �

1 � �3 tan �

�2

2�cos x � sin x�

1

2�cos x � �3 sin x�

sin �

�2

2�cos � � sin ��

�cos �sin � Use identities to write each expression as a function of x or . See Examples 3 and 5.

39. 40. 41.

42. 43. 44.

45. 46. 47.

48. 49. 50.

Use the given information to find (a) , (b) , (c) , and (d) thequadrant of . See Example 6.

51. and , s and t in quadrant I

52. and , s and t in quadrant II

53. and , s in quadrant II and t in quadrant IV

54. and , s in quadrant I and t in quadrant III

55. and , s and t in quadrant III

56. and , s in quadrant II and t in quadrant I


(Exercises 57–60)

The identities for and can be used to find exact values of expres-sions like cos 195� and cos 255�, where the angle is not in the first quadrant. WorkExercises 57–60 in order, to see how this is done.

57. By writing 195� as , use the identity for to express cos 195�as �cos 15�.

58. Use the identity for to find �cos 15�.

59. By the results of Exercises 57 and 58, .

60. Find each exact value using the method shown in Exercises 57–59.

(a) cos 255� (b)

Find each exact value. Use the technique developed in Relating Concepts Exer-cises 57–60.

61. sin 165� 62. tan 165� 63. sin 255�

64. tan 285� 65. 66. sin��13

12 �tan 11

12

cos 11

12

cos 195� �

cos�A � B�

cos�A � B�180� � 15�


sin t �4

5cos s � �

15

17

cos t � �3

5cos s � �

8

17

sin t � �12

13sin s �

3

5

sin t � �1

3sin s �

2

3

sin t �3

5cos s � �

1

5

sin t �5

13cos s �

3

5

s � ttan�s � t�sin�s � t�cos�s � t�

tan�

4� x�tan�x �

6 �tan�� 30��

tan�60� � ��sin�

4� x�sin�5

6� x�

sin�180� � ��sin�45� � ��cos�

2� x�

cos�x �3

2 �cos�180� � ��cos�90� � ��

�



71.

72.

79. 3 80. 163 and �163; no81. (a) 425 lb (c) 0�

1 � tan x

1 � tan x� tan�

4� x�

sin�

2� x� � cos x

9

9

67. Use the identity , and replace with , to derive theidentity .

68. Explain how the identities for , , and can be foundby using the sum identities given in this section.

69. Why is it not possible to use a method similar to that of Example 5(c) to find a for-mula for ?

70. Concept Check Show that if A, B, and C are angles of a triangle, then.

Graph each expression and use the graph to conjecture an identity. Then verify yourconjecture algebraically.

71. 72.

Verify that each equation is an identity.

73.

74.

75.

76.

77.

78.

Exercises 79 and 80 refer to Example 7.

79. How many times does the current oscillate in .05 sec?

80. What are the maximum and minimum voltages in this outlet? Is the voltage alwaysequal to 115 volts?

(Modeling) Solve each problem.

81. Back Stress If a person bends at thewaist with a straight back making anangle of degrees with the horizontal,then the force F exerted on the backmuscles can be modeled by the equation

,

where W is the weight of the person.(Source: Metcalf, H., Topics inClassical Biophysics, Prentice-Hall,1980.)

(a) Calculate F when lb and .(b) Use an identity to show that F is approximately equal to .(c) For what value of is F maximum?�

2.9W cos �� 30�W � 170

F �.6W sin�� 90��

sin 12�

�

sin�s � t�sin t

�cos�s � t�

cos t�

sin s

sin t cos t

sin�x � y�sin�x � y�

�tan x � tan y

tan x � tan y

sin�s � t�cos s cos t

� tan s � tan t

cos�� cos � sin �

� tan � � cot �

tan�x � y� � tan� y � x� �2�tan x � tan y�1 � tan x tan y

sin�x � y� � sin�x � y� � 2 sin x cos y

1 � tan x

1 � tan xsin�

2� x�

sin�A � B � C� � 0

tan�270� � ��

cot�A � B�csc�A � B�sec�A � B�cos A � sin�90� � A�

90� � A�cos�90� � �� sin �


7.4 Double-Angle Identities and Half-Angle IdentitiesDouble-Angle Identities ■ Product-to-Sum and Sum-to-Product Identities ■ Half-Angle Identities

Double-Angle Identities When in the identities for the sum of twoangles, these identities are called the double-angle identities. For example, toderive an expression for cos 2A, we let in the identity

Cosine sum identity (Section 7.3)

cos 2A � cos2 A � sin2 A

� cos A cos A � sin A sin A

cos 2A � cos�A � A�

cos A cos B � sin A sin B.cos�A � B� �B � A

A � B

TEACHING TIP A common error isto write as 2 cos A.cos 2A

60

0 .05

–60

For x = t, V = V1 + V2 = 30 sin 120�t + 40 cos 120�t

82. Back Stress Refer to Exercise 81.

(a) Suppose a 200-lb person bends at the waist so that . Estimate the forceexerted on the person’s back muscles.

(b) Approximate graphically the value of that results in the back muscles exertinga force of 400 lb.

83. Sound Waves Sound is a result of waves applying pressure to a person’s eardrum.For a pure sound wave radiating outward in a spherical shape, the trigonometricfunction defined by

can be used to model the sound pressure at a radius of r feet from the source, wheret is time in seconds, is length of the sound wave in feet, c is speed of sound in feetper second, and a is maximum sound pressure at the source measured in poundsper square foot. (Source: Beranek, L., Noise and Vibration Control, Institute of Noise Control Engineering, Washington, D.C., 1988.) Let ft and

ft per sec.

(a) Let lb per . Graph the sound pressure at distance ft from itssource in the window by Describe P at this distance.

(b) Now let and . Graph the sound pressure in the window byWhat happens to pressure P as radius r increases?

(c) Suppose a person stands at a radius r so that , where n is a positiveinteger. Use the difference identity for cosine to simplify P in this situation.

84. Voltage of a Circuit When the two voltages

and

are applied to the same circuit, the resulting voltage V will be equal to their sum.(Source: Bell, D., Fundamentals of Electric Circuits, Second Edition, RestonPublishing Company, 1981.)

(a) Graph the sum in the window by (b) Use the graph to estimate values for a and so that .(c) Use identities to verify that your expression for V is valid.

V � a sin�120t � ��60, 60.�0, .05

V2 � 40 cos 120 tV1 � 30 sin 120t

r � n��2, 2.

�0, 20]t � 10a � 3��.05, .05.�0, .05

r � 10ft2a � .4

c � 1026� � 4.9

�

P �a

r cos�2r

�� ct�

�

� � 45�


82. (a) 408 lb (b) 46.1�83. (a) The pressure P isoscillating.

(b) The pressure oscillates andamplitude decreases as rincreases.

(c)

84. (a)

(b) ; � � �5.353a � 50

P �a

n� cos ct

0 20

2

–2

For x = r,

P(r) = cos[ – 10,260]3r

2�r4.9

0 .05

.05

–.05

For x = t,

P(t) = cos[ – 1026t].410

20�4.9


7.4 Double-Angle Identities and Half-Angle Identities 633

TEACHING TIP Students might findit helpful to see each formula illus-trated with a concrete examplethat they can check. For instance,you might show that

� cos2 30� � sin2 30�.

cos 60� � cos�2 30°�

Looking Ahead to CalculusThe identities

and

can be rewritten as

and

These identities are used to integrate

the functions and

cos2 A.g�A� �

f �A� � sin2 A

cos2 A �1

2�1 � cos 2A�.

sin2 A �1

2�1 � cos 2 A�

cos 2A � 2 cos2 A � 1

cos 2A � 1 � 2 sin2 A

Two other useful forms of this identity can be obtained by substitutingeither or Replace with the expression to get

Fundamental identity (Section 7.1)

or replace with to get

Fundamental identity

We find sin 2A with the identity letting

Sine sum identity

Using the identity for we find tan 2A.

Tangent sum identity

Double-Angle Identities

EXAMPLE 1 Finding Function Values of 2� Given Information about �

Given and find and

Solution To find we must first find the value of sin �.

Simplify.

Choose the negative square root since sin � � 0. sin � � �4

5

sin2 � �16

25

cos � � 35sin2 � � cos2 � � 1; sin2 � � � 3

5�2

� 1

sin 2�,

tan 2�.cos 2�,sin 2�,sin � � 0,cos � � 35

tan 2A �2 tan A

1 � tan2 A

cos 2A � 2 cos2 A � 1 sin 2A � 2 sin A cos A

cos 2A � cos2 A � sin2 A cos 2A � 1 � 2 sin2 A

tan 2A �2 tan A

1 � tan2 A

�tan A � tan A

1 � tan A tan A

tan 2A � tan�A � A�

tan�A � B�,

sin 2A � 2 sin A cos A

� sin A cos A � cos A sin A

sin 2A � sin�A � A�

B � A.sin�A � B� � sin A cos B � cos A sin B,

cos 2A � 2 cos2 A � 1.

� cos2 A � 1 � cos2 A

� cos2 A � �1 � cos2 A� cos 2A � cos2 A � sin2 A

1 � cos2 Asin2 A

cos 2A � 1 � 2 sin2 A,

� �1 � sin2 A� � sin2 A

cos 2A � cos2 A � sin2 A

1 � sin2 Acos2 Asin2 A � 1 � cos2 A.cos2 A � 1 � sin2 A



Using the double-angle identity for sine,

Now we find using the first of the double-angle identities for cosine.(Any of the three forms may be used.)

The value of can be found in either of two ways. We can use the double-

angle identity and the fact that

Simplify. (Section R.5)

Alternatively, we can find by finding the quotient of and

Now try Exercise 9.

EXAMPLE 2 Verifying a Double-Angle Identity


Solution We start by working on the left side, using the hint from Section 7.1about writing all functions in terms of sine and cosine.

Quotient identity

Double-angle identity

The final step illustrates the importance of being able to recognize alterna-tive forms of identities.


2 cos2 x � 1 � cos 2xcos 2x � 2 cos2 x � 1, so � 1 � cos 2x

� 2 cos2 x

�cos x

sin x�2 sin x cos x�

cot x sin 2x �cos x

sin x sin 2x

cot x sin 2x � 1 � cos 2x

tan 2� �sin 2�

cos 2��

�2425

�7

25

�24

7

cos 2�.sin 2�tan 2�

tan 2� �2 tan �

1 � tan2 ��

2��43�

1 �169

��

83

�79

�24

7

tan � � sin �cos � �

�45

35

� �43.

tan 2�

cos 2� � cos2 � � sin2 � �9

25�

16

25� �

7

25

cos 2�,

sin � � �45 ; cos � � 3

5 � 2��4

5�� 3

5� � �24

25.

sin 2� � 2 sin � cos �



EXAMPLE 3 Simplifying Expressions Using Double-Angle Identities

Simplify each expression.

(a) (b)

Solution

(a) This expression suggests one of the double-angle identities for cosine:Substituting 7x for A gives

(b) If this expression were 2 sin 15° cos 15°, we could apply the identity forsin 2A directly since We can still apply the identitywith by writing the multiplicative identity element 1 as

Multiply by 1 in the form

Associative property (Section R.1)

with

(Section 5.3)


Identities involving larger multiples of the variable can be derived by re-peated use of the double-angle identities and other identities.

EXAMPLE 4 Deriving a Multiple-Angle Identity

Write sin 3x in terms of sin x.

Solution

Sine sum identity(Section 7.3)

Double-angle identities

Multiply.

Distributive property(Section R.1)

Combine terms.


� 3 sin x � 4 sin3 x

� 2 sin x � 2 sin3 x � sin x � sin3 x � sin3 x

cos2 x � 1 � sin2 x � 2 sin x�1 � sin2 x� � �1 � sin2 x� sin x � sin3 x

� 2 sin x cos2 x � cos2 x sin x � sin3 x

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

� sin 2x cos x � cos 2x sin x

sin 3x � sin�2x � x�

sin 30� � 12 �

1

2

1

2�

1

4

�1

2 sin 30�

A � 15�2 sin A cos A � sin 2A, �1

2 sin�2 15��

�1

2�2 sin 15� cos 15��

12 �2�. sin 15� cos 15� �

1

2� 2� sin 15� cos 15�

12 �2�.A � 15�

sin 2A � 2 sin A cos A.

cos2 7x � sin2 7x � cos 2�7x� � cos 14x.

cos 2A � cos2 A � sin2 A.

sin 15� cos 15�cos2 7x � sin2 7x



Looking Ahead to CalculusThe product-to-sum identities are used

in calculus to find integrals of func-

tions that are products of trigonometric

functions. One classic calculus text in-

cludes the following example:

Evaluate

The first solution line reads:

“We may write

”cos 5x cos 3x �1

2�cos 8x � cos 2x.

cos 5x cos 3xdx.

2000

–500

0 .05

For x = t,

W(t) =(163 sin 120�t)2

15

Figure 6

The next example applies a multiple-angle identity to answer a questionabout electric current.

EXAMPLE 5 Determining Wattage Consumption

If a toaster is plugged into a common household outlet, the wattage consumed isnot constant. Instead, it varies at a high frequency according to the model

where V is the voltage and R is a constant that measures the resistance of the toaster in ohms. (Source: Bell, D., Fundamentals of Electric Circuits,Fourth Edition, Prentice-Hall, 1998.) Graph the wattage W consumed by a typi-cal toaster with and in the window by

How many oscillations are there?

Solution Substituting the given values into the wattage equation gives

To determine the range of W, we note that sin 120 t has maximum value 1, sothe expression for W has maximum value The minimum value is 0.The graph in Figure 6 shows that there are six oscillations.


Product-to-Sum and Sum-to-Product Identities Because they make itpossible to rewrite a product as a sum, the identities for and

are used to derive a group of identities useful in calculus.Adding the identities for and gives

or

Similarly, subtracting from gives

Using the identities for and in the same way, we gettwo more identities. Those and the previous ones are now summarized.

Product-to-Sum Identities

(continued)

sin A sin B �12

[cos(A � B) � cos(A � B)]

cos A cos B �12

[cos(A � B) � cos(A � B)]

sin�A � B�sin�A � B�

sin A sin B �1

2�cos�A � B� � cos�A � B�.


cos A cos B �1

2�cos�A � B� � cos�A � B�.

cos�A � B� � cos�A � B� � 2 cos A cos B

cos�A � B� � cos A cos B � sin A sin B cos�A � B� � cos A cos B � sin A sin B

cos�A � B�cos�A � B�cos�A � B�

cos�A � B�

1632

15 � 1771.

W �V 2

R�

�163 sin 120t�2

15.

��500, 2000.�0, .05V � 163 sin 120tR � 15

W �V 2

R,



EXAMPLE 6 Using a Product-to-Sum Identity

Write cos 2� sin � as the sum or difference of two functions.

Solution Use the identity for cos A sin B, with and


From these new identities we can derive another group of identities that areused to write sums of trigonometric functions as products.

Sum-to-Product Identities

EXAMPLE 7 Using a Sum-to-Product Identity

Write as a product of two functions.

Solution Use the identity for with and

(Section 7.1)


sin�� sin � � �2 cos 3� sin �

� 2 cos 3� sin��

� 2 cos 6�

2 sin��2�

2 � sin 2� � sin 4� � 2 cos�2� � 4�

2 � sin�2� � 4�

2 �4� � B.2� � Asin A � sin B,

sin 2� � sin 4�

cos A � cos B � �2 sin�A � B2 � sin�A � B

2 � cos A � cos B � 2 cos�A � B

2 � cos�A � B2 �

sin A � sin B � 2 cos�A � B2 � sin�A � B

2 � sin A � sin B � 2 sin�A � B

2 � cos�A � B2 �

�1

2 sin 3� �

1

2 sin �

cos 2� sin � �1

2�sin�2� � �� sin�2� � ��

� � B.2� � A

cos A sin B �12

[sin(A � B) � sin(A � B)]

sin A cos B �12

[sin(A � B) � sin(A � B)]



Half-Angle Identities From the alternative forms of the identity for cos 2A,we derive three additional identities for and These areknown as half-angle identities.

To derive the identity for start with the following double-angle iden-tity for cosine and solve for sin x.

Add subtract cos 2x.

Divide by 2; take square roots. (Section 1.4)

Let so substitute.

The � sign in this identity indicates that the appropriate sign is chosen de-pending on the quadrant of For example, if is a quadrant III angle, wechoose the negative sign since the sine function is negative in quadrant III.

We derive the identity for using the double-angle identity

Add 1.

Rewrite; divide by 2.

Take square roots.

Replace x with

An identity for comes from the identities for and

We derive an alternative identity for using double-angle identities.

Double-angle identities

From this identity for we can also derive

tan A

2�

1 � cos A

sin A.

tan A2 ,

tan A

2�

sin A

1 � cos A

�sin 2� A

2 �1 � cos 2� A

2 �

tan A

2�

sin A2

cos A2

�2 sin A

2 cos A2

2 cos2 A2

tan A2

tan A

2�

sin A2cos A2

�

��1 � cos A

2

��1 � cos A

2

� ��1 � cos A

1 � cos A

cos A2 .sin A2tan A2

A2 . cos

A

2� ��1 � cos A

2

cos x � ��1 � cos 2x

2

cos2 x �1 � cos 2x

2

1 � cos 2x � 2 cos2 x

cos 2x � 2 cos2 x � 1.cos A

2

A2

A2 .

x � A2 ;2x � A, sin

A

2� ��1 � cos A

2

sin x � ��1 � cos 2x

2

2 sin2 x; 2 sin2 x � 1 � cos 2x

cos 2x � 1 � 2 sin2 x

sin A2 ,

tan A2 .cos A

2 ,sin A2 ,

Multiply by 2 in numer-ator and denominator.

cos A2



TEACHING TIP Point out that the

first identity for follows

directly from the cosine and sinehalf-angle identities; however, the

other two identities for are

more useful.

tan A2

tan A2

TEACHING TIP Have students com-pare the value of cos 15° inExample 8 to the value in Example 1(a) of Section 7.3,where we used the identity for thecosine of the difference of twoangles. Although the expressionslook completely different, they areequal, as suggested by a calculatorapproximation for both,.96592583.

Half-Angle Identities

N O T E The last two identities for do not require a sign choice. Whenusing the other half-angle identities, select the plus or minus sign according tothe quadrant in which terminates. For example, if an angle then

which lies in quadrant II. In quadrant II, and are negative,

while is positive.

EXAMPLE 8 Using a Half-Angle Identity to Find an Exact Value

Find the exact value of cos 15° using the half-angle identity for cosine.

Solution

Choose the positive square root.

Simplify the radicals. (Section R.7)


EXAMPLE 9 Using a Half-Angle Identity to Find an Exact Value

Find the exact value of tan 22.5° using the identity

Solution Since replace A with 45°.

Now multiply numerator and denominator by 2. Then rationalize the denominator.


�2��2 � 1�

2� �2 � 1

tan 22.5� ��2

2 � �2�

�2

2 � �2

2 � �2

2 � �2�

2�2 � 2

2

tan 22.5� � tan 45°

2�

sin 45�

1 � cos 45��

�22

1 � �22

22.5� � 12 �45��,

tan A2 � sin A

1 � cos A .

� �1 � �32

2� ��1 � �3

2 � 2

2 2�

�2 � �3

2

cos 15� � cos 1

2�30�� 1 � cos 30�

2

sin A2

tan A2cos A

2A2 � 162�,

A � 324°,A2

tan A2

tan A2

� ��1 � cos A1 � cos A

tan A2

�sin A

1 � cos Atan

A2

�1 � cos A

sin A

cos A2

� ��1 � cos A2

sinA2

� ��1 � cos A2



Figure 7

0�

3�2

�2

s2

s

EXAMPLE 10 Finding Functions of Given Information about s

Given with find and

Solution Since

and Divide by 2. (Section 1.7)

terminates in quadrant II. See Figure 7. In quadrant II, the values of andare negative and the value of is positive. Now use the appropriate half-

angle identities and simplify the radicals.

Notice that it is not necessary to use a half-angle identity for once we findand However, using this identity would provide an excellent check.


EXAMPLE 11 Simplifying Expressions Using the Half-Angle Identities

Simplify each expression.

(a) (b)

Solution

(a) This matches part of the identity for

Replace A with 12x to get

(b) Use the third identity for given earlier with to get


1 � cos 5�

sin 5�� tan

5�

2.

A � 5�tan A2

��1 � cos 12x

2� cos

12x

2� cos 6x.

cos A

2� ��1 � cos A

2

cos A2 .

1 � cos 5�

sin 5��1 � cos 12x

2

cos s2 .sin s2

tan s2

tan s

2�

sin s2cos s2

��66

��306

� ��5

5

cos s

2� ��1 �

23

2� �� 5

6� �

�30

6

sin s

2� �1 �

23

2� � 1

6�

�6

6

sin s2tan s2

cos s2s2

3

4�

s

2� ,

3

2� s � 2

tan s2 .sin s2 ,cos s2 ,32 � s � 2,cos s � 2

3 ,

s2



1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. 13. 14.

15. 16.

17. 18.

19. 20.

21.22.

23.

24.25.

26.4 tan x cos2 x � 2 tan x

1 � tan2 x� sin 2x

cos4 x � sin4 x � cos 2xcos 4x � 8 cos4 x � 8 cos2 x � 1

tan 3x �3 tan x � tan3 x

1 � 3 tan2 x

4 sin3 x cos xsin 4x � 4 sin x cos3 x �cos 3x � 4 cos3 x � 3 cos x

1

16 sin 59°

1

4 cos 94.2°

1

4 tan 68°

1

2 tan 102°

�2

4�

�2

2

�2

2�3

2�3

3

�3

2sin 2� �

2�66

25

cos 2� � �19

25;

sin 2� � �4�55

49

cos 2� �39

49;

sin 2x �15

17cos 2x � �

8

17;

sin 2x �4

5cos 2x � �

3

5;

sin 2� � �120

169cos 2� �

119

169;

sin 2� � �4�21

25

cos 2� �17

25;

sin x ��6

6cos x � �

�30

6;

sin x ��102

12cos x � �

�42

12;

sin � � ��2

4cos � � �

�14

4;

sin � ��5

5cos � �

2�5

5;

7.4 Exercises

Use identities to find values of the sine and cosine functions for each angle measure. SeeExample 1.

1. , given and terminates in quadrant I

2. , given and terminates in quadrant III

3. given and

4. given and

5. , given and

6. , given and

7. given and

8. given and

9. given and

10. , given and

Use an identity to write each expression as a single trigonometric function value or as asingle number. See Example 3.

11. 12. 13.

14. 15. 16.

17. 18. 19.

20.

Express each function as a trigonometric function of x. See Example 4.

21. 22. 23. 24.


25. 26.

Verify that each equation is an identity. See Example 2.

27. 28. sec 2x �sec2 x � sec4 x

2 � sec2 x � sec4 x�sin x � cos x�2 � sin 2x � 1

4 tan x cos2 x � 2 tan x

1 � tan2 xcos4 x � sin4 x

cos 4xtan 3xsin 4xcos 3x

1

8 sin 29.5° cos 29.5°

1

4�

1

2 sin2 47.1°

tan 34°

2�1 � tan2 34°�tan 51°

1 � tan2 51°

cos2

8�

1

22 cos2 67

1

2

°� 11 � 2 sin2 22

1

2°

1 � 2 sin2 15°2 tan 15°

1 � tan2 15°cos2 15° � sin2 15°

sin � � 0cos � ��3

52�

cos � � 0sin � � ��5

72�,

sin x � 0tan x �5

32x,

cos x � 0tan x � 22x,

sin � � 0cos � � �12

132�

cos � � 0sin � �2

52�

2� x � cos 2x �

2

3x,

2� x � cos 2x � �

5

12x,

�cos 2� �3

4�

�cos 2� �3

5�



41.42.

43.

44.

45.46.47.48.49.50.

51. 52.

53. 54.

55. 56.

59. 60. 61. 3

62.

63.

64.

65. 66.�5

5��7

�50 � 15�10

10

�50 � 10�5

10

��50 � 20�6

10

�13

4�10

4

�2 � �3

2�

�2 � �3

2

2 � �3��2 � �3

2

��2 � �3

2�2 � �2

2

2 cos 6x sin 3x2 cos 6x cos 2x2 cos 98.5° sin 3.5°�2 sin 11.5° cos 36.5°2 cos 6.5x cos 1.5x�2 sin 3x sin x

1

2 cos x �

1

2 cos 9x

5

2 cos 5x �

5

2 cos x

sin 225° � sin 55°sin 160° � sin 44°

9

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

Write each expression as a sum or difference of trigonometric functions. See Example 6.

41. 42.

43. 44.

Write each expression as a product of trigonometric functions. See Example 7.

45. 46. 47.

48. 49. 50.

Use a half-angle identity to find each exact value. See Examples 8 and 9.

51. 52. 53.

54. 55. 56.

57. Explain how you could use an identity of this section to find the exact value of(Hint: .)

58. The identity can be used to find and

the identity can be used to find . Show thatthese answers are the same, without using a calculator. (Hint: If and and then )

Find each of the following. See Example 10.

59. given with

60. given with

61. given with

62. given with

63. given , with

64. given with

65. given with

66. given with 90° � � � 180°tan � � ��5

2,cot

�

2,

180° � � � 270°tan � ��7

3,tan

�

2,

2� x � cot x � �3,cos

x

2,

0 � x �

2tan x � 2sin

x

2,

180° � � � 270°sin � � �1

5,cos

�

2,

90° � � � 180°sin � �3

5,tan

�

2,

2� x � cos x � �

5

8,sin

x

2,

0 � x �

2cos x �

1

4,cos

x

2,

a � b.a2 � b2,b � 0a � 0

tan 22.5° � �2 � 1tan A2 � sin A1 � cos A

tan 22.5° � �3 � 2�2,tan A2 � ��1 � cos A1 � cos A

7.5 � 12�1

2��30�sin 7.5°.

sin 165°cos 165°tan 195°

cos 195°sin 195°sin 67.5°

sin 9x � sin 3xcos 4x � cos 8xsin 102° � sin 95°

sin 25° � sin��48°�cos 5x � cos 8xcos 4x � cos 2x

sin 4x sin 5x5 cos 3x cos 2x

2 cos 85° sin 140°2 sin 58° cos 102°

sin 2x

2 sin x� cos2

x

2� sin2

x

2sin2

x

2�

tan x � sin x

2 tan x

cot2 x

2�

�1 � cos x�2

sin2 xsec2

x

2�

2

1 � cos x

cot x � tan x

cot x � tan x� cos 2xtan x � cot x � 2 csc 2x

cos 2x �1 � tan2 x

1 � tan2 xsin 2x cos 2x � sin 2x � 4 sin3 x cos x

sin 4x � 4 sin x cos x � 8 sin3 x cos x2 cos 2x

sin 2x� cot x � tan x

1 � cos 2x

sin 2x� cot xsin 4x � 4 sin x cos x cos 2x


Use an identity to write each expression with a single trigonometric function. SeeExample 11.

67. 68. 69.

70. 71. 72.

73. Use the identity to derive the equivalent identity by multiplying both the numerator and denominator by

74. Consider the expression

(a) Why can’t we use the identity for to express it as a function of x alone?(b) Use the identity to rewrite the expression in terms of sine and

cosine.(c) Use the result of part (b) to show that

(Modeling) Mach Number An airplane flying fasterthan sound sends out sound waves that form a cone, asshown in the figure. The cone intersects the ground toform a hyperbola. As this hyperbola passes over a par-ticular point on the ground, a sonic boom is heard at thatpoint. If is the angle at the vertex of the cone, then

where m is the Mach number for the speed of the plane. (We assume .) The Machnumber is the ratio of the speed of the plane and the speed of sound. Thus, a speed ofMach 1.4 means that the plane is flying at 1.4 times the speed of sound. In Ex-ercises 75–78, one of the values or m is given. Find the other value.

75. 76. 77. 78.

(Modeling) Solve each problem. See Example 5.

79. Railroad Curves In the United States, circular rail-road curves are designated by the degree of curvature,the central angle subtended by a chord of 100 ft. Seethe figure. (Source: Hay, W. W., Railroad Engineering,John Wiley & Sons, 1982.)

(a) Use the figure to write an expression for (b) Use the result of part (a) and the third half-angle

identity for tangent to write an expression for

(c) If what is the measure of angle to the nearest degree?

80. Distance Traveled by a Stone The distance Dof an object thrown (or propelled) from height h(feet) at angle with initial velocity v is modeledby the formula

See the figure. (Source: Kreighbaum, E. and K. Barthels, Biomechanics, Allyn &Bacon, 1996.) Also see the Chapter 5 Quantitative Reasoning.

(a) Find D when that is, when the object is propelled from the ground.(b) Suppose a car driving over loose gravel kicks up a small stone at a velocity of

36 ft per sec (about 25 mph) and an angle . How far will the stone travel?� � 30°

h � 0;

D �v2 sin � cos � � v cos � ��v sin ��2 � 64h

32.

�

�b � 12,

tan �4 .

cos �2 .

� � 60°� � 30°m �5

4m �

3

2

�

m � 1

sin �

2�

1

m,

�

tan�2 � x� � �cot x.

tan � � sin �cos �

tan�A � B�tan�

2 � x�.1 � cos A.

tan A2 � 1 � cos Asin Atan A2 � sin A

1 � cos A

sin 158.2°

1 � cos 158.2°

1 � cos 59.74°

sin 59.74°�1 � cos 165°

1 � cos 165°

�1 � cos 147°

1 � cos 147°�1 � cos 76°

2�1 � cos 40°

2


67. 68.69. 70.71. 72.

74. (a) is undefined, so it

cannot be used.

(b)

75. 84° 76. 106° 77. 3.9

78. 2 79. (a)

(b) (c) 54°

80. (a)

(b) approximately 35 ft

D �v2 sin 2�

32

tan �

4�

b

50

cos �

2�

R � b

R

tan�

2� x� �

sin�

2� x�

cos�

2� x�

tan

2

tan 79.1°tan 29.87°cot 82.5°tan 73.5°

cos 38°sin 20°

50 50

b

Rθ2

θ2

h

θ

D

�


81. Wattage Consumption Refer to Example 5. Use an identity to determine values ofa, c, and so that Check your answer by graphing both expres-sions for W on the same coordinate axes.

82. Amperage, Wattage, and Voltage Amperage is a measure of the amount of elec-tricity that is moving through a circuit, whereas voltage is a measure of the forcepushing the electricity. The wattage W consumed by an electrical device can bedetermined by calculating the product of the amperage I and voltage V. (Source:Wilcox, G. and C. Hesselberth, Electricity for Engineering Technology, Allyn &Bacon, 1970.)

(a) A household circuit has voltage when an incandescentlightbulb is turned on with amperage Graph the wattage

consumed by the lightbulb in the window by (b) Determine the maximum and minimum wattages used by the lightbulb.(c) Use identities to determine values for a, c, and so that (d) Check your answer by graphing both expressions for W on the same coordinate axes.(e) Use the graph to estimate the average wattage used by the light. For how many

watts do you think this incandescent lightbulb is rated?

W � a cos��t� � c.�

��50, 300.�0, .05W � VII � 1.23 sin�120t�.

V � 163 sin�120t�

W � a cos��t� � c.�


81. , ,

82. (a)

(b) maximum: 200.49 watts;minimum: 0 watts(c) , ,

(e) 100.245 wattsc � 100.245� � 240a � �100.245

300

–50

0 .05

For x = t, W = VI =(163 sin 120�t)(1.23 sin 120�t)

� � 240c � 885.6a � �885.6

These summary exercises provide practice with the various types of trigonometric iden-tities presented in this chapter. Verify that each equation is an identity.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.tan � � cot �

tan � � cot �� 1 � 2 cos2 �

cot s � tan s

cos s � sin s�

cos s � sin s

sin s cos s

tan� x

2�

4 � � sec x � tan xtan 4� �2 tan 2�

2 � sec2 2�

sin s

1 � cos s�

1 � cos s

sin s� 2 csc s

tan2 t � 1

tan t csc2 t� tan t

cos 2x �2 � sec2 x

sec2 xcos x �

1 � tan2 x21 � tan2 x2

csc4 x � cot4 x �1 � cos2 x

1 � cos2 x

sin � � tan �

1 � cos �� tan �

1 � tan2 �

2�

2 cos �

1 � cos �

sin�x � y�cos�x � y�

�cot x � cot y

1 � cot x cot y

1

sec t � 1�

1

sec t � 1� 2 cot t csc tcot � � tan � �

2 cos2 � � 1

sin � cos �

2

1 � cos x� tan2

x

2� 1sin 2� �

2 tan �

1 � tan2 �

1 � sin t

cos t�

1

sec t � tan t

sin t

1 � cos t�

1 � cos t

sin t

sec� � x� � �sec xtan x

2� csc x � cot x

csc � cos2 � � sin � � csc �tan � � cot � � sec � csc �

Summary Exercises on Verifying Trigonometric Identities


23. 24.

25. 26.

27. 28.1

2 cot

x

2�

1

2 tan

x

2� cot x

2�sin x � sin3 x�cos x

� sin 2x

csc t � 1

csc t � 1� �sec t � tan t�2cos4 x � sin4 x

cos2 x� 1 � tan2 x

2 cos2 x

2 tan x � tan x � sin x

tan�x � y� � tan y

1 � tan�x � y� tan y� tan x

7.5 Inverse Circular Functions 645

Inverse Functions We first discussed inverse functions in Section 4.1. Wegive a quick review here for a pair of inverse functions f and

1. If a function f is one-to-one, then f has an inverse function

2. In a one-to-one function, each x-value corresponds to only one y-value andeach y-value corresponds to only one x-value.

3. The domain of f is the range of and the range of f is the domain of

4. The graphs of f and are reflections of each other about the line

5. To find from follow these steps.

Step 1 Replace with y and interchange x and y.

Step 2 Solve for y.

Step 3 Replace y with

In the remainder of this section, we use these facts to develop the inversecircular (trigonometric) functions.

Inverse Sine Function From Figure 8 and the horizontal line test, we seethat does not define a one-to-one function. If we restrict the domain tothe interval which is the part of the graph in Figure 8 shown in color,this restricted function is one-to-one and has an inverse function. The range of

is so the domain of the inverse function will be and its range will be

Figure 8

y

1

(0, 0)–1

–2 y = sin x

Restricted domain

x–2� 2�–� �– 3�

2

3�2

– �2

�2

[– , ]2�

2�

(– , –1)2�

( , 1)2�

��2 ,

2 .��1, 1,��1, 1,y � sin x

��2 ,

2 ,y � sin x

f �1�x�.

f�x�f�x�,f �1�x�

y � x.f �1

f �1.f �1,

f �1.

f �1.

7.5 Inverse Circular FunctionsInverse Functions ■ Inverse Sine Function ■ Inverse Cosine Function ■ Inverse Tangent Function ■

Remaining Inverse Circular Functions ■ Inverse Function Values

Looking Ahead to CalculusThe inverse circular functions are used

in calculus to solve certain types of re-

lated-rates problems and to integrate

certain rational functions.

TEACHING TIP Mention that the inter-

val contains enough of

the graph of the sine function toinclude all possible values of y. Whileother intervals could also be used,this interval is an accepted conventionthat is adopted by scientific calcula-tors and graphing calculators.

�

2,

2 �



Algebraic Solution

(a) The graph of the function defined by (Figure 9) includes the point Thus,

Alternatively, we can think of as “y is the number in whose sine is ” Then we can write the given equation as Since and is in the range of the arcsine function,

(b) Writing the equation in the formshows that This can be veri-

fied by noticing that the point is on thegraph of

(c) Because is not in the domain of the inversesine function, does not exist.

Graphing Calculator Solution

To find these values with a graphing calculator, wegraph and locate the points with X-values

and Figure 10(a) shows that when Similarly, Figure 10(b) shows

that when

(a) (b)

Figure 10

Since does not exist, a calculator will givean error message for this input.

sin�1��2�

2�

–2

–1 1

–

2�

2�

–1 1

Y � �2 � �1.570796.X � �1,

Y � 6 � .52359878.

X � 12 ,�1.1

2

Y1 � sin�1 X

sin�1��2��2

y � sin�1 x.��1, �

2 �

y � �2 .sin y � �1

y � sin�1��1�

y � 6 .

6sin 6 � 1

2

sin y � 12 .

12 .��

2 ,

2 y � arcsin 12

arcsin 1

2�

6.

�12 ,

6 �.y � arcsin x

Reflecting the graph of on the restricted domain across the line gives the graph of the inverse function, shown in Figure 9. Some key points are labeled on the graph. The equation of the inverse of isfound by interchanging x and y to get This equation is solved for y bywriting (read “inverse sine of x”). As Figure 9 shows, the domain of

is while the restricted domain of is therange of An alternative notation for is

Inverse Sine Function

or means that for

We can think of or as “y is the number in the inter-val whose sine is x.” Just as we evaluated by writing it in ex-ponential form as (Section 4.3), we can write as to evaluate it. We must pay close attention to the domain and range intervals.

EXAMPLE 1 Finding Inverse Sine Values

Find y in each equation.

(a) (b) (c) y � sin�1��2�y � sin�1��1�y � arcsin 1

2

sin y � xy � sin�1 x2y � 4y � log2 4��

2 ,

2 y � arcsin xy � sin�1 x

�2 � y �

2 .x � sin y,y � arcsin xy � sin�1 x

arcsin x.sin�1 xy � sin�1 x.��

2 ,

2 ,y � sin x,��1, 1,y � sin�1 xy � sin�1 x

x � sin y.y � sin x

y � xy � sin x


C A U T I O N In Example 1(b), it is tempting to give the value of as since Notice, however, that is not in the range of the inverse

sine function. Be certain that the number given for an inverse function value isin the range of the particular inverse function being considered.

32sin 3

2 � �1.32 ,

sin�1��1�

1

0

y = sin–1 x or y = arcsin x

–1

(0, 0)

y

x

(1, )2�

(–1, – ) –

�2

�2

�

3( , )√32

�

4( , )√22

�

4(– , – )√22

�

3(– , – )√32

�

612( , )

�

612(– , – )

2�

Figure 9


Our observations about the inverse sine function from Figure 9 lead to thefollowing generalizations.

INVERSE SINE FUNCTIONy � sin�1x or y � arcsin x

Domain: Range:

Figure 11

• The inverse sine function is increasing and continuous on its domain

• Its x-intercept is 0, and its y-intercept is 0.

• Its graph is symmetric with respect to the origin; it is an odd function.

Inverse Cosine Function The function (or ) is de-fined by restricting the domain of the function to the interval asin Figure 12, and then interchanging the roles of x and y. The graph of

is shown in Figure 13. Again, some key points are shown on thegraph.

Figure 12 Figure 13

Inverse Cosine Function

or means that for 0 � y � .x � cos y,y � arccos xy � cos�1 x

y

x10

y = cos–1 x or y = arccos x

(1, 0)–1

(–1, �)�

�

6( , )√32

�

4( , )√22

�

312( , )

2�

3

(– , )√22

3�

4

(– , )√32

5�

6

12(– , )

(0, )�2

y

–1

0

y = cos xRestricted domain

[0, �]

(0, 1)

(�, –1)

1

x( , 0)�2

�– �2

�2

y � cos�1 x

�0, y � cos xy � arccos xy � cos�1 x

��1, 1.

–

2�

2�

–1 1

y = sin–1 x

–

x

y

y = sin–1 x

–1 0 1

–

�2

�2

��2 ,

2 ��1, 1


x y

0 0

1 2

4

�22

�4�

�22

�2�1

TEACHING TIP In problems likeExample 1(a), use phrases such as “y is a value in radians between

and whose sine is equal

to to help students understand

the meaning of inverse sine.Illustrate the answer graphically by

graphing and in

the interval then

determine the point of intersection.

�

2,

2 �,

y �12

y � sin x

12

�

2�

2



–1

–.5

1

5�4

These screens support the re-sults of Example 2, since

≈ 2.3561945.3�4

–1

–.5

1

5�4

x y

0

1 0

4

�22

2

34�

�22

�1

EXAMPLE 2 Finding Inverse Cosine Values

Find y in each equation.

(a) (b)

Solution

(a) Since the point lies on the graph of in Figure 13 on the previous page, the value of y is 0. Alternatively, we can think of

as “y is the number in whose cosine is 1,” or Then since and 0 is in the range of the arccosine function.

(b) We must find the value of y that satisfies where y is in the in-terval the range of the function The only value for y thatsatisfies these conditions is Again, this can be verified from the graph inFigure 13.


Our observations about the inverse cosine function lead to the followinggeneralizations.

INVERSE COSINE FUNCTIONy � cos�1 x or y � arccos x

Domain: Range:

Figure 14

• The inverse cosine function is decreasing and continuous on its domain

• Its x-intercept is 1, and its y-intercept is

• Its graph is not symmetric with respect to the y-axis or the origin.

Inverse Tangent Function Restricting the domain of the function to the open interval yields a one-to-one function. By interchanging theroles of x and y, we obtain the inverse tangent function given by or

Figure 15 shows the graph of the restricted tangent function.Figure 16 gives the graph of y � tan�1 x.y � arctan x.

y � tan�1 x��

2 ,

2 �y � tan x

2 .

��1, 1.

�

0–1 1

y = cos–1 x

x

y

y = cos–1 x

–1 0 1

�

�0, ��1, 1

34 .

y � cos�1 x.�0, ,cos y � ��2

2 ,

cos 0 � 1y � 0,cos y � 1.�0, y � arccos 1

y � arccos x�1, 0�

y � cos�1��2

2 �y � arccos 1


Figure 15 Figure 16

y

x–1 (0, 0)

y = tan–1 x or y = arctan x

1

–1

1 2

–2�

6( , )√33

(1, )

(–1, – )

(√3 , )

�

3(–√3 , – )

�

6(– , – )√33

–

�2

�2

4�

4�

3�

y

x0

y = tan x

(0, 0)

Restricted domain

( , 1)4�

(– , –1)4�

– �2

�2

(– , )2�

2�


Inverse Tangent Function

or means that for

INVERSE TANGENT FUNCTIONy � tan�1 x or y � arctan x

Domain: Range:

Figure 17

• The inverse tangent function is increasing and continuous on its domain

• Its x-intercept is 0, and its y-intercept is 0.

• Its graph is symmetric with respect to the origin; it is an odd function.

• The lines and are horizontal asymptotes.

Remaining Inverse Circular Functions The remaining three inversetrigonometric functions are defined similarly; their graphs are shown inFigure 18. All six inverse trigonometric functions with their domains and rangesare given in the table on the next page.

y � �2y �

2

��, ��.

–

2�

2�

–3 3

y = tan–1 x

x

y

y = tan–1 x

–1–2 0 1 2

–

�2

�2

��2 ,

2 ��, ��

�2 � y �

2 .x � tan y,y � arctan xy � tan�1 x

x y

0 0

1 4

6

�33

�6�

�33

�4�1

The first three screens showthe graphs of the three re-maining inverse circular func-tions. The last screen showshow they are defined.

–4 40

�

–

2�

2�

–4 4

�

0–4 4

Figure 18



Range

Inverse Quadrants of theFunction Domain Interval Unit Circle

I and IV

I and II

I and IV

I and II

* I and II

* I and IV��2 ,

2 , y � 0��, �1 � �1, ��y � csc�1 x

�0, , y � 2��, �1 � �1, ��y � sec�1 x

�0, ��, ��y � cot�1 x

��2 ,

2 ��, ��y � tan�1 x

�0, ��1, 1y � cos�1 x

��2 ,

2 ��1, 1y � sin�1 x

Inverse Function Values The inverse circular functions are formally de-fined with real number ranges. However, there are times when it may be conve-nient to find degree-measured angles equivalent to these real number values. It isalso often convenient to think in terms of the unit circle and choose the inversefunction values based on the quadrants given in the preceding table.

EXAMPLE 3 Finding Inverse Function Values (Degree-Measured Angles)

Find the degree measure of in the following.

(a) (b)

Solution

(a) Here must be in but since must be in quadrant I. Thealternative statement, leads to

(b) Write the equation as For is in quadrant I or II. Because 2 is positive, is in quadrant I and since Note that 60°the degree equivalent of is in the range of the inverse secant function.


The inverse trigonometric function keys on a calculator give results in the proper quadrant for the inverse sine, inverse cosine, and inverse tangentfunctions, according to the definitions of these functions. For example, on a cal-culator, in degrees, and

Finding and with a calculator is not as straightfor-ward, because these functions must be expressed in terms of and

respectively. If for example, then which must bewritten as a cosine function as follows:

If then or and y � cos�1 1

x.cos y �

1

x,

1

cos y� xsec y � x,

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

csc�1 xsec�1 x,cot�1 x,cos�1��.5� � 120�.

tan�1��1� � �45�,sin�1��.5� � �30�,sin�1 .5 � 30�,

3 ��

sec 60� � 2.� � 60�,��sec�1 x,sec � � 2.

� � 45�.tan � � 1,�1 � 0,��90�, 90��,�

� � sec�1 2� � arctan 1

�

*The inverse secant and inverse cosecant functions are sometimes defined with different ranges. We useintervals that match their reciprocal functions (except for one missing point).



Figure 19

y

x0

3

2�

� = tan–123

√13

Figure 20

In summary, to find we find Similar statements apply to and There is one additional consideration with Since we take theinverse tangent of the reciprocal to find inverse cotangent, the calculator givesvalues of inverse cotangent with the same range as inverse tangent,which is not the correct range for inverse cotangent. For inverse cotangent, theproper range must be considered and the results adjusted accordingly.

EXAMPLE 4 Finding Inverse Function Values with a Calculator

(a) Find y in radians if

(b) Find in degrees if

Solution

(a) With the calculator in radian mode, enter as to getSee Figure 19.

(b) Set the calculator to degree mode. A calculator gives the inverse tangentvalue of a negative number as a quadrant IV angle. The restriction on therange of arccotangent implies that must be in quadrant II, so enter

as

As shown in Figure 19,


EXAMPLE 5 Finding Function Values Using Definitions of the Trigonometric Functions

Evaluate each expression without using a calculator.

(a) (b)

Solution

(a) Let so The inverse tangent function yields valuesonly in quadrants I and IV, and since is positive, is in quadrant I. Sketch

in quadrant I, and label a triangle, as shown in Figure 20. By thePythagorean theorem, the hypotenuse is . The value of sine is the quo-tient of the side opposite and the hypotenuse, so

(Section 5.3)

(b) Let Then, Since for a negativevalue of x is in quadrant II, sketch A in quadrant II, as shown in Figure 21.


tan�cos�1��5

13�� tan A � �12

5

cos�1 xcos A � �5

13 .A � cos�1��5

13�.

sin�tan�1 3

2� � sin � �3

�13�

3�13

13.

�13�

�32

tan � � 32 .� � tan�1 32,

tan�cos�1��5

13��sin�tan�1 3

2�

� � 109.4990544�.

tan�1� 1

�.3541� � 180�.arccot��.3541�

�

y � �.3398369095.sin�1��

13�csc�1��3�

� � arccot��.3541�.�

y � csc�1��3�.

��2 ,

2 �,

cot�1 x.cot�1 x.csc�1 xcos�1 1x .sec�1 x,

y

x0

13

–5

12

A

A = cos–1(– )513

Figure 21



y

x0

13

B

2√2

y

x0

2√3

1

A

Figure 22

y

x0

25

B

√21

Figure 23

TEACHING TIP Point out that theequations

andare true wherever

they are defined. However,

and are true onlyfor values of x in the restricteddomains of the sine, cosine, andtangent functions.

tan�1�tan x� � xcos�1�cos x� � x,sin�1�sin x� � x,

tan�tan�1x� � xcos�cos�1x� � x,

sin�sin�1x� � x,

EXAMPLE 6 Finding Function Values Using Identities

Evaluate each expression without using a calculator.

(a) (b)

Solution

(a) Let and so and Sketch both A and B in quadrant I, as shown in Figure 22. Now, use the co-sine sum identity.

(Section 7.3)

(1)

From Figure 22,

Substitute these values into equation (1) to get

(b) Let Then, from the double-angle tangent identity,

(Section 7.4)

Since Sketch a triangle in quadrant I, find thelength of the third side, and then find tan B. From the triangle in Figure 23,

and


While the work shown in Examples 5 and 6 does not rely on a calculator, wecan support our algebraic work with one. By entering from Example 6(a) into a calculator, we get the approximation .1827293862, the

same approximation as when we enter (the exact value we obtainedalgebraically). Similarly, we obtain the same approximation when we evaluate

and , supporting our answer in Example 6(b).4�21

17tan�2 arcsin 25�

2�2 � �36

cos�arctan �3 � arcsin 13�

tan�2 arcsin 2

5� �2� 2

�21�1 � � 2

�21�2�

4�21

1 �4

21�

4�21

17.

tan B � 2�21

,

sin B � 25 .arcsin 25 � B,

tan�2 arcsin 2

5� � tan 2B �2 tan B

1 � tan2 B.

arcsin 25 � B.

cos�arctan �3 � arcsin 1

3� �1

2

2�2

3�

�3

2

1

3�

2�2 � �3

6.

sin�arcsin 1

3� � sin B �1

3. sin�arctan �3 � � sin A �

�3

2,

cos�arcsin 1

3� � cos B �2�2

3, cos�arctan �3 � � cos A �

1

2,

� sin�arctan �3 � sin�arcsin 1

3� cos�arctan �3 � arcsin

1

3� � cos�arctan �3 � cos�arcsin 1

3� cos�A � B� � cos A cos B � sin A sin B

sin B � 13 .tan A � �3B � arcsin 13 ,A � arctan �3

tan�2 arcsin 2

5�cos�arctan �3 � arcsin 1

3�



y

x0

u, u > 0

u, u < 0

1��

√u2 + 1

√u2 + 1

Figure 24

EXAMPLE 7 Writing Function Values in Terms of u

Write each trigonometric expression as an algebraic expression in u.

(a) (b)

Solution

(a) Let so Here, u may be positive or negative. Sincesketch in quadrants I and IV and label two triangles,

as shown in Figure 24. Since sine is given by the quotient of the side oppo-site and the hypotenuse,

The result is positive when u is positive and negative when u is negative.

(b) Let so To find use the identity


EXAMPLE 8 Finding the Optimal Angle of Elevation of a Shot Put

The optimal angle of elevation a shot-putter should aim for to throw the great-est distance depends on the velocity v of the throw and the initial height h of theshot. See Figure 25. One model for that achieves this greatest distance is

(Source: Townend, M. Stewart, Mathematics in Sport, Chichester, EllisHorwood Limited, 1984.)

Figure 25

Suppose a shot-putter can consistently throw the steel ball with andper sec. At what angle should he throw the ball to maximize distance?

Solution To find this angle, substitute and use a calculator in degree mode.


h � 6.6, v � 42� � arcsin�� 422

2�422� � 64�6.6�� 41.9�

v � 42 fth � 6.6 ft

h

D

�

� � arcsin�� v2

2v2 � 64h�.

�

�

cos�2 sin�1 u� � cos 2� � 1 � 2 sin2 � � 1 � 2u2

1 � 2 sin2 �.cos 2� �cos 2�,sin � � u.� � sin�1 u,

sin�tan�1 u� � sin � �u

�u2 � 1�

u�u2 � 1

u2 � 1.

��2 � tan�1 u �

2,

tan � � u.� � tan�1 u,

cos�2 sin�1 u�sin�tan�1 u�



Concept Check Complete each statement.

1. For a function to have an inverse, it must be .

2. The domain of equals the of .

3. The range of equals the of .

4. The point lies on the graph of . Therefore, the point lies on the graph of .

5. If a function f has an inverse and , then .

6. How can the graph of be sketched if the graph of f is known?

Concept Check In Exercises 7–10, write short answers.

7. Consider the inverse sine function, defined by or .

(a) What is its domain? (b) What is its range?(c) Is this function increasing or decreasing?(d) Why is not defined?

8. Consider the inverse cosine function, defined by or .

(a) What is its domain? (b) What is its range?(c) Is this function increasing or decreasing?

(d) . Why is not equal to ?

9. Consider the inverse tangent function, defined by or .

(a) What is its domain? (b) What is its range?(c) Is this function increasing or decreasing?(d) Is there any real number x for which arctan x is not defined? If so, what is it (or

what are they)?

10. Give the domain and range of the three other inverse trigonometric functions, asdefined in this section.

(a) inverse cosecant function (b) inverse secant function(c) inverse cotangent function

11. Concept Check Is calculated as or as ?

12. Concept Check For positive values of a, is calculated as . How iscalculated for negative values of a?

Find the exact value of each real number y. Do not use a calculator. See Examples 1and 2.

13. 14. 15.

16. 17. 18.

19. 20. 21.

22. 23. 24.

25. 26. 27.

28. 29. 30.

31. 32. y � csc�1 �2y � arcsec 2�3

3

y � arccot��3 �y � csc�1��2�y � sec�1��2 �

y � cot�1��1�y � arcsin��2

2 �y � arccos��3

2 �y � cos�1��

1

2�y � sin�1 �2

2y � tan�1��1�

y � arccos 0y � arcsin��3

2 �y � arctan 0

y � cos�1 1

2y � sin�1��1�y � arctan��1�

y � cos�1��1�y � tan�1 1y � sin�1 0

cot�1 atan�1 1acot�1 a

1cos�1 acos�1 1asec�1 a

y � arctan xy � tan�1 x

�43arccos��

12�Arccos��

12� � 2

3

y � arccos xy � cos�1 x

arcsin��2�

y � arcsin xy � sin�1 x

f �1

f �1��1� �f �� 1

y � tan x�4 , 1�

y � cos xy � cos�1 x

y � sin xy � arcsin x

1. one-to-one 2. range3. domain

4. ; or

5.6. Sketch the reflection of thegraph of f across the line .

7. (a) (b)

(c) increasing (d) �2 is not in thedomain.8. (a) (b)(c) decreasing

(d) is not in the range.

9. (a) (b)

(c) increasing (d) no10. (a) ;

(b) ;

(c) ;

11. 12. Find

(or 180°).

13. 0 14. 15. 16.

17. 18. 19. 0

20. 21. 22.

23. 24. 25.

26. 27. 28.

29. 30. 31.

32. 33. �45� 34. 120�

35. �60� 36. �45� 37. 120�

4

6

5

6�

6

3

4

3

4�

4

5

6

2

3

4

�

4

2�

3

3�

2

�

4

4

tan�1 1

a�

cos�1 1

a

�0, ��, ��

0,

2 � � �

2, �

��, �1 � �1, ��

�

2, 0� � �0,

2 ��, �1 � �1, ��

��

2,

2 ��, ��

�4

3

�0, ��1, 1

�

2,

2 ��1, 1

y � x

y � arctan x

y � tan�1 x�1,

4 �

7.5 Exercises



38. �30� 39. 120� 40. �90�41. �7.6713835�42. 97.671207�43. 113.500970�44. 51.1691219�45. 30.987961� 46. 29.506181�47. .83798122 48. .9601269849. 2.3154725 50. 2.460522151. 1.1900238 52. 1.108230353. 54.

55. 56.

57.

58. 1.003 is not in the domain of.

59. The domain of is.

60. In both cases, the result is x. Ineach case, the graph is a straightline bisecting quadrants I and III(i.e., the line ).61. It is the graph of .

62. It does not agree because therange of the inverse tangent

function is , not

, as was the case in Exercise 61.

10

–10

–10 10

��, ��

��

2,

2 �

10

–10

–10 10

y � xy � x

��, ��y � tan�1 x

y � sin�1 x

–2 2x

y

0

�

y = arcsec x12�

2

–1

10x

y

�2

–

y = arccsc 2x

x

y

y = sec–1 x

–10 1

�

–2

2x

y

y = csc–1 x

�2

�2

–

0

x

y

y = cot–1 x

–10 1

�

*The authors wish to thank Carol Walker of Hinds Community College for making a suggestion onwhich these exercises are based.

Give the degree measure of . Do not use a calculator. See Example 3.

33. 34. 35.

36. 37. 38.

39. 40.

Use a calculator to give each value in decimal degrees. See Example 4.

41. 42.

43. 44.

45. 46.

Use a calculator to give each real number value. (Be sure the calculator is in radianmode.) See Example 4.

47. 48.

49. 50.

51. 52.

Graph each inverse function as defined in the text.

53. 54. 55.

56. 57.

58. Explain why attempting to find on your calculator will result in an errormessage.

59. Explain why you are able to find on your calculator. Why is this situ-ation different from the one described in Exercise 58?


(Exercises 60–62)*

60. Consider the function defined by and its inverse .Simplify and . What do you notice in each case? What would thegraph look like in each case?

61. Use a graphing calculator to graph in the standard viewing win-dow, using radian mode. How does this compare to the graph you described inExercise 60?

62. Use a graphing calculator to graph in the standard viewing window,using radian and dot modes. Why does this graph not agree with the graph you foundin Exercise 61?

y � tan�1�tan x�

y � tan�tan�1 x�

f �1� f �x�f � f �1�x�f �1�x� � x � 2

3f �x� � 3x � 2

tan�1 1.003

sin�1 1.003

y � arcsec 1

2xy � arccsc 2x

y � sec�1 xy � csc�1 xy � cot�1 x

y � arccos .44624593y � arcsin .92837781

y � sec�1��1.2871684�y � cot�1��.92170128�y � arcsin .81926439y � arctan 1.1111111

� � cot�1 1.7670492� � csc�1 1.9422833

� � arcsin .77900016� � arccos��.39876459�� cos�1��.13348816�� sin�1��.13349122�

� � csc�1��1�� sec�1��2�

� � csc�1��2�� cot�1��3

3 �� arcsin��2

2 �� arcsin��

�3

2 �� arccos��1

2�� arctan��1�

�



63. 64. 65.

66. 67. 68.

69. 70.

71. 72. 73. 2

74. 75. 76.

77.

78.

79. .89442719180. .968245836681. .123439981182. .716386406 83.

84. 85.

86. 87.

88. 89.

90. 91.

92.

93. (a) 45� (b)94. (a) 113� (b) 84� (c) 60�(d) 47�

� � 45°

3�9 � u2

9 � u2

2�4 � u2

4 � u2�u2 � 5

u

u�2

2

u�u2 � 9

u2 � 9

�u2 � 4

u�1 � u2

u

�1 � u2�1 � u2

u

�1 � u2

48 � 25�3

39

�10 � 3�30

20

�16

65

63

65�2

�3

5

4�6

25

��15

7�

7

25

7

8

120

169

5�6

12

�5

5�15

4�7

3

Give the exact value of each expression without using a calculator. See Examples 5and 6.

63. 64. 65.

66. 67. 68.

69. 70. 71.

72. 73. 74.

75. 76.

77. 78.

Use a calculator to find each value. Give answers as real numbers.

79. 80.

81. 82.

Write each expression as an algebraic (nontrigonometric) expression in u, . SeeExample 7.

83. 84. 85.

86. 87. 88.

89. 90.

91. 92.


93. Angle of Elevation of a Shot Put Refer to Example 8.

(a) What is the optimal angle when ?(b) Fix h at 6 ft and regard as a function of v. As v gets larger and larger, the graph

approaches an asymptote. Find the equation of that asymptote.

94. Angle of Elevation of a Plane Suppose an airplane flying faster than sound goesdirectly over you. Assume that the plane is flying at a constant altitude. At theinstant you feel the sonic boom from the plane, the angle of elevation to the planeis given by

,

where m is the Mach number of the plane’s speed. (The Mach number is the ratio ofthe speed of the plane and the speed of sound.) Find to the nearest degree for eachvalue of m.

(a) (b) (c) (d) m � 2.5m � 2m � 1.5m � 1.2

�

� � 2 arcsin 1

m

�h � 0

csc�arctan �9 � u2

u �sec�arccot �4 � u2

u �sec�cos�1

u

�u2 � 5�tan�sin�1

u

�u2 � 2�

cos�tan�1 3

u�sin�sec�1 u

2�cot�arcsin u�

cos�arcsin u�tan�arccos u�sin�arccos u�

u � 0

cot�arccos .58236841�tan�arcsin .12251014�sin�cos�1 .25�cos�tan�1 .5�

tan�cos�1 �3

2� sin�1��

3

5��sin�sin�1 1

2� tan�1��3��

cos�sin�1 3

5� cos�1

5

13�cos�tan�1 5

12� tan�1

3

4�csc�csc�1 �2 �sec�sec�1 2�cos�2 tan�1��2��

sin�2 cos�1 1

5�tan�2 cos�1 1

4�cos�2 arctan 4

3�cos�2 sin�1

1

4�sin�2 tan�1 12

5 �sec�sin�1��1

5��cos�tan�1��2��sin�arccos

1

4�tan�arccos 3

4�



95. Observation of a Painting A painting 1 m high and 3 m from the floor will cut off an angle to anobserver, where

.

Assume that the observer is x meters from the wallwhere the painting is displayed and that the eyes ofthe observer are 2 m above the ground. (See the fig-ure.) Find the value of for the following values of x. Round to the nearest degree.

(a) 1 (b) 2 (c) 3(d) Derive the formula given above. (Hint: Use the identity for . Use

right triangles.)(e) Graph the function for with a graphing calculator, and determine the distance

that maximizes the angle.(f) The idea in part (e) was first investigated in 1471 by the astronomer

Regiomontanus. (Source: Maor, E., Trigonometric Delights, PrincetonUniversity Press, 1998.) If the bottom of the picture is a meters above eyelevel and the top of the picture is b meters above eye level, then the optimum

value of x is meters. Use this result to find the exact answer to part (e).

96. Landscaping Formula A shrub is planted in a 100-ft-wide space between build-ings measuring 75 ft and 150 ft tall. The location of the shrub determines how muchsun it receives each day. Show that if is the angle in the figure and x is the distanceof the shrub from the taller building, then the value of (in radians) is given by

.

97. Communications Satellite Coverage The figure shows a stationary communica-tions satellite positioned 20,000 mi above the equator. What percent of the equatorcan be seen from the satellite? The diameter of Earth is 7927 mi at the equator.

�

100 ft

75 ft

150 ft

x

�

� � � arctan� 75

100 � x� � arctan�150

x ��

�

�ab

�

tan��

�

� � tan�1� x

x2 � 2��

95. (a) 18� (b) 18� (c) 15�(e) 1.4142151 m (Note: Due tothe computational routine, theremay be a discrepancy in the lastfew decimal places.)

(f)97. about 44.7%

�2

1

–.5

0 10

Radian mode

y = tan–1( )xx2 + 2

1

1

3

x

θα

2



7.6 Trigonometric EquationsSolving by Linear Methods ■ Solving by Factoring ■ Solving by Quadratic Methods ■ Solving by UsingTrigonometric Identities ■ Equations with Half-Angles ■ Equations with Multiple Angles ■ Applications

Looking Ahead to CalculusThere are many instances in calculus

where it is necessary to solve trigono-

metric equations. Examples include

solving related-rates problems and

optimization problems.

(a)

(b)

Figure 26

x

y

0

� = 210°

�' = 30°

x

y

0

� = 330°

�' = 30°

Earlier in this chapter, we studied trigonometric equations that were identities.We now consider trigonometric equations that are conditional; that is, equationsthat are satisfied by some values but not others.

Solving by Linear Methods Conditional equations with trigonometric (orcircular) functions can usually be solved using algebraic methods and trigono-metric identities.

EXAMPLE 1 Solving a Trigonometric Equation by Linear Methods

Solve over the interval

Solution Because sin � is the first power of a trigonometric function, we usethe same method as we would to solve the linear equation

Subtract 1. (Section 1.1)

Divide by 2.

To find values of � that satisfy we observe that � must be ineither quadrant III or IV since the sine function is negative only in these twoquadrants. Furthermore, the reference angle must be 30° since Thegraphs in Figure 26 show the two possible values of �, 210° and 330°. The solu-tion set is

Alternatively, we could determine the solutions by referring to Figure 11 inSection 6.2 on page 546.


Solving by Factoring

EXAMPLE 2 Solving a Trigonometric Equation by Factoring


Solution

Subtract sin x.

Factor. (Section R.4)

or Zero-factor property (Section 1.4)

or or

The solution set is


�0�, 45�, 180�, 225��.

x � 225�x � 45�x � 180�x � 0�

tan x � 1

tan x � 1 � 0sin x � 0

sin x�tan x � 1� � 0

sin x tan x � sin x � 0

sin x tan x � sin x

�0�, 360��.sin x tan x � sin x

�210�, 330��.

sin 30� � 12 .

sin � � �12 ,

sin � � �1

2

2 sin � � �1

2 sin � � 1 � 0

2x � 1 � 0.

�0�, 360��.2 sin � � 1 � 0


7.6 Trigonometric Equations 659

C A U T I O N There are four solutions in Example 2. Trying to solve the equa-tion by dividing each side by sin x would lead to just which wouldgive or The other two solutions would not appear. The miss-ing solutions are the ones that make the divisor, sin x, equal 0. For this reason,we avoid dividing by a variable expression.

Solving by Quadratic Methods In Section 1.6, we saw that an equation inthe form where u is an algebraic expression, is solved byquadratic methods. The expression u may also be a trigonometric function, as inthe equation

EXAMPLE 3 Solving a Trigonometric Equation by Factoring


Solution This equation is quadratic in form and can be solved by factoring.

Factor.

or Zero-factor property

or

The solutions for over the interval are and To solve over that interval, we use a scientific calculator set in

radian mode. We find that This is a quadrant IVnumber, based on the range of the inverse tangent function. (Refer to Figure 11in Section 6.2 on page 546.) However, since we want solutions over the interval

we must first add to �1.1071487, and then add 2.

The solutions over the required interval form the solution set

Exact Approximate valuesvalues to the nearest tenth


EXAMPLE 4 Solving a Trigonometric Equation Using the Quadratic Formula

Find all solutions of

Solution We multiply the factors on the left and subtract 1 to get the equationin standard quadratic form.

(Section 1.4)

Since this equation cannot be solved by factoring, we use the quadratic formula,with and cot x as the variable.c � �1,b � 3,a � 1,

cot2 x � 3 cot x � 1 � 0

cot x�cot x � 3� � 1.

�

4,

5

4, 2.0, 5.2 �.

x � �1.1071487 � 2 � 5.1760366

x � �1.1071487 � � 2.0344439

�0, 2�,

tan�1��2� � �1.1071487.tan x � �2

x � 54 .x �

4�0, 2�tan x � 1

tan x � �2 tan x � 1

tan x � 2 � 0 tan x � 1 � 0

�tan x � 1� �tan x � 2� � 0

tan2 x � tan x � 2 � 0

�0, 2�.tan2 x � tan x � 2 � 0

tan2 x � tan x � 2 � 0.

au2 � bu � c � 0,

x � 225�.x � 45�tan x � 1,



TEACHING TIP Unlike the cosinefunction, equations of the form

will not contain a sec-

ond value for x between and

Remind students that other

solutions to are foundusing period radians.

y � tan x

2.

�

2

y � tan x

*We usually give solutions of equations as solution sets, except when we ask for all solutions of atrigonometric equation.

Divide by rationalize thedenominator. (Section R.7)

2�3;

TEACHING TIP Point out inExample 5 that our first goal is torewrite the equation in terms of asingle trigonometric function.

Use a calculator.

We cannot find inverse cotangent values directly on a calculator, so we use the fact that and take reciprocals to get

To find all solutions, we add integer multiples of the period of the tangentfunction, which is , to each solution found above. Thus, all solutions of theequation are written as

where n is any integer.*


Solving by Using Trigonometric Identities Recall that squaring bothsides of an equation, such as will yield all solutions but mayalso give extraneous values. (In this equation, 0 is a solution, while �3 is extra-neous. Verify this.) The same situation may occur when trigonometric equationsare solved in this manner.

EXAMPLE 5 Solving a Trigonometric Equation by Squaring


Solution Since the tangent and secant functions are related by the identitysquare both sides and express in terms of

(Section R.3)

Pythagorean identity (Section 7.1)

Subtract

The possible solutions are and Now check them. Try first.

Left side:

(Section R.7)

Right side: Not equalsec x � sec 5

6� �

2�3

3

tan x � �3 � tan 5

6� �3 � �

�3

3� �3 �

2�3

3

56

116 .5

6

tan x � �1

�3� �

�3

3

3 � tan2 x. 2�3 tan x � �2

tan2 x � 2�3 tan x � 3 � 1 � tan2 x

�x � y�2 � x2 � 2xy � y2 tan2 x � 2�3 tan x � 3 � sec2 x

tan x � �3 � sec x

tan2 x.sec2 x1 � tan2 x � sec2 x,

�0, 2�.tan x � �3 � sec x

�x � 4 � x � 2,

�.2940013018 � n and 1.276795025 � n,

x � �.2940013018 or x � 1.276795025.

tan x � �.3027756377 or tan x � 3.302775638

cot x � 1tan x ,

cot x � �3.302775638 or cot x � .3027756377

cot x ��3 � �9 � 4

2�

�3 � �13

2

Quadratic formula with

(Section 1.4)c � �1b � 3,a � 1,

TEACHING TIP Explain that when-ever possible, answers should be given in exact form, such as

rather than as decimal

approximations.

11

6,



0

4

–4

2�

y = tan x + √3 – sec x

Dot mode; radian mode

The graph shows that on theinterval [0, 2�), the only x-intercept of the graph of y = tan x + √3 – sec x is5.7595865, which is an ap-

proximation for , the sol-

ution found in Example 5.

11�6

The check shows that is not a solution. Now check

Left side:

Right side: Equal

This solution satisfies the equation, so is the solution set.


The methods for solving trigonometric equations illustrated in the examplescan be summarized as follows.

Solving a Trigonometric Equation

1. Decide whether the equation is linear or quadratic in form, so you candetermine the solution method.

2. If only one trigonometric function is present, first solve the equation forthat function.

3. If more than one trigonometric function is present, rearrange the equa-tion so that one side equals 0. Then try to factor and set each factor equalto 0 to solve.

4. If the equation is quadratic in form, but not factorable, use the quadraticformula. Check that solutions are in the desired interval.

5. Try using identities to change the form of the equation. It may be helpfulto square both sides of the equation first. If this is done, check for extra-neous solutions.

Some trigonometric equations involve functions of half-angles or multiple angles.

Equations with Half-Angles

EXAMPLE 6 Solving an Equation Using a Half-Angle Identity

Solve

(a) over the interval and (b) give all solutions.

Solution

(a) Write the interval as the inequality

The corresponding interval for is

Divide by 2. (Section 1.7) 0 �x

2� .

x2

0 � x � 2.

�0, 2�

�0, 2�,

2 sin x

2� 1

�116 �

sec 11

6�

2�3

3

tan 11

6� �3 � �

�3

3� �3 �

2�3

3

116 .5

6



TEACHING TIP As a slight variation of the problem in Example 6,

replace with u, solve

for u, and then multi-ply the solutions by 2 to find x.2 sin u � 1

x2

The x-intercepts are the solu-tions found in Example 6.

Using Xscl = makes it

possible to support the exactsolutions by counting the tickmarks from 0 on the graph.

0

4

–4

2�

y = 2 sin – 1x2

�3

To find all values of over the interval that satisfy the given equation,first solve for

Divide by 2.

The two numbers over the interval with sine value are and so

Multiply by 2.

The solution set over the given interval is

(b) Since this is a sine function with period 4, all solutions are given by theexpressions

where n is any integer.


Equations with Multiple Angles

EXAMPLE 7 Solving an Equation with a Double Angle


Solution First change to a trigonometric function of x. Use the identityso the equation involves only cos x. Then factor.

Substitute; double-angle identity (Section 7.4)

Subtract cos x.

Factor.

Zero-factor property

Over the required interval,

The solution set is


�0, 23 , 4

3 �.

x �2

3or x �

4

3or x � 0.

cos x � �1

2or cos x � 1

2 cos x � 1 � 0 or cos x � 1 � 0

�2 cos x � 1� �cos x � 1� � 0

2 cos2 x � cos x � 1 � 0

2 cos2 x � 1 � cos x

cos 2x � cos x

cos 2x � 2 cos2 x � 1cos 2x

�0, 2�.cos 2x � cos x

3� 4n and

5

3� 4n,

�3 , 5

3 �.

x �

3or x �

5

3.

x

2�

6or

x

2�

5

6

56 ,

612�0, �

sin x

2�

1

2

2 sin x

2� 1

sin x2 .�0, �x

2



TEACHING TIP Students should beaware that may have asmany as 2a solutions from 0° to360°.

y � sin ax

Figure 27

.006

0 .04

–.006

y = .004 sin(300�x)

C A U T I O N In the solution of Example 7, cos 2x cannot be changed to cos xby dividing by 2 since 2 is not a factor of cos 2x.

The only way to change cos 2x to a trigonometric function of x is by using oneof the identities for cos 2x.

EXAMPLE 8 Solving an Equation Using a Multiple-Angle Identity


Solution The identity is useful here.

(Section 7.4)

Divide by 2.

From the given interval the interval for 2� is List all solutions over this interval.

or Divide by 2.

The final two solutions for 2� were found by adding 360° to 60° and 120°, re-spectively, giving the solution set


Applications Music is closely related to mathematics.

EXAMPLE 9 Describing a Musical Tone from a Graph

A basic component of music is a pure tone. The graph in Figure 27 models thesinusoidal pressure in pounds per square foot from a pure tone at time

in seconds.

(a) The frequency of a pure tone is often measured in hertz. One hertz is equalto one cycle per second and is abbreviated Hz. What is the frequency f inhertz of the pure tone shown in the graph?

(b) The time for the tone to produce one complete cycle is called the period.Approximate the period T in seconds of the pure tone.

(c) An equation for the graph is Use a calculator to esti-mate all solutions to the equation that make over the interval�0, .02.

y � .004y � .004 sin�300x�.

x � ty � P

�30�, 60�, 210�, 240��.

� � 30�, 60�, 210�, 240�

2� � 60�, 120�, 420�, 480�

0� � 2� � 720�.0� � � � 360�,

sin 2� ��3

2

2 sin � cos � � sin 2� 2 sin 2� � �3

4 � 2 2 2�2 sin � cos �� 3

4 sin � cos � � �3

2 sin � cos � � sin 2�

�0�, 360��.4 sin � cos � � �3

cos 2x

2� cos x


Figure 29

.005

–.005

0 .01

P = P1 + P2 + P3 + P4 + P5


Figure 28

.007

–.007

0 .02

Y2 = .004

Y1 = .004 sin(300�X)

Solution

(a) From the graph in Figure 27 on the previous page, we see that there are6 cycles in .04 sec. This is equivalent to per sec. The puretone has a frequency of

(b) Six periods cover a time of .04 sec. One period would be equal to or

(c) If we reproduce the graph in Figure 27 on a calculator as and also graph asecond function as we can determine that the approximate valuesof x at the points of intersection of the graphs over the interval are

The first value is shown in Figure 28.


A piano string can vibrate at more than one frequency when it is struck. Itproduces a complex wave that can mathematically be modeled by a sum of sev-eral pure tones. If a piano key with a frequency of is played, then the corre-sponding string will not only vibrate at but it will also vibrate at the higherfrequencies of is called the fundamental frequencyof the string, and higher frequencies are called the upper harmonics. The human ear will hear the sum of these frequencies as one complex tone. (Source:Roederer, J., Introduction to the Physics and Psychophysics of Music, SecondEdition, Springer-Verlag, 1975.)

EXAMPLE 10 Analyzing Pressures of Upper Harmonics

Suppose that the A key above middle C is played. Its fundamental frequency isand its associated pressure is expressed as

The string will also vibrate at

The corresponding pressures of these upper harmonics are

The graph of

shown in Figure 29, is “saw-toothed.”

(a) What is the maximum value of P?

(b) At what values of x does this maximum occur over the interval �0, .01?

P � P1 � P2 � P3 � P4 � P5,

P4 �.002

4 sin�3520t�, and P5 �

.002

5 sin�4400 t�.

P2 �.002

2 sin�1760t�, P3 �

.002

3 sin�2640 t�,

f2 � 880, f3 � 1320, f4 � 1760, f5 � 2200, . . . Hz.

P1 � .002 sin�880 t�.

f1 � 440 Hz,

f1. . . .nf1,. . . ,4 f1,3 f1,2 f1,f1

f1

.0017, .0083, and .015.

�0, .02Y2 � .004,

Y1

.006 sec.T � .046 � 1

150

f � 150 Hz.

6.04 � 150 cycles



8. Use an identity to rewrite as anequation with one trigonometricfunction.

9. 10. 135°, 180°,

225°, 270° 11.

12. 13.

14. 15. 16.

17.

18.

19. 20.

21.

22.

23.24.25.26.27.28. 29.30.31.32.33.34.35.36. �78.0°, 282.0°�

�53.6°, 126.4°, 187.9°, 352.1°��45°, 135°, 225°, 315°��0°, 45°, 135°, 180°, 225°, 315°��0°, 90°, 180°, 270°��0°, 30°, 150°, 180°��90°, 270°�

�45°, 225°��0°, 180°��45°, 135°, 225°, 315°��60°, 135°, 240°, 315°��90°, 210°, 330°��0°, 45°, 225°��30°, 210°, 240°, 300°�

�0,2

3,4

3 ��7

6,3

2,11

6 ��

6,

2,3

2,11

6 ��

�

4,3

4,7

6,11

6 ��

4,2

3,5

4,5

3 �0�0��

3,5

3 ��

6,5

6 ��

2 ��3

4,7

4 �

3, ,

4

3

Figure 30

.005

–.005

0 .01

P = P1 + P2 + P3 + P4 + P5 Solution

(a) A graphing calculator shows that the maximum value of P is approximately.00317. See Figure 30.

(b) The maximum occurs at .00246, .00474, .00701, and .00928.Figure 30 shows how the second value is found; the others are foundsimilarly.


x � .000188,

Concept Check Refer to the summary box on solving a trigonometric equation. Decideon the appropriate technique to begin the solution of each equation. Do not solve theequation.

1. 2.

3. 4.

5. 6.

7. 8.

Concept Check Answer each question.

9. Suppose you are solving a trigonometric equation for solutions over the interval , and your work leads to What are the corresponding values

of x?

10. Suppose you are solving a trigonometric equation for solutions over the interval, and your work leads to 60°, 75°, 90°. What are the corre-

sponding values of ?

Solve each equation for exact solutions over the interval . See Examples 1–4.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

Solve each equation for exact solutions over the interval . See Examples 2–5.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36. 4 cos2 � � 4 cos � � 19 sin2 � � 6 sin � � 1

cos2 � � sin2 � � 0sec2 � tan � � 2 tan �

sin2 � cos2 � � 02 tan2 � sin � � tan2 � � 0

sin2 � cos � � cos �csc2 � � 2 cot � � 0

cos2 � � sin2 � � 1tan � � cot � � 0

tan � � 1 � �3 � �3 cot �2 sin � � 1 � csc �

�tan � � 1� �cos � � 1� � 0�cot � � �3 � �2 sin � � �3 � � 0

�0°, 360°�

2 cos2 x � cos x � 1�2 sin2 x � 3 sin x � 1

2 cos2 x � �3 cos x � 0cos2 x � 2 cos x � 1 � 0

�csc x � 2� �csc x � �2 � � 0�cot x � 1� ��3 cot x � 1� � 0

sec2 x � 2 � �1tan2 x � 3 � 0

2 sec x � 1 � sec x � 32 sin x � 3 � 4

sin x � 2 � 32 cot x � 1 � �1

�0, 2�

�

13 � � 45°,�0°, 360°�

83 .2,2x � 2

3 ,�0, 2�

cos2 x � sin2 x � 1tan x � cot x � 0

tan2 x � 4 tan x � 2 � 09 sin2 x � 5 sin x � 1

2 cos2 x � cos x � 15 sec2 x � 6 sec x

sin x � 2 � 32 cot x � 1 � �1

7.6 Exercises

1. Solve the linear equation for cot x. 2. Solve the linear equation for sin x. 3. Solve thequadratic equation for sec x by factoring. 4. Solve the quadratic equation for cos x by thezero-factor property. 5. Solve the quadratic equation for sin x using the quadratic formula.6. Solve the quadratic equation for tan x using the quadratic formula. 7. Use an identityto rewrite as an equation with one trigonometric function.



37.38.39. 40.41.42.43.

where n isany integer

44. ,

, where n is any integer

45.

where n

is any integer 46.

where n is any integer

47.where n is any

integer 48.

where n is anyinteger 49.


where n is anyinteger 51.52.

55.

56.

57.

58.

59.

60.

61.

62.

63. �

2,3

2 ��

12,5

12,13

12,17

12 ��3

8,5

8,11

8,13

8 ��

18,7

18,13

18,19

18,25

18,31

18 ��

18,7

18,13

18,19

18,25

18,31

18 ��0,

3,2

3, ,

4

3,5

3 ��

2,7

6,11

6 ��

3,2

3,4

3,5

3 ��

12,11

12,13

12,23

12 ��0, .3760�

�.6806, 1.4159�251.6° � 360° n,

71.6° � 360° n,315° � 360° n,135° � 360° n,

108.4° � 180° n,45° � 180° n,

318.2° � 360° n,221.8° � 360° n,

90° � 360° n,326.4° � 360° n,

33.6° � 360° n,

4

3� 2n,

2

3� 2n,

5

3� 2n,

4

3� 2n,

2

3� 2n,

3� 2n,

5

3� 2n

� 2n,

3� 2n

5.8 � 2n,3.6 � 2n,2.3 � 2n,.9 � 2n,

�114.3°, 335.7°��57.7°, 159.2°�

�68.5°, 291.5°�0��38.4°, 218.4°, 104.8°, 284.8°��149.6°, 329.6°, 106.3°, 286.3°�

9

9

37. 38.

39. 40.

41. 42.

Determine all solutions of each equation in radians ( for x) or degrees ( for �) to thenearest tenth as appropriate. See Example 4.

43. 44.

45. 46.

47. 48.

49. 50.

The following equations cannot be solved by algebraic methods. Use a graphing calcu-lator to find all solutions over the interval . Express solutions to four decimalplaces.

51. 52.

53. Explain what is wrong with the following solution.Solve over the interval

or

The solutions are and

54. The equation has no solution over the interval . Usingthis information, what can be said about the graph of

over this interval? Confirm your answer by actually graphing the function over theinterval.

Solve each equation for exact solutions over the interval See Examples 6–8.

55. 56. 57.

58. 59. 60.

61. 62. 63.

64. 65. 66.

67. 68. 69.

70. sec x

2� cos

x

2

sin x

2� cos

x

2sin2

x

2� 2 � 08 sec2

x

2� 4

cos 2x � cos x � 0sin x � sin 2xtan 4x � 0

sin x

2� �2 � sin

x

22�3 sin 2x � �3�2 cos 2x � �1

cot 3x � �33 tan 3x � �3sin 3x � 0

sin 3x � �1cos 2x � �1

2cos 2x �

�3

2

�0, 2�.

y � cot x

2� csc

x

2� 1

�0, 2�cot x2 � csc x2 � 1 � 0

54 .

4

� �5

4� �

4

tan � � 1

tan 2�

2�

2

2

tan 2� � 2

�0, 2�.tan 2� � 2

x3 � cos2 x �1

2x � 1x2 � sin x � x3 � cos x � 0

�0, 2�

sec2 � � 2 tan � � 42 tan �

3 � tan2 �� 1

3 sin2 � � sin � � 25 sec2 � � 6 sec �

2 cos2 x � 5 cos x � 2 � 04 cos2 x � 1 � 0

2 cos2 x � cos x � 13 sin2 x � sin x � 1 � 0

2 sin � � 1 � 2 cos �cot � � 2 csc � � 3

2 cos2 � � 2 cos � � 1 � 0sin2 � � 2 sin � � 3 � 0

3 cot2 � � 3 cot � � 1 � 0tan2 � � 4 tan � � 2 � 0



64.

65.

66. 67. 68.

69. 70.

71.72.73. 74.75. 76.77.78.79.

where n is any integer80.

where n is any integer81.where n is any integer82.

, where n is anyinteger 83.


where n is anyinteger85.

where n is any integer86.

where n is any integer87. (a) .00164 and .00355(b)(c) outward88. (a) 3 beats per sec

.01

.15 1.15

–.01

For x = t,P(t) = .005 sin 440�t + .005 sin 446�t

�.00164, .00355

300° � 360° n,180° � 360° n,60° � 360° n,0° � 360° n,

330° � 360° n,270° � 360° n,210° � 360° n,150° � 360° n,

90° � 360° n,30° � 360° n,

292.5° � 360° n,202.5° � 360° n,112.5° � 360° n,

22.5° � 360° n,258.2° � 360° n,

191.8° � 360° n,78.2° � 360° n,11.8° � 360° n,

289.5° � 360° n70.5° � 360° n,

300° � 360° n,60° � 360° n,

270° � 360° n,225° � 360° n,90° � 360° n,45° � 360° n,

180° � 360° n,150° � 360° n,30° � 360° n,0° � 360° n,

�60°, 90°, 270°, 300°��30°, 150°, 270°�

�300°��120°, 240°��180°��0°�

�75°, 105°, 255°, 285°��15°, 45°, 135°, 165°, 255°, 285°�

�0��

2 �0�0��0,

2

3,4

3 ��0,

3, ,

5

3 ��0,

4,

2,3

4, ,

5

4,3

2,7

4 � Solve each equation in Exercises 71–78 for exact solutions over the interval .In Exercises 79–86, give all exact solutions. See Examples 6–8.

71. 72. 73.

74. 75. 76.

77. 78. 79.

80. 81. 82.

83. 84. 85.

86.

(Modeling) Solve each problem. See Examples 9 and 10.

87. Pressure on the Eardrum No musical instrument can generate a true pure tone. Apure tone has a unique, constant frequency and amplitude that sounds rather dull anduninteresting. The pressures caused by pure tones on the eardrum are sinusoidal. Thechange in pressure P in pounds per square foot on a person’s eardrum from a puretone at time t in seconds can be modeled using the equation

where f is the frequency in cycles per second, and is the phase angle. When P ispositive, there is an increase in pressure and the eardrum is pushed inward; when Pis negative, there is a decrease in pressure and the eardrum is pushed outward.(Source: Roederer, J., Introduction to the Physics and Psychophysics of Music,Second Edition, Springer-Verlag, 1975.) A graph of the tone middle C is shown inthe figure.

(a) Determine algebraically the values of t forwhich over

(b) From the graph and your answer in part (a),determine the interval for which over

(c) Would an eardrum hearing this tone bevibrating outward or inward when

88. Hearing Beats in Music Musicians sometimes tune instruments by playing thesame tone on two different instruments and listening for a phenomenon known asbeats. Beats occur when two tones vary in frequency by only a few hertz. When thetwo instruments are in tune, the beats disappear. The ear hears beats because thepressure slowly rises and falls as a result of this slight variation in the frequency.This phenomenon can be seen using a graphing calculator. (Source: Pierce, J., TheScience of Musical Sound, Scientific American Books, 1992.)

(a) Consider two tones with frequencies of 220 and 223 Hz and pressuresand respectively. Graph the pressure

felt by an eardrum over the 1-sec interval How manybeats are there in 1 sec?

(b) Repeat part (a) with frequencies of 220 and 216 Hz.(c) Determine a simple way to find the number of beats per second if the frequency

of each tone is given.

�.15, 1.15.P � P1 � P2

P2 � .005 sin 446t,P1 � .005 sin 440t

P � 0?

�0, .005.P � 0

�0, .005.P � 0

�

P � A sin�2 f t � ��,

sin � � sin 2� � 0

2 cos2 2� � 1 � cos 2�4 cos 2� � 8 sin � cos �2 � sin 2� � 4 sin 2�

cos � � sin2 �

2csc2

�

2� 2 sec �sin 2� � 2 cos2 �

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

2�3 cos �

2� �32�3 sin

�

2� 3sin

�

2� 1

cos �

2� 1�2 cos 2� � �3�2 sin 3� � 1 � 0

�0°, 360°�

.005

–.005

0 .005

For x = t,P(t) = .004 sin[2�(261.63)t + ]�

7


89. Pressure of a Plucked String If a string with a fundamental frequency of 110 Hzis plucked in the middle, it will vibrate at the odd harmonics of 110, 330, 550, . . . Hzbut not at the even harmonics of 220, 440, 660, . . . Hz. The resulting pressure Pcaused by the string can be modeled by the equation

(Source: Benade, A., Fundamentals of Musical Acoustics, Dover Publications, 1990;Roederer, J., Introduction to the Physics and Psychophysics of Music, SecondEdition, Springer-Verlag, 1975.)

(a) Graph P in the window by (b) Use the graph to describe the shape of the sound wave that is produced.(c) See Exercise 87. At lower frequencies, the inner ear will hear a tone only when

the eardrum is moving outward. Determine the times over the interval when this will occur.

90. Hearing Difference Tones Small speakers like those found in older radios andtelephones often cannot vibrate slower than 200 Hz—yet 35 keys on a piano havefrequencies below 200 Hz. When a musical instrument creates a tone of 110 Hz, italso creates tones at 220, 330, 440, 550, 660, . . . Hz. A small speaker cannot repro-duce the 110-Hz vibration but it can reproduce the higher frequencies, which arecalled the upper harmonics. The low tones can still be heard because the speaker pro-duces difference tones of the upper harmonics. The difference between consecutivefrequencies is 110 Hz, and this difference tone will be heard by a listener. We canmodel this phenomenon using a graphing calculator. (Source: Benade, A.,Fundamentals of Musical Acoustics, Dover Publications, 1990.)

(a) In the window by graph the upper harmonics represented by thepressure

(b) Estimate all t-coordinates where P is maximum.(c) What does a person hear in addition to the frequencies of 220, 330, and 440 Hz?(d) Graph the pressure produced by a speaker that can vibrate at 110 Hz and above.

91. Daylight Hours in New Orleans The seasonal variation in length of daylight canbe modeled by a sine function. For example, the daily number of hours of daylightin New Orleans is given by

where x is the number of days after March 21 (disregarding leap year). (Source:Bushaw, Donald et al., A Sourcebook of Applications of School Mathematics.Copyright © 1980 by The Mathematical Association of America.)

(a) On what date will there be about 14 hr of daylight?(b) What date has the least number of hours of daylight?(c) When will there be about 10 hr of daylight?

(Modeling) Alternating Electric Current The study of alternating electric current requires the solutions of equations of the form

for time t in seconds, where i is instantaneous current in amperes, is maximum cur-rent in amperes, and f is the number of cycles per second. (Source: Hannon, R. H., Basic

Imax

i � Imax sin 2ft,

h �35

3�

7

3 sin

2x

365,

P �1

2 sin�2 �220�t �

1

3 sin�2 �330�t �

1

4 sin�2 �440�t.

��1, 1,�0, .03

�0, .03

��.005, .005.�0, .03

�.003

7 sin 1540t.P � .003 sin 220t �

.003

3 sin 660t �

.003

5 sin 1100t


88. (b) 4 beats per sec

(c) The number of beats is equalto the absolute value of thedifference in the frequencies ofthe two tones.89. (a)

(b) The graph is periodic, and thewave has “jagged square” topsand bottoms. (c) This will occurwhen t is in one of these intervals:

90. (a)

(b) .0007576, .009847, .01894,.02803 (c) 110 Hz

1

0 .03

–1

For x = t,

P(t) = sin[2�(220)t] +

sin[2�(330)t] +

sin[2�(440)t]

121314

�.0227, .0273�.�.0136, .0182�,�.0045, .0091�,

.005

0 .03

–.005

For x = t,P(t) = .003 sin 220�t +

sin 660�t +

sin 1100�t +

sin 1540�t

.0033

.0035

.0037

.01

.15 1.15

–.01

For x = t,P(t) = .005 sin 440�t + .005 sin 432�t


Technical Mathematics with Calculus, W. B. Saunders Company, 1978.) Find the small-est positive value of t, given the following data.

92. 93.

94. 95.


96. Accident Reconstruction The model

is used to reconstruct accidents in which a vehicle vaults into the air after hitting anobstruction. is velocity in feet per second of the vehicle when it hits, D is distance(in feet) from the obstruction to the landing point, and h is the difference in height(in feet) between landing point and takeoff point. Angle is the takeoff angle, theangle between the horizontal and the path of the vehicle. Find to the nearest degreeif , and

97. Electromotive Force In an electric circuit, let

model the electromotive force in volts at t seconds. Find the smallest positive valueof t where for each value of V.

(a) (b) (c)

98. Voltage Induced by a Coil of Wire A coil of wire rotating in a magnetic fieldinduces a voltage modeled by

where t is time in seconds. Find the smallest positive time to produce each voltage.

(a) 0 (b)

99. Movement of a Particle A particle moves along a straight line. The distance ofthe particle from the origin at time t is modeled by

Find a value of t that satisfies each equation.

(a) (b)

100. Explain what is wrong with the following solution for all x over the interval of the equation

Divide by sin x.

Add 1.

The solution set is �2 �.

x �

2

sin x � 1

sin x � 1 � 0

sin2 x � sin x � 0

sin2 x � sin x � 0.�0, 2�

s�t� �3�2

2s�t� �

2 � �3

2

s�t� � sin t � 2 cos t.

10�3

e � 20 sin�t

4�

2 �,

V � .25V � .5V � 0

0 � t �12

V � cos 2t

h � 2.D � 80,V0 � 60�

�

V0

.342D cos � � h cos2 � �16D2

V 20

f � 60i �1

2Imax,f � 60i � Imax,

f � 120Imax � 100,i � 50,f � 60Imax � 100,i � 40,


90. (d)

91. (a) 91.3 days after March 21,on June 20 (b) 273.8 days afterMarch 21, on December 19(c) 228.7 days after March 21, onNovember 4, and again after318.8 days, on February 292. .001 sec 93. .0007 sec94. .004 sec 95. .0014 sec

96. 14° 97. (a) sec

(b) sec (c) .21 sec

98. (a) 2 sec (b) sec

99. (a) One such value is

(b) One such value is

4.

3.

31

3

1

6

1

4

2

0 .03

–2

For x = t,P(t) = sin[2�(110)t] +

sin[2�(220)t] +

sin[2�(330)t] +

sin[2�(440)t]

121314

9



7.7 Equations Involving Inverse Trigonometric FunctionsSolving for x in Terms of y Using Inverse Functions ■ Solving Inverse Trigonometric Equations

Until now, the equations in this chapter have involved trigonometric functions ofangles or real numbers. Now we examine equations involving inverse trigono-metric functions.

Solving for x in Terms of y Using Inverse Functions

EXAMPLE 1 Solving an Equation for a Variable Using Inverse Notation

Solve for x.

Solution We want cos 2x alone on one side of the equation so we can solve for2x, and then for x.

Divide by 3.

Definition of arccosine (Section 7.5)

Multiply by

Now try Exercise 7.

Solving Inverse Trigonometric Equations

EXAMPLE 2 Solving an Equation Involving an Inverse Trigonometric Function

Solve

Solution First solve for arcsin x, and then for x.

Divide by 2.

Definition of arcsine (Section 7.5)

(Section 6.2)

Verify that the solution satisfies the given equation. The solution set is


�1�.

x � 1

x � sin

2

arcsin x �

2

2 arcsin x �

2 arcsin x � .

12 . x �

1

2 arccos

y

3

2x � arccos y

3

y

3� cos 2x

y � 3 cos 2x

y � 3 cos 2x


7.7 Equations Involving Inverse Trigonometric Functions 671

Figure 31

0

2 1u

x

y

√3

EXAMPLE 3 Solving an Equation Involving Inverse Trigonometric Functions

Solve

Solution Let Then and for u in quadrant I, the equationbecomes

Alternative form

Sketch a triangle and label it using the facts that u is in quadrant I and

See Figure 31. Since and the solution set is Check.


EXAMPLE 4 Solving an Inverse Trigonometric Equation Using an Identity

Solve

Solution Isolate one inverse function on one side of the equation.

Add arccos x. (1)

Definition of arcsine

Let so by definition.

Substitute. (2)

Sine sum identity (Section 7.3)

Substitute this result into equation (2) to get

(3)

From equation (1) and by the definition of the arcsine function,

Subtract (Section 1.7)

Since we must have Thus, and wecan sketch the triangle in Figure 32. From this triangle we find that �1 � x2.

sin u �x � 0,0 � arccos x �

3 .0 � arccos x � ,

6 . �

2

3� arccos x �

3.

�

2� arccos x �

6�

2

sin u cos

6� cos u sin

6� x.

sin�u �

6 � � sin u cos

6� cos u sin

6

sin�u �

6 � � x

0 � u � u � arccos x,

sin�arccos x �

6 � � x

arcsin x � arccos x �

6


6


6.

��32 �.x �

�32 ,x � cos u,

sin u � 12 .

cos u � x.

cos�1 x � u

sin u � 12sin�1 12 � u.

cos�1 x � sin�1 1

2.

Figure 32

x

y

1

u

x0

√1 – x2


Concept Check Answer each question.

1. Which one of the following equations has solution 0?

A. B. C.

2. Which one of the following equations has solution ?

A. B. C.


A. B. C.


A. B. C. arcsin��1

2� � xarccos��1

2� � xarctan �3

3� x

�6

arccos��2

2 � � xarcsin �2

2� xarctan 1 � x

34

arctan �3

3� xarccos��

�2

2 � � xarcsin �2

2� x

4

arcsin 0 � xarccos 0 � xarctan 1 � x


Now substitute into equation (3) using and

(3)

Multiply by 2.

Subtract x.

Square both sides. (Section 1.6)

Distributive property (Section R.1)

Add

Quotient rule (Section R.7)

To check, replace x with in the original equation:

as required. The solution set is


��32 �.

arcsin �3

2� arccos

�3

2�

3�

6�

6,

�32

x ��3

2

x � � 3

4

3x2. 3 � 4x2

3 � 3x2 � x2

3�1 � x2� � x2

��3��1 � x2 � x

��1 � x2��3 � x � 2x

��1 � x2� �3

2� x

1

2� x

sin u cos

6� cos u sin

6� x

cos u � x.cos 6 � �32 ,

sin 6 � 12 ,sin u � �1 � x2,

7.7 Exercises

1. C 2. A 3. C 4. C

5.

6.

7.

8.

9. x �1

2 arctan

y

3

x � arcsec 12y

x �1

3 arccot 2y

x � arcsin 4y

x � arccos y

5

Solve for x; Choose the positivesquare root because x � 0.



10.

11.

12.

13.

14.

15.

16.

17.18.

19.

20.

23. 24.

25. 26.

27. 28. 29.

30. 31. 32.

33. 34.

35. 36.

37. 38.39.

40.

41.

42. �2.2824135��4.4622037�

2

–8

–2 2

Y = arcsin X – arccos X – �6

�0��0�

��5

5 ��1

2�0�� 1

2��3

2 ��0�� 3

4�� 4

5��12

5 �� 3

5��3�3 � 2

6 �� 3�

0��2�2 �x � arccos� y � 4

3 �x � arcsin� y � 4

2 �x � arccot� y � 1�x � arcsin� y � 2�

x �1

2�1 � arctan y�

x � �3 � arccos y

x �1

5 arccot

y

3

x �1

5 arccos��

y

2�x � 3 arcsin��y�

x � 4 arccos y

6

x � 2 arcsin y

3

2

–8

–2 2

Y1 = arcsin X – arccos X �6

Y2 =

9

9

Solve each equation for x. See Example 1.

5. 6. 7.

8. 9. 10.

11. 12. 13.

14. 15. 16.

17. 18. 19.

20.

21. Refer to Exercise 17. A student attempting to solve this equation wrote as the firststep , inserting parentheses as shown. Explain why this is incorrect.

22. Explain why the equation cannot have a solution. (No work isrequired.)

Solve each equation for exact solutions. See Examples 2 and 3.

23. 24.

25. 26.

27. 28.

29. 30.

Solve each equation for exact solutions. See Example 4.

31. 32.

33. 34.

35. 36.

37. 38.

39. Provide graphical support for the solution in Example 4 by showing that the graph

of has x-intercept .

40. Provide graphical support for the solution in Example 4 by showing that the x-coordinate of the point of intersection of the graphs of

and is .

The following equations cannot be solved by algebraic methods. Use a graphing calcu-lator to find all solutions over the interval . Express solutions to as many decimalplaces as your calculator displays.

41. 42. sin�1�.2x� � 3 � ��x�arctan x�3 � x � 2 � 0

�0, 6

�32 � .8660254Y2 �

6

Y1 � arcsin X � arccos X

�32 � .8660254y � arcsin x � arccos x �

6

sin�1 x � tan�1 x � 0cos�1 x � tan�1 x �

2

arcsin 2x � arcsin x �

2arcsin 2x � arccos x �

6

arccos x � 2 arcsin �3

2�

3arccos x � 2 arcsin

�3

2�

sin�1 x � tan�1 �3 �2

3sin�1 x � tan�1 1 � �

4

cot�1 x � tan�1 4

3cos�1 x � sin�1

3

5

arctan x � arccos 5

13arcsin x � arctan

3

4

arccos�y �

3 � �

62 arccos� y �

3 � � 2

4 � 4 tan�1 y � 4

3 cos�1

y

4�

sin�1 x � cos�1 2

y � sin�x � 2�

y � 4 � 3 cos x

y � 2 sin x � 4y � cot x � 1y � sin x � 2

y � tan�2x � 1�y � cos�x � 3�y � 3 cot 5x

y � �2 cos 5xy � �sin x

3y � 6 cos

x

4

y � 3 sin x

2y � 3 tan 2x6y �

1

2 sec x

2y � cot 3x4y � sin xy � 5 cos x




43. Tone Heard by a Listener When two sources located at different positions pro-duce the same pure tone, the human ear will often hear one sound that is equal to thesum of the individual tones. Since the sources are at different locations, they willhave different phase angles . If two speakers located at different positions producepure tones and , where ,

, then the resulting tone heard by a listener can be written as, where

and .

(Source: Fletcher, N. and T. Rossing, The Physics of Musical Instruments, SecondEdition, Springer-Verlag, 1998.)

(a) Calculate A and if , , , and . Alsofind an expression for if .

(b) Graph and on the same coordinate axes over the interval . Are the two graphs the same?

44. Tone Heard by a Listener Repeat Exercise 43, with , ,, , and .

45. Depth of Field When a large-view camera is used to take a picture of an object thatis not parallel to the film, the lens board should be tilted so that the planes contain-ing the subject, the lens board, and the film intersect in a line. This gives the best“depth of field.” See the figure. (Source: Bushaw, Donald et al., A Sourcebook ofApplications of School Mathematics. Copyright © 1980 by The MathematicalAssociation of America.)

(a) Write two equations, one relating , x, and z, and the other relating , x, y, and z.(b) Eliminate z from the equations in part (a) to get one equation relating , , x,

and y.(c) Solve the equation from part (b) for .(d) Solve the equation from part (b) for .

46. Programming Language for Inverse Functions In Visual Basic, a widely usedprogramming language for PCs, the only inverse trigonometric function available isarctangent. The other inverse trigonometric functions can be expressed in terms ofarctangent as follows.

(a) Let . Solve the equation for x in termsof u.

(b) Use the result of part (a) to label the three sides ofthe triangle in the figure in terms of x.

(c) Use the triangle from part (b) to write an equationfor tan u in terms of x.

(d) Solve the equation from part (c) for u.

u � arcsin x

��

��

Subject Lens

Film

z

y

x

��

f � 300�2 � 6A2 � .001

�1 � 7A1 � .0025

�0, .01Y2 � P1 � P2Y1 � P

f � 220P � A sin�2ft � ��2 � .61A2 � .004�1 � .052A1 � .0012�

� � arctan�A1 sin �1 � A2 sin �2

A1 cos �1 � A2 cos �2�

A � ��A1 cos �1 � A2 cos �2�2 � �A1 sin �1 � A2 sin �2�2

P � A sin�2ft � ��2 �

4

�4 � �1P2 � A2 sin�2ft � �2�P1 � A1 sin�2ft � �1�

�

43. (a) , ;

(b) The two graphs are the same.

44. (a) , ;

(b) The two graphs are the same.

45. (a) ;

(b)

(c)

(d)

46. (a) ,

(b)

(c) tan u �x�1 � x2

1 � x2

x

y

0u

x1

√1 – x2

�

2� u �

2x � sin u

� � arctan��x � y� tan �

x �� arctan�x tan �

x � y �x

tan ��

x � y

tan �

tan � �x � y

ztan � �

x

z

P � .0035 sin�600t � .47�� .470A � .0035

P � .00506 sin�440t � .484�� .484A � .00506

x

y

0u

.006

–.006

0 .01

For x = t,P(t) = .00506 sin(440�t + .484)P1(t) + P2(t) = .0012 sin(440�t + .052) + .004 sin(440�t + .61)

.006

–.006

0 .01

For x = t,P(t) = .0035 sin(600�t + .47)

P1(t) + P2(t) = .0025 sin(600�t + ) +

.001 sin(600�t + )�7

�6


47. Alternating Electric Current In the study of alternating electric current, instanta-neous voltage is modeled by

,

where f is the number of cycles per second, is the maximum voltage, and t istime in seconds.

(a) Solve the equation for t.(b) Find the smallest positive value of t if , , and . Use a

calculator.

48. Viewing Angle of an ObserverWhile visiting a museum, MarshaLanglois views a painting that is 3 fthigh and hangs 6 ft above the ground.See the figure. Assume her eyes are 5 ft above the ground, and let x be thedistance from the spot where she isstanding to the wall displaying thepainting.

(a) Show that , the viewing anglesubtended by the painting, isgiven by

.

(b) Find the value of x for each value of .

(i) (ii)

(c) Find the value of for each value of x.(i) (ii)

49. Movement of an Arm In the exercises for Section 6.3 we found the equation

,

where t is time (in seconds) and y is the angle formed by a rhythmically moving arm.

(a) Solve the equation for t.(b) At what time(s) does the arm form an angle of .3 radian?

50. The function is not found on graphing calculators. However, with somemodels it can be graphed as

.

(This formula appears as in the screen here.) Use the formula to obtain the graph of in thewindow by .�0, ��4, 4

y � sec�1 xY1

y �

2� ��x � 0� � �x � 0��

2� tan�1��x2 � 1� ��

y � sec�1 x

y �1

3 sin

4t

3

x � 3x � 4�

� �

8� �

6

�

� � tan�1� 4

x � � tan�1� 1

x �

�

f � 100e � 5Emax � 12

Emax

e � Emax sin 2ft


(d)

47. (a)

(b) .00068 sec48. (b) (i) approximately .94 or4.26 (ii) approximately .60 or6.64 (c) (i) approximately .54(ii) approximately .61

49. (a)

(b) .27 sec50.

0

Radian mode

–4 4

�

y = sec–1 x

t �3

4 arcsin 3y

t �1

2f arcsin

e

Emax

u � arctan x�1 � x2

1 � x2

x ft

5 ft6 ft

3 ftθ


Chapter 7 Summary

NEW SYMBOLS

inverse sine of xinverse cosine of xinverse tangent of x

inverse cotangent of xinverse secant of xinverse cosecant of xcsc�1 x (arccsc x)

sec�1 x (arcsec x)cot�1 x (arccot x)

tan�1 x (arctan x)cos�1 x (arccos x)sin�1 x (arcsin x)

QUICK REVIEW

C O N C E P T S E X A M P L E S

7.1 Fundamental Identities

Reciprocal Identities

Quotient Identities

Pythagorean Identities

Negative-Angle Identities

If is in quadrant IV and find , , and

cos is positive in quadrant IV.

sin�� sin � �3

5

� cos � � ��16

25�

4

5

cos2 � � 1 �9

25�

16

25

��3

5�2

� cos2 � � 1

sin2 � � cos2 � � 1

csc � �1

sin ��

1

�35

� �5

3

sin��.cos �csc �sin � � �

35 ,�

cot�� cot �sec�� sec �csc�� csc �

tan�� tan �cos�� cos �sin�� sin �

1 � cot2 � � csc2 �

tan2 � � 1 � sec2 �sin2 � � cos2 � � 1

cot � �cos �

sin �tan � �

sin �

cos �

csc � �1

sin �sec � �

1

cos �cot � �

1

tan �

7.2 Verifying Trigonometric Identities

See the box titled Hints for Verifying Identities on page 613.

7.3 Sum and Difference Identities

Cofunction Identities

Sum and Difference Identities

Find a value of such that .

Find the exact value of

��6

4�

�2

4�

�6 � �2

4

��3

2

�2

2�

1

2

�2

2

� cos 30° cos 45° � sin 30° sin 45°

cos��15°� � cos�30° � 45°�

cos��15°�.

� � 12°

90° � � � 78°

cot�90° � �� cot 78°

tan � � cot 78°

tan � � cot 78°�





csc�90° � �� sec � tan�90° � �� cot �

sec�90° � �� csc � sin�90° � �� cos �

cot�90° � �� tan � cos�90° � �� sin �




Sum and Difference Identities Write in terms of .

tan 4 � 1 �1 � tan �

1 � tan �

tan�

4� ��

tan 4 � tan �

1 � tan 4 tan �

tan �tan�4 � ��

tan�A � B� �tan A � tan B

1 � tan A tan B

tan�A � B� �tan A � tan B

1 � tan A tan B

7.4 Double-Angle Identities and Half-Angle Identities

Double-Angle Identities

Product-to-Sum Identities

Sum-to-Product Identities

Half-Angle Identities

Given and find .

Sketch a triangle in quadrant II and use it to find : .

Write as the difference of two functions.

Write as a product of two functions.

Find the exact value of tan 67.5°.

We choose the last form with

�2 � �2

�2or �2 � 1�

1 ��22

�22

2

2

tan 67.5° � tan 135°

2�

1 � cos 135°

sin 135°�

1 � ��22 �

�22

A � 135°.

� 2 cos 2� cos �

� 2 cos 2� cos��

� 2 cos�4�

2 � cos��2�

2 � cos � � cos 3� � 2 cos�� 3�

2 � cos�� 3�

2 �cos � � cos 3�

�1

2 cos 3� �

1

2 cos �

�1

2cos��3��

1

2 cos �

�1

2�cos��3�� cos �

sin�� sin 2� �1

2�cos�� 2�� cos�� 2��

sin�� sin 2�

� 2�12

13��5

13� � �120

169

sin 2� � 2 sin � cos �

sin � � 1213

sin �

sin 2�sin � � 0,cos � � �5

13

tan A

2�

1 � cos A

sin A

tan A

2�

sin A

1 � cos Atan

A

2� ��1 � cos A

1 � cos A

sin A

2� ��1 � cos A

2cos

A

2� ��1 � cos A

2

cos A � cos B � �2 sin�A � B

2 � sin�A � B

2 � cos A � cos B � 2 cos�A � B

2 � cos�A � B

2 � sin A � sin B � 2 cos�A � B

2 � sin�A � B

2 � sin A � sin B � 2 sin�A � B

2 � cos�A � B

2 �

cos A sin B �1

2�sin�A � B� � sin�A � B�

sin A cos B �1

2�sin�A � B� � sin�A � B�

sin A sin B �1

2�cos�A � B� � cos�A � B�

cos A cos B �1

2�cos�A � B� � cos�A � B�

tan 2A �2 tan A

1 � tan2 A

sin 2A � 2 sin A cos A cos 2A � 2 cos2 A � 1

cos 2A � 1 � 2 sin2 A cos 2A � cos2 A � sin2 A

Rationalize the denominator; simplify.

y

x

12

–5

13�

CHAPTER 7 Summary 677

The sign is chosen based on thequadrant of A2.��



7.5 Inverse Circular Functions

See page 649 for graphs of the other inverse circular(trigonometric) functions.

Evaluate Write as Then , because

and is in the range of

Evaluate

Let . Then Since is nega-

tive in quadrant IV, sketch a triangle as shown.

We want . From the triangle,

sin u � �35 .

sin�tan�1��34� � sin u

y

x

–35

4u

tan�1 xtan u � �34 .u � tan�1��3

4�sin�tan�1��3

4�.

cos�1 x.2cos 2 � 0

y � 2cos y � 0.y � cos�1 0

y � cos�1 0.

x

y

y = tan–1 x

–1–2 0 1 2

–

�2

�2

(1, )4�

(–1, – )4�

x

y

y = cos–1 x

–1 0 1

�

(0, )2�

(–1, �)

(1, 0)

x

y

y = sin–1 x

–1 0 1

–

�2

�2

(1, )2�

(–1, – )2�

Range

Inverse QuadrantsFunction Domain Interval of the Unit Circle

I and IV

I and II

I and IV

I and II

I and II

I and IV��2 ,

2 , y � 0��, �1 � �1, ��y � csc�1 x

�0, , y � 2��, �1 � �1, ��y � sec�1 x

�0, ��, ��y � cot�1 x

��2 ,

2 ��, ��y � tan�1 x

�0, ��1, 1y � cos�1 x

��2 ,

2 ��1, 1y � sin�1 x

7.6 Trigonometric Equations

Solving a Trigonometric Equation

1. Decide whether the equation is linear or quadratic inform, so you can determine the solution method.

2. If only one trigonometric function is present, first solvethe equation for that function.

3. If more than one trigonometric function is present, re-arrange the equation so that one side equals 0. Then try tofactor and set each factor equal to 0 to solve.

4. If the equation is quadratic in form, but not factorable,use the quadratic formula. Check that solutions are in thedesired interval.

5. Try using identities to change the form of the equation. Itmay be helpful to square both sides of the equation first.If this is done, check for extraneous solutions.

Solve over the interval .Use a linear method.

Another solution over is

The solution set is .�60°, 240°�

� � 60° � 180° � 240°.

�0°, 360°�

� � 60°

tan � � �3

tan � � �3 � 2�3

�0°, 360°�tan � � �3 � 2�3



CHAPTER 7 Review Exercises 679

Concept Check For each expression in Column I, choose the expression fromColumn II that completes an identity.

1. B 2. A 3. C 4. F 5. D

6. E 7. 1 8.

9. 10.

11.

12.

13. E 14. B 15. J 16. A17. I 18. C 19. H 20. D

21. G 22. B 23.

I

24.

IV 25.

26.�6

3

1

2

4 � 9�11

12�11 � 3;

12�11 � 3

50;

4 � 9�11

50;

4 � 3�15

4�15 � 3;

4�15 � 3

20;

4 � 3�15

20;

sec x � ��41

4csc x �

�41

5;

cot x � �4

5;cot��x� �

3

4

tan x � �4

3;sin x � �

4

5;

1 � cos �

sin �

1

cos2 �

cos2 �

sin �

Chapter 7 Review Exercises

I

1. 2.

3. 4.

5. 6.

II

A. B.

C. D.

E. F.cos x

sin x

1

cos2 x

1

cot2 x

sin x

cos x

1

cos x

1

sin x

sec2 x �tan2 x �

cot x �tan x �

csc x �sec x �

I

13. cos 210° 14. sin 35°

15. 16.

17. cos 35° 18. cos 75°

19. sin 75° 20. sin 300°

21. cos 300° 22. cos��55°�

�sin 35°tan��35°�

Use identities to write each expression in terms of and , and simplify.

7. 8.

9. 10.

11. Use the trigonometric identities to find sin x, tan x, and given andx is in quadrant IV.

12. Given where , use the trigonometric identities to find cot x,csc x, and sec x.

Concept Check For each expression in Column I, use an identity to choose an expres-sion from Column II with the same value.

2 � x � tan x � �

54,

cos x � 35cot��x�,

csc � � cot �tan2 ��1 � cot2 ��

cot �

sec �sec2 � � tan2 �

cos �sin �

For each of the following, find and the quadrant of.

23. x and y in quadrant III

24. x in quadrant I, y in quadrant IV

Find each of the following.

25. , given with

26. , given , with

2� y � cos 2y � �

1

3sin y

90° � � � 180°cos � � �1

2,cos

�

2

cos y �4

5,sin x �

1

10,

cos y � �4

5,sin x � �

1

4,

x � ytan�x � y�,cos�x � y�,sin�x � y�,

II

A. B.

C. D.

E.

F.

G.

H.

I. J. cot 125°cos��35°�sin 15° cos 60° � cos 15° sin 60°

cos2 150° � sin2 150°

cot��35°�cos 150° cos 60° � sin 150° sin 60°

2 sin 150° cos 150°�1 � cos 150°

2

cos 55°sin��35°�




27. 28.


29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

Give the exact real number value of y. Do not use a calculator.

45. 46. 47.

48. 49. 50.

51. 52. 53.

Give the degree measure of . Do not use a calculator.

54. 55. 56.

Use a calculator to give the degree measure of .

57. 58. 59.

60. 61. 62.

Evaluate the following without using a calculator.

63. 64. 65.

66. 67. 68.

69. 70. 71. cos�csc�1��2��cos�arctan 3�sin�arccos 3

4�cos�1�cos 0�tan�1�tan

4 �arcsec�sec �

arccos�cos 3

4 �sin�arcsin��3

2 ��cos�arccos��1��

� � csc�1 7.4890096� � arcsec 3.4723155� � cot�1 4.5046388

� � cos�1 .80396577� � sin�1��.66045320�� arctan 1.7804675

�

� � tan�1 0� � arcsin��3

2 �� arccos 1

2

�

y � arccot��1�y � arccsc 2�3

3y � sec�1��2�

y � arctan �3

3y � cos�1��

�2

2 �y � arcsin��1�

y � tan�1��3 �y � arccos��1

2�y � sin�1 �2

2

sin3 � � sin � � cos2 � sin �2 cos3 x � cos x �cos2 x � sin2 x

sec x

sec2 � � 1 �sec 2� � 1

sec 2� � 1tan � cos2 � �

2 tan � cos2 � � tan �

1 � tan2 �

2 cos2 � � 1 �1 � tan2 �

1 � tan2 �2 tan x csc 2x � tan2 x � 1

csc A sin 2A � sec A � cos 2A sec Atan � sin 2� � 2 � 2 cos2 �

2 cot x

tan 2x� csc2 x � 21 � tan2 � � 2 tan � csc 2�

2 tan B

sin 2B� sec2 B2 cos A � sec A � cos A �

tan A

csc A

sin 2x

sin x�

2

sec x

sin2 x

2 � 2 cos x� cos2 x

2

2 cos3 x � cos x �cos2 x � sin2 x

sec xsin2 x � sin2 y � cos2 y � cos2 x

1 � cos 2x

sin 2x�

sin 2x � sin x

cos 2x � cos x

27.

28.

45. 46. 47.

48. 49. 50.

51. 52. 53.

54. 60° 55. 56. 0°57. 60.67924514°58. �41.33444556°59. 36.4895081°60. 12.51631252°61. 73.26220613°62. 7.673567973°

63. �1 64. 65.

66. 67. 68. 0 69.

70. 71. 72.

73.

74. 75.

76.

77. 3.605240263��.463647609,

�

2,3

2 �1

u�1 � u2

294 � 125�6

92

9

7�3

2�10

10

�7

4

4

3

4�

�3

2

�60°

3

4

3

2

3

6

3

4�

2

�

3

2

3

4

1 � cos 2x

sin 2x� tan x

�sin 2x � sin x

cos 2x � cos x� cot

x

2


CHAPTER 7 Review Exercises 681

72. 73.

Write each of the following as an algebraic (nontrigonometric) expression in u.

74. 75.

Solve each equation for solutions over the interval .

76. 77. 78.

79. 80. 81.

Give all solutions for each equation.

82. 83. 84.

Solve each equation for solutions over the interval . If necessary, express solu-tions to the nearest tenth of a degree.

85. 86.

87. 88.

89. 90.

Solve each equation in Exercises 91–97 for x.

91. 92. 93.

94. 95. 96.

97.

98. Solve for t in terms of d.


99. Viewing Angle of an Observer A 10-ft-wide chalk-board is situated 5 ft from the left wall of a classroom.See the figure. A student sitting next to the wall x feetfrom the front of the classroom has a viewing angle of

radians.

(a) Show that the value of is given by the functiondefined by

(b) Graph with a graphing calculator to estimate the value of x that maximizesthe viewing angle.

f �x�

f �x� � arctan�15

x � � arctan� 5

x �.

�

�

d � 550 � 450 cos�

50t�

arccos x � arctan 1 �11

12

arccos x � arcsin 2

7

4

3 arctan

x

2� 5y � 4 sin x � 3

2y � tan�3x � 2�y � 3 cos x

24y � 2 sin x

5 cot2 � � cot � � 2 � 03 cos2 � � 2 cos � � 1 � 0

2 sin 2� � 1sin 2� � cos 2� � 1

2 tan2 � � tan � � 1sin2 � � 3 sin � � 2 � 0

�0°, 360°�

4 sin x cos x � �3cos 2x � cos x � 0sec x

2� cos

x

2

tan2 2x � 1 � 0sec4 2x � 4tan x � cot x

3 sin2 x � 5 sin x � 2 � 02 tan x � 1 � 0sin2 x � 1

�0, 2�

tan�arcsec �u2 � 1

u �cos�arctan u

�1 � u2�

tan�arcsin 3

5� arccos

5

7�sec�2 sin�1��1

3��78.

79.

80.

81.

82. , where n is anyinteger

83. , ,

, where n is any integer

84. , ,

, , where n

is any integer 85. {270°}86. {45°, 153.4°, 225°, 333.4°}87. {45°, 90°, 225°, 270°}88. {15°, 75°, 195°, 255°}89. {70.5°, 180°, 289.5°}90. {53.5°, 118.4°, 233.5°,298.4°} 91.

92.

93.

94.

95. 96. 97.

98.

99. (b) 8.6602567 ft; There maybe a discrepancy in the final digits.

1

0 20

–1

f (x) = arctan( ) – arctan( )15x

5x

t �50

arccos�d � 550

450 ��

1

2��3�5

7 �0�

x � arcsin�5y � 3

4 �x � � 1

3 arctan 2y� �

2

3

x � 2 arccos y

3

x � arcsin 2y

4

3� 2n

7

6� 2n

3� 2n

6� 2n

5

3� 2n

� 2n

3� 2n

0 � 2n

7

8,9

8,11

8,13

8,15

8 ��

8,3

8,5

8,

13

8,15

8 ��

8,3

8,5

8, 7

8,9

8,11

8,

�

4,3

4,5

4,7

4 �2.411864997�

�.7297276562,

2,

5 10

x

�



100. Snell’s Law Recall Snell’s law fromExercises 102 and 103 of Section 5.3:

where is the speed of light in one medium,is the speed of light in a second medium,

and and are the angles shown in the fig-ure. Suppose a light is shining up through water into the air as in the figure. As increases, approaches 90°, at which pointno light will emerge from the water. Assume the ratio in this case is .752. Forwhat value of does ? This value of is called the critical angle forwater.

101. Snell’s Law Refer to Exercise 100. What happens when is greater than thecritical angle?

102. British Nautical Mile The Britishnautical mile is defined as the length ofa minute of arc of a meridian. SinceEarth is flat at its poles, the nauticalmile, in feet, is given by

where is the latitude in degrees. See the figure. (Source: Bushaw, Donald et al.,A Sourcebook of Applications of School Mathematics. Copyright © 1980 by TheMathematical Association of America.)

(a) Find the latitude between 0° and 90° at which the nautical mile is 6074 ft.(b) At what latitude between 0° and 180° is the nautical mile 6108 ft?(c) In the United States, the nautical mile is defined everywhere as 6080.2 ft. At

what latitude between 0° and 90° does this agree with the British nautical mile?

103. The function is not found on graphingcalculators. However, with some models it can begraphed as

(This formula appears as in the screen here.) Use the formula to obtain the graph

of in the window by .

104. (a) Use the graph of to approximate .

(b) Use the inverse sine key of a graphing calculator to approximate .sin�1 .4

sin�1 .4y � sin�1 x

��2,

2 ��4, 4y � csc�1 x

Y1

y � ��x � 0� � �x � 0��

2� tan�1��x2 � 1� ��.

y � csc�1 x

�

L � 6077 � 31 cos 2�,

�1

�1�2 � 90°�1

c1

c2

�2�1

�2�1

c2

c1

c1

c2

�sin �1

sin �2

,

100. 48.8° 101. The light beamis completely underwater.102. (a) 42.2° (b) 90°(c) 48.0°103.

104. (a), (b)

In both cases,sin�1 .4 � .41151685.

–1 1

�2

Radian mode

–4 4

y = csc–1 x�2

�2

–

Air

Water

�1

�2

A nautical mile isthe length on anyof the meridianscut by a centralangle of measure1 minute.

1. Given , use trigonometric identities to find and

2. Express in terms of and , and simplify.

3. Find and if x is inquadrant III, and y is in quadrant II.

4. Use a half-angle identity to find sin��22.5°�.

cos y � �25 ,sin x � �

13 ,tan�x � y�,cos�x � y�,sin�x � y�,

cos xsin xtan2 x � sec2 x

cos x.sin x32 � x � 2tan x � �

56 ,

Chapter 7 Test

1.

2. �1cos x �6�61

61

sin x � �5�61

61;


CHAPTER 7 Test 683


5. 6.


7. 8.

9. Use an identity to write each expression as a trigonometric function of alone.

(a) (b)

10. Voltage The voltage in common household current is expressed aswhere is the angular speed (in radians per second) of the genera-

tor at the electrical plant and t is time (in seconds).

(a) Use an identity to express V in terms of cosine.(b) If , what is the maximum voltage? Give the smallest positive value of

t when the maximum voltage occurs.

11. Graph and indicate the coordinates of three points on the graph. Givethe domain and range.

12. Find the exact value of y for each equation.

(a) (b)

(c) (d)

13. Find each exact value.

(a) (b)

14. Write as an algebraic (nontrigonometric) expression in u.

15. Solve for solutions over the interval .

16. Solve for solutions over the interval . Expressapproximate solutions to the nearest tenth of a degree.

17. Solve for solutions over the interval .

18. Solve , giving all solutions in radians.

19. Solve each equation for x.

(a) (b)

20. (Modeling) Movement of a Runner’s Arm A runner’s arm swings rhythmicallyaccording to the model

where y represents the angle between the actual position of the upper arm and thedownward vertical position and t represents time in seconds. At what times over theinterval is the angle y equal to 0?�0, 3�

y �

8 cos�t �

1

3��,

arcsin x � arctan 4

3y � cos 3x

2�3 sin x

2� 3

�0, 2�cos x � cos 2x

�0°, 360°�csc2 � � 2 cot � � 4

�0°, 360°�sin2 � � cos2 � � 1

tan�arcsin u�

sin�2 cos�1 1

3�cos�arcsin 2

3�y � arcsec��2�y � tan�1 0

y � sin�1��3

2 �y � arccos��1

2�

y � sin�1 x,

� � 120

�V � 163 sin �t,

sin� � ��cos�270° � ��

cos 2A �cot A � tan A

csc A sec Asec2 B �

1

1 � sin2 B

cot x

2� cot xsec x � sin x tan x

3.

4.

5.

6.

9. (a) (b)

10. (a)

(b) 163 volts; sec

11.

12. (a) (b)

(c) 0 (d)

13. (a) (b)

14. 15. {90°, 270°}

16. {18.4°, 135°, 198.4°, 315°}

17.

18. , , where

n is any integer

19. (a)

(b)

20. sec, sec, sec17

6

11

6

5

6

� 4

5�x �

1

3 arccos y

4

3� 4n

2

3� 4n

�0,2

3,4

3 �

u�1 � u2

1 � u2

4�2

9�5

3

2

3

�

3

2

3

y

x1–1

(0, 0) y = sin–1 x

�2

�2

–�2(–1, – )

�2(1, )

�

2,

2 ��1, 1;

1

240

V � 163 cos�

2� �t�

�sin ��sin �

cot x

2� cot x � csc x

sec x � sin x tan x � cos x

��2 � �2

2

tan�x � y� �2�2 � 4�21

8 � �42

cos�x � y� �4�2 � �21

15;

sin�x � y� �2 � 2�42

15;



Chapter 7 Quantitative ReasoningHow can we determine the amount of oil in a submerged storage tank?The level of oil in a storage tank buried in the ground can be found in much thesame way a dipstick is used to determine the oil level in an automobile crankcase.The person in the figure on the left has lowered a calibrated rod into an oil storagetank. When the rod is removed, the reading on the rod can be used with the dimen-sions of the storage tank to calculate the amount of oil in the tank.

Suppose the ends of the cylindrical storage tank in the figure are circles ofradius 3 ft and the cylinder is 20 ft long. Determine the volume of oil in the tank ifthe rod shows a depth of 2 ft. (Hint: The volume will be 20 times the area of theshaded segment of the circle shown in the figure on the right.)

3 ft

2 ft

20 165 ft3�9 arctan��8 � � �8 �


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Trigonometric Identities and Equationsavijitmaths.weebly.com/.../5/19956057/book1-_trigonometric_identities.pdf · 606 CHAPTER 7 Trigonometric Identities and Equations 7.1 Fundamental