Trigonometric Identities and Equation

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 sincos

sin cos

sin cos

tan :tancossin

cos cos .

cos x

rand cos

x

r,

sin sin .

sin sin

sin yr

and sin yr

,

x, yx, y

sin() = = sin yr

O x

y

yx

(x, y)

yr

(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

349 sec

349

sec2

259

1 sec2

tan 53 532 1 sec2 tan2 1 sec2

cotsin 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

11 tan2 x cos x

11 tan2 x

cos2 x

11 tan2 x

1

sec2 x

1 tan2 x sec2 x.

tan 53 ; cot 1

53

35

cot 1

tan

cot 1

tan

sin 534

34

33434 53 sin 1

sec tan sin cos tan sin

tan sin cos

From part (a),tan 53

1sec

334

33434 ;



Figure 2 Write 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 85tan s 0sec s

114

sec s 0tan s 7

2tan s 0coss

55

cot s 13

cos s 34

tanx sin x .6cos x .8cot x tan x 1.6

cosx cos x .65tanx tan x 2.6

1. 2.6 2. .65 3. .625

4. .75 5. 6.

7. 8.

9. 10. 58

105

11

7711

25

5

310

107

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 xy2 = 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 47

7

cot 37

7tan

73

cos 34

sin 7

4

csc 54

sec 53

cot 34

tan 43

cos 35

csc 53

sec 54

tan 34

cos 45sin

35

sec 521

21cot

212

tan 221

21cos

215

sin 25

csc 17sec 17

4

cot 4cos 417

17

sin 17

17

csc 5612

sec 5cot 612

tan 26sin 26

5

csc 32

sec 35

5

cot 52

tan 25

5

cos 53

f x f xf x f xf x f x

Relating ConceptsFor individual or collaborative investigation

(Exercises 1217)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 1217 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?14. Complete the statement: .15. Is the function defined by even or odd?16. Complete the statement: .17. 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 14

sin 0sec 43

cos 0sin 45sin 0cot

43

csc 52

tan 14

cos 15sin

23

4

4

2 2

4

4

2 2

4

4

2 2

f x f xf x f x

f x tan xtanx f x cos x

cosx f x sin x

sinx

f x f xf 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 1tan x sec2 x 1

cot x 1 sin2 x

sin x

sin x 1 cos2 x

tan 22p 4

p

sin 2x 1

x 1

csc 415

15sec 4

cot 1515tan 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 x31. 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 students 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 students 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 x46. 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 1sin2 csc2 1cot2 1 tan2 cos csc sec cot sin cot sin

sec p 4

ptan

sin cos xx 1

sin cos 1sin2 cos2 1

1 cot2 csc2cos2 x

csc2 x cot2 x sin2 x1 sin2 x

sinxsec x

csc x

1sec2 x

sec2 x 1

sin2 xcos2 x

tan x cos x

1

sin xcos x

tan2 x 1

sec2 xcosx 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 sin3xcos2xsin4x

cos4xsin2x

sin2x

22 89

22 89

256 6012

256 6012

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 6771)In Chapter 6 we graphed functions defined by

with the assumption that . To see what happens when , work Ex-ercises 6771 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 6770 to rewrite the following with a positive value

of 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. cosx 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 sin3xcos2xsin4x

y cos4xy cos4x

y sin2xy sin2x

b 0b 0

y c a f bx d

sin x cos xsec xcsc x 3

sec x tan xsin xcos x

15

tansec

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 incalculus to simplify trigonometric ex-pressions, determine derivatives oftrigonometric functions, and changethe 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 14.

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

1sin

1

cos

cos

sin cos

sin sin cos

1sin

1cos ,

sin x 12.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 1tan2 x 1 tan2 x 1

cot2 x 1tan2 x tan2 x tan2 x

1tan2 x

tan2 x1 cot2 x tan2 x tan2 x cot2 x

tan2 x1 cot2 x 1

1 sin2 x

cos ssin s cot s;

sin ssin s 1 cot s 1

cos s

sin s

sin ssin s

csc s 1sin s csc scos s sin s 1

sin scos s sin s

cot s 1 csc scos 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 x1 sin x

x y x y x2 y2 1 sin2 x

cos x1 sin x

1 sin xcos x

1 sin x 1 sin x

cos x1 sin x

1 sin x.

cos x

1 sin x

1 sin xcos x

1cos2 t sec

2 t; 1sin2 t csc

2 t sec2 t csc2 t

1

cos2 t

1sin2 t

tan t sin tcos t ; cot t cos tsin t

sin tcos t

1

sin t cos t

cos t

sin t

1sin t cos t

ab a

1b tan t

1sin t cos t

cot t 1

sin t cos t

a bc

ac

bc

tan t cot tsin t cos t

tan t

sin t cos t

cot tsin t cos t

tan t cot tsin t cos t

sec2 t csc2 t



(Section R.4)x2 2xy y2 x y2

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 14: 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

sec tan cos sec tan cos

sec tan 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.


Et

ksin2 cos2 1 (Here 2Ft). k1

ksin22Ft cos22Ft k sin22Ft k cos22Ft

Et Lt Ct

Et Lt Ct.

Ct k cos22Ft,

Lt k sin22Ft

1 sin 1 sin

1 sin cos

1 sin cos

sec tan sec tan

1cos

sin cos

1cos

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 3368, verify that each trigonometric equation is an identity. SeeExamples 15.

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 ssin2 1 cot2 1cos2 tan2 1 1

tan2 1sec

sec 1 sin2

cos cos

tan sec

sin cot csc

cos

1tan2

cot tan

sin2 xcos2 x

sin x csc xcsc2 t 1sec2 x 1

csc sec

cot sin tan

cos

cot sin tan cos

sin3 cos3 sin3 x cos3 xcot4 x 3 cot2 x 2cos4 x 2 cos2 x 14 tan2 tan 32 sin2 x 3 sin x 1tan x cot x2 tan x cot x2sin x 12 sin x 12sec2 1sin2 1

sin cos 21

1 cos x

11 cos x

1 tan s2 2 tan s

1 sin t2 cos2 tcos

sin

sin 1 cos

cos x

sec x

sin xcsc x

1sin 1

1

sin 11

csc2

1sec2

cos sec csc

tan scot s csc ssec x

csc x

csc x

sec xcot

1cot

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 xcot2 x 2cot2 x 2 cot2 x 1cos2 x 124 tan 3 tan 12 sin x 1 sin x 1

sec 1 sec 1sin 1 sin 11 2 sin cos

2 cot x csc x2 cos xsin2 x

sec2 s2 2 sin t

csc 1

sin

2 sec2 2

cos2

1 cot 1 sec s

1sin x cos x

1sin 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-

dents reasoning.70. An equation that is an identity has an infinite number of solutions. If an equation has

an infinite number of solutions, is it necessarily an identity? Explain.

0 1 12 90cos sin 1

sin4 cos4 sin2 cos2

1

sec csc cos sin cot tan

sec tan 2 1 sin 1 sin

1 cos x1 cos x

1 cos x1 cos x

4 cot x csc x

1 sin x cos x2 21 sin x 1 cos x

tan2 t 1sec2 t

tan t cot ttan t cot t

cot2 t 11 cot2 t

1 2 sin2 t

sec4 s tan4 ssec2 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

1tan sec

1

tan sec 2 tan

cot 1cot 1

1 tan 1 tan

1 cos x1 cos x

cot x csc x2

tan s1 cos s

sin s

1 cos s cot s sec s csc s

1sec tan

sec tan

11 sin

1

1 sin 2 sec2

sec tan 2 1sec 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 Lamberts law, the

intensity 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

,Ct 3 sin26,000,000t

Lt 3 cos26,000,000t

sin2t

P ky2

y 4 cos2t

0

I k cos2

sin x 1 cos2 x

cos t 1 sin2 tcsc t 1 cot2 tcos2 s cos ssincsc s 1

11 sin s

1

1 sin s sec2 s

tan s cot stan s cot s

2 sin2 s

1 cot2 s sec2 s

sec2 s 12 5 cos s

sin s 2 csc s 5 cot s

tan sin cos cos 1

sin tan

csc cot sec 1sec 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 16k1 sin22t

P 16k cos22t

cos2 1 2nI k1 sin2

sin x 0.

tan sin cos sec

cos 1sin 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 .Et

107t 0Yscl 1

Xscl 1071, 40, 106Et Lt Ct

7.3 Sum and Difference Identities 621

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

4

0 106

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 cosA B, sinA 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) SP (1, 0)

Q (cos B, sin B

A B.

B A.cosA B.

cos A cos B cos

2 cos 0 0 1 1.

cosA B cos 2 0 cos 2 0,B 0,A 2cos A cos B.

cosA B

(b) Let , , and . for all inputs.(c) Et 3

Y3 3Y3 EtY2 CtY1 Lt



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 93cos 5

12

cos(A B) cos A cos B sin A sin Bcos(A B) cos A cos B sin A sin B

cosA B cos A cos B sin A sin B

cos A cos B sin Asin B cos A cosB sin A sinB

cosA B cosA B

cosA B.A BA BcosA B,

cosA B cos A cos B sin A sin B. 2 2 cosA 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.cos2A B 2 cosA B 1 sin2A B

cos A cos B2 sin A sin B2.

cosA B 12 sinA B 02PR 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

= .512

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.

cosA B sin . 0 cos 1 sin

cos90 cos 90 cos sin 90 sin cosA 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 cos87 93

6 2

4

32

22

12

22

cos

6 cos

4 sin

6 sin

4

6

2

12 ;

4

3

12 cos

5

12

cos 6 4

6 24

22

32

22

12

cos 45 cos 30 sin 45 sin 30 cos 15 cos45 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 cos180 cos 180 cos sin 180 sin

cos180

cosA B

cosA BcosA B

4

sec 4 sec sec 2 34 sec

csc 3

4

sec

120 90 30

cos90 cos30 sin cos30

65 90 25

tan90 tan 25 cot tan 25

tan90 cot .

csc 3

4

sec sin cos30cot 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.

tanB tan BB

tan sin cos tanA B tan A tan B

1 tan A tan B

sin Acos A

sin Bcos B

1 sin Acos A

sin Bcos B

sin A cos Bcos A cos B

cos A sin Bcos A cos B

cos A cos Bcos A cos B

sin A sin Bcos A cos B

tanA B

sin A cos B cos A sin B1

cos A cos B sin A sin B1

1cos A cos B

1cos A cos B

1cos A cos B .

sin A cos B cos A sin Bcos A cos B sin A sin B

.

tanA B sinA BcosA B

tanA B,

sin(A B) sin A cos B cos A sin Bsin(A B) sin A cos B cos A sin B

sinA B sin A cos B cos A sin B sin A cosB cos A sinB

sinA B sinA B

sinA B.sinA BsinA B

sinA B sin A cos B cos A sin B.

cos90 A cos B sin90 A sin B cos90 A B

sinA B cos90 A B

A Bsin cos90 ,


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)


32

sin 120 sin120

sin 40 cos 160 cos 40 sin 160 sin40 160

2 3

22 3

2

4 23

2

3 3 1 3

1 3

3 11 3

1 31 3

3 1

1 3 1

tan 3 tan

4

1 tan 3 tan

4

tan 7

12

tan 3 4

6 24

22

32

22

12

sin 45 cos 30 cos 45 sin 30 sin 75 sin45 30

sin 40 cos 160 cos 40 sin 160tan 7

12

tan(A B) tan A tan B1 tan A tan B

tan(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

tanA B

43

125

1 43 125

1615

1 4815

16156315

1663

tan B 125 .tan A 43

tanA B,

2065

3665

1665

sinA B 45 513 351213

sinA B.sin B 1213 .

cos A 0. cos A 35

1625 . cos

2 A

925

sin A 45 1625 cos

2 A 1

sin2 A cos2 A 1

sinA B

A BtanA BsinA B

B 32 .cos B 5

13 ,

2 A ,

sin A 45 ,

A B

sin 0 cos 1 sin

sin180 sin 180 cos cos 180 sin

tan45 tan 45 tan 1 tan 45 tan

1 tan 1 tan

12

cos 32

sin .

sin30 sin 30 cos cos 30 sin sinA B,

sin180 tan45 sin30



(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 sec

so 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 the

same 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


Vt 163 cost 2 163 sin t. 2 ,

cosx 2 cos 2 x cos 2 x sin x.

200

0 .05

200

For x = t,V(t) = 163 sin 120t

200, 200Vt 163 sin t 163 sin 120 t.

120 radians 602

Vt 163 sin t.Vt 163 cost

200, 200.0, .05

Vt 163 sin t,

A BtanA B 1663,sinA B

1665


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

I II1. 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 tan551 tan 80 tan55

tan 80 tan 551 tan 80 tan 55

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

tan712sin712sin 12tan

12tan

5

12

sin 5

12

cot 10 tan2 20sin3 15 cos 25sin cos2 10tan cot45 2

72 cot 1833 sin 57

6 sin 23 6cot 3

tan 174 3sec 146 42cot 9

10

sin 5

8

sin 2

5cos

12

cos 7

9

cos 2

9

sin 7

9

sin 2

9

cos 40 cos 50 sin 40 sin 50

cos 12cos 712105 60 45105 60 45

cos105cos15

cos x cos y sin x sin ycos x sin y sin x cos ysin x cos y cos x sin ysinx y sin x cos y cos x sin ysinx y sin x sin y cos x cos ycosx y cos x cos y sin x sin ycosx 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

22

2 3

6 24

6 24

2 32 3

6 24

803

1003

cot84 3

csc56 42tan25 cos 8 cos 10sin

5

12

6 24

2 64

2 64

2 64

6 24

6 24

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) II56. (a) (b) (c) (d) III

58.

59.

60. (a)

(b)

61. 62.

63. 64.

65. 66. 6 24

2 3

2 36 2

4

2 36 2

4

6 24

2 64

6 24

6 24

3677

8485

7785

7736

1385

3685

6316

3365

1665

85 5220 210

42 59

210 29

86 34 66

86 325

4 6625

6316

3365

1665

1 tan x1 tan x

3 tan x 13 tan x

3 tan 13 tan

3 tan 1 3 tan

22

cos x sin x

12 cos x 3 sin xsin

22

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 5760)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 5760 in order, to see how this is done.

57. By writing 195 as , use the identity for to express cos 195as 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 5759.

(a) cos 255 (b)

Find each exact value. Use the technique developed in Relating Concepts Exer-cises 5760.

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

64. tan 285 65. 66. sin1312 tan 1112

cos 11

12

cos 195 cosA B

cosA B180 15

cosA BcosA B

sin t 45cos s

1517

cos t 35cos s

817

sin t 1213

sin s 35

sin t 13

sin s 23

sin t 35cos s

15

sin t 513

cos s 35

s ttans tsins tcoss t

tan 4 xtanx 6 tan 30tan60 sin 4 xsin56 xsin180 sin45 cos 2 xcosx 32 cos180 cos90



71.

72.

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

1 tan x1 tan x

tan 4 xsin 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 always

equal to 115 volts?

(Modeling) Solve each problem.81. Back Stress If a person bends at the

waist 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 30W 170

F .6W sin 90

sin 12

sins tsin t

coss t

cos t

sin ssin t cos t

sinx ysinx y

tan x tan ytan x tan y

sins tcos s cos t

tan s tan t

cos cos sin

tan cot

tanx y tan y x 2tan x tan y1 tan x tan y

sinx y sinx y 2 sin x cos y

1 tan x1 tan x

sin 2 x

sinA B C 0

tan270

cotA BcscA BsecA Bcos A sin90 A

90 Acos90 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 cosA A

cos A cos B sin A sin B.cosA 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 120t + 40 cos 120t

82. Back Stress Refer to Exercise 81.(a) Suppose a 200-lb person bends at the waist so that . Estimate the force

exerted on the persons back muscles.(b) Approximate graphically the value of that results in the back muscles exerting

a force of 400 lb.83. Sound Waves Sound is a result of waves applying pressure to a persons 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 its

source in the window by Describe P at this distance.(b) Now let and . Graph the sound pressure in the window by

What happens to pressure P as radius r increases?(c) Suppose a person stands at a radius r so that , where n is a positive

integer. 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 sin120t 60, 60.0, .05

V2 40 cos 120 tV1 30 sin 120t

r n2, 2.

0, 20]t 10a 3.05, .05.0, .05

r 10ft2a .4c 1026

4.9

P a

r cos2r ct

45


82. (a) 408 lb (b) 46.183. (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 2r4.9

0 .05

.05

.05

For x = t,P(t) = cos[ 1026t].410 204.9


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 cos2 30

Looking Ahead to CalculusThe identities

and

can be rewritten as

and

These identities are used to integratethe functions and

cos2 A.gA f A sin2 A

cos2 A 12

1 cos 2A.

sin2 A 12

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 45

sin2 1625

cos 35sin2 cos2 1; sin2 352 1sin 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 tanA A

tanA B,

sin 2A 2 sin A cos A sin A cos A cos A sin A

sin 2A sinA A

B A.sinA 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 A1 sin2 A

cos2 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 x2 sin x cos x

cot x sin 2x cos x

sin x sin 2x

cot x sin 2x 1 cos 2x

tan 2 sin 2cos 2

2425

7

25

247

cos 2.sin 2tan 2

tan 2 2 tan

1 tan2

2431 169

83

79

247

tan sin cos

45

35

43.

tan 2

cos 2 cos2 sin2 9

25 1625

725

cos 2,

sin 45 ; cos 35 2 45 35 2425 .

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 xcos2 x 1 sin2 x 2 sin x1 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 sin2x x

sin 30 12 12

12

14

12

sin 30

A 152 sin A cos A sin 2A, 12

sin2 15

12

2 sin 15 cos 15

12 2. sin 15 cos 15

12

2 sin 15 cos 15

12 2.A 15

sin 2A 2 sin A cos A.

cos2 7x sin2 7x cos 27x cos 14x.

cos 2A cos2 A sin2 A.

sin 15 cos 15cos2 7x sin2 7x



Looking Ahead to CalculusThe product-to-sum identities are usedin calculus to find integrals of func-tions that are products of trigonometricfunctions. One classic calculus text in-cludes the following example:

Evaluate

The first solution line reads:We may write

cos 5x cos 3x 12

cos 8x cos 2x.

cos 5x cos 3xdx.

2000

500

0 .05

For x = t,

W(t) = (163 sin 120t)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)]

sinA BsinA B

sin A sin B 12

cosA B cosA B.

cosA BcosA B

cos A cos B 12

cosA B cosA B.

cosA B cosA B 2 cos A cos B cosA B cos A cos B sin A sin B

cosA B cos A cos B sin A sin BcosA BcosA B

cosA BcosA B

163215 1771.

W V 2

R

163 sin 120t2

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 62

sin22 sin 2 sin 4 2 cos2 42 sin2 42

4 B.2 Asin A sin B,

sin 2 sin 4

cos A cos B 2 sinA B2 sinA B2 cos A cos B 2 cosA B2 cosA B2 sin A sin B 2 cosA B2 sinA B2 sin A sin B 2 sinA B2 cosA B2

12

sin 3 12

sin

cos 2 sin 12

sin2 sin2

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 A2

1 cos A

sin A.

tan A2 ,

tan A2

sin A

1 cos A

sin 2 A2

1 cos 2 A2

tan A2

sin A2cos

A2

2 sin A2 cos

A2

2 cos2 A2

tan A2

tan A2

sin A2cos

A2

1 cos A21 cos A2

1 cos A1 cos Acos

A2 .sin

A2tan

A2

A2 . cos

A2

1 cos A2 cos x 1 cos 2x2 cos2 x

1 cos 2x2

1 cos 2x 2 cos2 x

cos 2x 2 cos2 x 1.cos

A2

A2

A2 .

x A2 ;2x A, sin A2

1 cos A2 sin x 1 cos 2x2

2 sin2 x; 2 sin2 x 1 cos 2x cos 2x 1 2 sin2 x

sin A2 ,

tan A2 .cos A2 ,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.


22 1

2 2 1

tan 22.5 22 2

2

2 2

2 22 2

22 2

2

tan 22.5 tan 452

sin 45

1 cos 45 22

1 22

22.5 12 45,

tan A2 sin A

1 cos A .

1 322

1 32 22 2

2 3

2

cos 15 cos 12

30 1 cos 302

sin A2

tan A2cos A2

A2 162,

A 324,A2

tan A2

tan A2

1 cos A1 cos A tan A2 sin A1 cos A tan A2 1 cos Asin Acos

A2

1 cos A2 sin A2 1 cos A2



Figure 7

0

32

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 5sin 5 tan

52

.

A 5tan A2

1 cos 12x2 cos 12x2 cos 6x.

cos A2

1 cos A2cos

A2 .

1 cos 5sin 51 cos 12x2

cos s2 .sin

s2

tan s2

tan s

2

sin s2cos

s2

66

306

55

cos s

2 1 232 56 306

sin s

2 1 232 16 66

sin s2tan s2

cos s2

s2

3

4

s

2 ,

3

2

s 2

tan s2 .sin s2 ,cos

s2 ,

3

2 s 2,cos s

23 ,

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

116

sin 5914

cos 94.2

14

tan 6812

tan 102

24

22

22

32

33

32

sin 2 266

25

cos 2 1925;

sin 2 455

49

cos 2 3949

;

sin 2x 1517

cos 2x 8

17;

sin 2x 45cos 2x

35 ;

sin 2 120169

cos 2 119169

;

sin 2 421

25

cos 2 1725;

sin x 6

6cos x

306

;

sin x 102

12cos x

4212

;

sin 2

4cos

144

;

sin 5

5cos 25

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.

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

25. 26.

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

27. 28. sec 2x sec2 x sec4 x

2 sec2 x sec4 xsin x cos x2 sin 2x 1

4 tan x cos2 x 2 tan x1 tan2 x

cos4 x sin4 x

cos 4xtan 3xsin 4xcos 3x

18

sin 29.5 cos 29.5

14

12

sin2 47.1tan 34

21 tan2 34tan 51

1 tan2 51

cos2

8

12

2 cos2 6712

11 2 sin2 2212

1 2 sin2 152 tan 151 tan2 15cos

2 15 sin2 15

sin 0cos 3

52

cos 0sin 5

72,

sin x 0tan x 53

2x,

cos x 0tan x 22x,

sin 0cos 1213

2

cos 0sin 252

2 x cos 2x

23

x,

2 x cos 2x

512

x,

cos 2 34

cos 2 35



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. 557

50 151010

50 10510

50 206

10

134

104

2 32

2 3

2

2 32 3

2

2 3

22 2

2

2 cos 6x sin 3x2 cos 6x cos 2x2 cos 98.5 sin 3.52 sin 11.5 cos 36.52 cos 6.5x cos 1.5x2 sin 3x sin x

12

cos x 12

cos 9x

52

cos 5x 52

cos x

sin 225 sin 55sin 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 180tan 52

,cot

2,

180 270tan 7

3,tan

2,

2 x cot x 3,cos

x

2,

0 x

2tan x 2sin

x

2,

180 270sin 15 ,cos

2,

90 180sin 35 ,tan

2,

2 x cos x

58

,sin x

2,

0 x

2cos x

14

,cos x

2,

a b.a2 b2,b 0a 0

tan 22.5 2 1tan A2 sin A

1 cos A

tan 22.5 3 22,tan A2 1 cos A1 cos A7.5 121230sin 7.5.

sin 165cos 165tan 195cos 195sin 195sin 67.5

sin 9x sin 3xcos 4x cos 8xsin 102 sin 95sin 25 sin48cos 5x cos 8xcos 4x cos 2x

sin 4x sin 5x5 cos 3x cos 2x2 cos 85 sin 1402 sin 58 cos 102

sin 2x2 sin x

cos2 x

2 sin2

x

2sin2

x

2

tan x sin x2 tan x

cot2 x

2

1 cos x2

sin2 xsec2

x

2

21 cos x

cot x tan xcot x tan x

cos 2xtan x cot x 2 csc 2x

cos 2x 1 tan2 x1 tan2 x

sin 2x cos 2x sin 2x 4 sin3 x cos x

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

cot x tan x

1 cos 2xsin 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 cant 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 7578, 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 30m 54

m 32

m 1

sin

2

1m

,

tan2 x cot x.

tan sin cos

tanA Btan2 x.

1 cos A.tan A2

1 cos Asin Atan

A2

sin A1 cos A

sin 158.21 cos 158.2

1 cos 59.74sin 59.741 cos 1651 cos 165

1 cos 1471 cos 1471 cos 7621 cos 402


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

b50

cos

2

R bR

tan 2 x sin 2 xcos 2 x

tan

2

tan 79.1tan 29.87cot 82.5tan 73.5

cos 38sin 20

50 50b

R2

2

h

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 incandescent

lightbulb is turned on with amperage Graph the wattageconsumed 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 cost c.

50, 300.0, .05W VII 1.23 sin120t.

V 163 sin120t

W a cost 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 120t)(1.23 sin 120t)

240

c 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 scos s sin s

cos s sin s

sin s cos s

tan x2 4 sec x tan xtan 4 2 tan 22 sec2 2sin s

1 cos s

1 cos ssin s

2 csc stan2 t 1tan t csc2 t

tan t

cos 2x 2 sec2 x

sec2 xcos x

1 tan2 x21 tan2 x2

csc4 x cot4 x 1 cos2 x1 cos2 x

sin tan 1 cos

tan

1 tan2

2

2 cos 1 cos

sinx ycosx y

cot x cot y

1 cot x cot y

1sec t 1

1

sec t 1 2 cot t csc tcot tan

2 cos2 1sin cos

21 cos x

tan2 x

2 1sin 2

2 tan 1 tan2

1 sin tcos t

1

sec t tan tsin t

1 cos t

1 cos tsin 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. 12

cot x

2

12

tan x

2 cot x

2sin x sin3 xcos x

sin 2x

csc t 1csc t 1

sec t tan t2cos4 x sin4 x

cos2 x 1 tan2 x

2 cos2 x

2 tan x tan x sin x

tanx y tan y1 tanx 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 and

each 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

32

32

2

2

[ , ]2 2

( , 1)2

( , 1)2

2 , 2 .1, 1,1, 1,y sin x

2 , 2 ,y sin x

f 1x.

fxfx,f 1x

y x.f 1f 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 usedin calculus to solve certain types of re-lated-rates problems and to integratecertain 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.

sin12

2

2

1 1

2

2

1 1

Y 2 1.570796.X 1,Y 6 .52359878.

X 12 ,1.12

Y1 sin1 X

sin122

y sin1 x.1, 2

y 2 .sin y 1y sin11

y 6 .

6sin

6

12

sin y 12 .

12 .2 , 2

y arcsin 12

arcsin 12

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 sin12y sin11y arcsin 12

sin y xy sin1 x2y 4y log2 42 , 2

y arcsin xy sin1 x

2 y

2 .x sin y,y arcsin xy sin1 x

arcsin x.sin1 xy sin1 x.2 , 2 ,y sin x,1, 1,y sin1 x

y sin1 xx sin y.

y sin xy x

y 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.

3

2sin

3

2 1.

3

2 ,

sin11

1

0

y = sin1 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 sin1x 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 cos1 x

y

x

10

y = cos1 x or y = arccos x

(1, 0)1

(1, )

6( , )32

4( , )22

312( , )

23

( , )22 34( , )32 56

12( , )(0, )2

y

1

0

y = cos xRestricted domain

[0, ]

(0, 1)

(, 1)

1

x( , 0)2

2

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Trigonometric Identities and Equation