Trigonometric Identities and Equations · Pythagorean identity p. 313 Pitagoras identidad ... sm 9...

Trigonometric Identities and Equations

: Then

O In Chapter 4, youlearned to graph trigonometric functions and to solve right and oblique triangles.

-Now

O In Chapter 5, you will:

■ Use and verify trigonometric identities.

- Solve trigonometric equations.

■ Use sum and difference identities to evaluate trigonometric expressions and solve equations.

■ Use double-angle, power- reducing, half-angle, and product-sum identities to evaluate trigonometric expressions and solve equations.

: Why? a

O BUSINESS Musicians tune their instruments by listening for a beat, which is an interference between two sound waves with slightly different frequencies. The sum of the sound waves can be represented using a trigonometric equation.

PREREAD Using what you know about trigonometric functions, make a prediction of what you w ill learn in Chapter 5.

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Self-CheckPractice Worksheets

Get Ready for the ChapterDiagnose Readiness You have tw o options fo r checking Prerequisite Skills.

Textbook Option Take the Quick Check below.

QuickCheck

Solve each equation by factoring. (Lesson 0-3)

1. x 2 + 5 x — 24 = 0 2. x 2 — 11 x + 28 = 0

3. 2x2 — 9x — 5 = 0 4. 15x2 4- 26x + 8 = 0

5. 2x3 - 2x2 — 12x= 0 6. 12x3 + 78x2 - 42* = 0

7. ROCKETS A rocket is projected vertically into the air. Its vertical distance in feet after t seconds is represented by s(t) = - 1 6 f 2 + 1921. Find the amount of time that the rocket is in the air. (Lesson 0-3)

Find the missing side lengths and angle measures of each triangle. (Lessons 4-1 and 4-7)

8 .35

Find the exact value of each expression. (Lesson 4-3)7tt12. cot 420°

14. s e c ^

16. c s c y

13. cos-

15. tan 480°

17. sin 510°

2 Online Option Take an online se lf-check Chapter Readiness Quiz a t c o n n e c tE D .m ca ra w -h ill.co m .

NewVocabulary

English Espanol

trig on om e tric iden tity p. 312 identidad trigonom etrica

rec iproca l iden tity p. 312 identidad reciproca

quo tien t iden tity p. 312 cociente de identidad

Pythagorean iden tity p. 313 Pitagoras identidad

o dd -e ven -ide n tity p. 314 im par-inc luso -de identidad

cofunction p. 314 co funcion

verify an iden tity p. 320 ve rifica r una identidad

sum iden tity p. 337 sum a de identidad

reduction iden tity p. 340 identidad de reduccion

doub le -ang le iden tity p. 346 dob le-angu lo de la identidad

pow er-reduc ing iden tity p. 347 poder-reduc ir la identidad

ha lf-ang le iden tity p. 348 m edio angulo identidad

ReviewVocabularyextraneous so lu tion p. 91 solucion extrana a solution that does not satisfy the original equationquadranta l angle p. 243 angulo cuadranta an angle e in standard position that has a terminal side that lies on one of the coordinate axes

u n it c irc le p. 247 c ircu lo un ita rio a circle of radius 1 centered at the origin

period ic fun c tio n p. 250 func ion period ica a function with range values that repeat at regular intervalstrig o n o m e tric fun c tions p. 220 func iones trigonom etricas Let e be any angle in standard position and point P(x, y) be a point on the terminal side of 9. Let r represent the nonzero distance from Pto theorigin or \r\ = \J x 2 + y 2 i= 0. Then the trigonometric functions of 9 are as follows.

sin 9 = ■

csc 9 = j , y ± 0

cos 9 = j

sec 9 = j , x 0

tan 9 = x 0

co te = | , y # 0

311

NewVocabularyidentitytrigonometric identity cofunction

1 Basic Trigonometric Identities An equation is an identity if the left side is equal to the

right side for all values of the variable for which both sides are defined. Consider the equations below.

= x + 3

sin x = 1 — cos x

This is an identity since both sides of the equation are defined and equal for all x such that x # 3.

This is nofan identity. Both sides of this equation are defined and equal for certain values, such as when x = 0, but not for other values for which both sides are defined, such as when * = •?-.

• Many physics and engineering applications, such as determining the path of an aircraft, involve trigonometric functions. These functions are made more flexible if you can change the trigonometric expressions involved from one form to an equivalent but more convenient form You can do this by using trigonometric identities.

Trigonometric identities are identities that involve trigonometric functions. You already know a few basic trigonometric identities. The reciprocal and quotient identities below follow directly from the definitions of the six trigonometric functions introduced in Lesson 4-1.

• You foundtrigonometric values using the unit circle.(Lesson 4-3)

1 Identify and use basic trigonometric identities to find trigonometric values.

2 Use basic trigonometric identities to simplify and rewrite trigonometric expressions.

KeyConcept Reciprocal and Quotient Identities

Reciprocal Identities Quotient Identities

sin 9 = — rCSC 9

cos 9 = —— tan 9 = —j— sec 9 cot 9tan f l = M

cos 9

csc 9 = - 4 —sin 9sec 9 = — cot 9 = —I —cos 9 tan 9

cot 0 = “ * ! sin 9v ....... J

312 I Lesson 5-1

You can use these basic trigonometric identities to find trigonometric values. As with any fraction, the denominator cannot equal zero.

Use Reciprocal and Quotient Identities

a. If csc 9 = —, find sin 9. 42 V5b. If cot x = —— and sin x = --- , find cos x.

5V5

sin 9 = ■c s c 9

17_4

47

Reciprocal Identity

CSC 9 -

. „ _ cos xcot X = — -------sm x

2 _ cos*

Divide. = cos x

GuidedPracticec

1A. If sec x = —, find cos x.

5V5 V53

2 V5 5V5 3

_2 _15

1B. If csc /3 = -y- and sec (3 = find tan /3.

Quotient Identity

cot x = — s i n x = - Y -5V5

Multiply each side by

Simplify.

25

V5

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ReadingM athPowers of Trigonometric Functions sin2 0 is read as sine squared theta and interpreted as the square of the quantity sin 0.

Recall from Lesson 4-3 that trigonometric functions can be defined on a unit circle as shown. Notice that for any angle 9, sine and cosine are the directed lengths of the legs of a right triangle with hypotenuse 1. We can apply the Pythagorean Theorem to this right triangle to establish another basic trigonometric identity.

(sin 0)2 + (cos 0)2 Pythagorean Theorem

Simplify.

While the signs of these directed lengths may change depending on the quadrant in which the triangle lies, notice that because these lengths are squared, the equation above holds true for any value of 9. This equation is one of three Pythagorean identities.

KeyConcept Pythagorean Identities

sin2 0 + cos2 0 = 1 tan2 0 + 1 = sec2 0 cot2 f l + 1 = csc2 0

You will prove the remaining two Pythagorean Identities in Exercises 69 and 70.

Notice the shorthand notation used to represent powers of trigonometric functions: sin2 9 = (sin 9)2, cos2 9 = (cos 9)2, tan2 9 = (tan 9)2, and so on.

StudyTipChecking Answers It is beneficial to confirm your answers using a different identity than the ones you used to solve the problem, as in Example 2, so that you do not make the same mistake twice.

| 1 2 f f i 3 3 Use Pythagorean Identities

If tan 0 = —8 and sin 0 > 0, find sin 0 and cos 0.

Use the Pythagorean Identity that involves tan 9.

tan2 0 + 1 = sec2 9

(—8 ) 2 + 1 = sec2 9

65 = sec2 9

±V65^ = sec 0

+ V 6 5 = ^ — cos 6

^ V 65± — = cos 9 65

Pythagorean Identity

tan 0 = - 8

Simplify.

Take the square root of each side.

Reciprocal Identity

Solve for cos 0.

Since tan 9 = is negative and sin 9 is positive, cos 9 must be negative. So, cos 9 = —-~P-.

You can then use this quotient identity again to find sin 9.

tan 0 =

- 8 =

8V 65 _ 65

CHECK

s in t) cos 0

s in 8 V65

65

Quotient Identity

tan 0 = — 8 and cos 9 -

Multiply each side by


V6S65

V 65 65 '

. . 8V65 , „ V65sin 0 = ■■■■ and cos 9 = — r r - 65 bb

64 J _ .65 65 ' 1 ✓ Simplify.

p GuidedPracticeFind the value of each expression using the given information.

2A. csc 9 and tan 9; cot 9 = —3> cos 9 < 0 2B. cot x and sec x; sin x = —, cos x >0O

ssnsraBBnaasnasmsmsssmnassEmBconnectED.mcgraw-hill.com | 313

Another set of basic trigonometric identities involve cofunctions. A trigonometric function/is a cofunction of another trigonometric function g i f f (a) = g(/3) when a and /3 are complementary angles. In the right triangle shown, angles a and (3 are complementary angles. Using the right triangle ratios, you can show that the following statements are true.

sin a = cos /3 = cos (90° — a) = j

tan a = cot (3 = cot (90° — a) = y

sec a = csc (3 = csc (90° — a) = y

From these statements, we can write the following cofunction identities, which are valid for all real numbers, not just acute angle measures.

StudyTipWriting Cofunction Identities Each of the cofunction identities can also be written in terms of degrees. For example, sin 6 = cos (90° - 0).

KeyConcept Cofunction Identities

sin 0 — cos ( j - I£|cmooII e03 sec 0 = csc l j -

cos 9 = sin ^ - o'j cot 0 = tan ^ j - o'j O cn 0 II C/5 CD O 1

You will prove these identities for any angle in Lesson 5-3.

You have also seen that each of the basic trigonometric functions— sine, cosine, tangent, cosecant, secant, and cotangent—is either odd or even. Using the unit circle, you can show that the following statements are true.

sin a = y sin (—a) = — ycos a = x cos (—a) = x

Recall from Lesson 1-2 that a function/is even if for every x in the domain o f f , f ( —x) = f(x ) and odd if for every x in the domain o f f , f { —x) = —f(x). These relationships lead to the following odd-even identities.

KeyConcept Odd-Even Identities

sin ( - 0 ) = -s in 0

csc ( - 0 ) = -c s c 0

cos (-0) = cos 0

sec (-0) = sec 0

tan (-0) = - ta n 0

cot (—0) = - c o te

You can use cofunction and odd-even identities to find trigonometric values.

P E S E H ® Use Cofunction and Odd-Even Identities

If tan 0 = 1.28, find cot 0 — y j .

cot I# — y j = cot £ — — ffj j

—cot ( ? - )= —tan 6 =(—1.28

Odd-Even Identity

Cofunction Identity

tan 0 = 1.28

p GuidedPractice3. If sin x = —0.37, find cos — y)-

314 | Lesson 5-1 | T rig o n o m e tric Id en tities

TechnologyTipGraphing Reciprocal Functions When using a calculator to graph a reciprocal function, such as y = csc x, you can enter the reciprocal of the function.

P lo t i P lo t i M o tJ\Y iB l/s in < X >\Vz=^Yj =nVh=^Ye=■■■V 6 =\V ? =

2 Simplify and Rewrite Trigonometric Expressions To simplify a trigonometricexpression, start by rewriting it in terms of one trigonometric function or in terms of sine and

cosine only.

Simplify by Rewriting Using Only Sine and Cosine

Simplify csc 9 sec 9 — cot 9.

Solve A lgebra ically

csc 6 sec 6 — cot 1 _ c o s 9 s i n 9

c o s 9s m 9 c o s (

1___________s i n 9 c o s 9 s i n 9

1 c o s 2 9s m 9 c o s 9 s m 9 c o s I

l — cos2 es i n 9 c o s 9

sin2 6 s i n 9 c o s 9s i n 9c o s 9 or tan 9

R ew rite in te rm s o f sine and cosine using Reciprocal and Quotient identities.

M ultip ly.

R ew rite fra c tio n s using a com m on denom inator.

Subtract.


D ivide the num era to r and denom ina to r by s in 0.

S upport G raphically The graphs of y = csc 9 sec 6 — cot 9 and y = tan 9 appear to be identical.

u \ I LJ / /

y = c s c 9 s e c 9 — c o t 9 [

[ — 2 tt, 2 -it] s d y b y I— 2 , 2 ] s c l: 0 .5

p GuidedPractice

4. Simplify sec x — tan x sin x.

Some trigonometric expressions can be simplified by applying identities and factoring.

WatchOut!G raphing While the graphical approach shown in Examples 4 and 5 can lend support to the equality of two expressions, it cannot be used to prove that two expressions are equal. It is impossible to show that the graphs are identical over their entire domain using only the portion of the graph shown on your calculator.

“A: 11" f.j ■> Simplify by Factoring

Simplify sin2 x cos x — sin ( y ~ *)•

Solve A lgebra ically

Cofunction Identity

Factor —cos x from each term .

= —cos x (1 — sin2 x) C om m utative Property

= —cos x (cos2 x ) or —cos3 x Pythagorean Identity

sin2 x cos x — sin ( y ~ * ) = sin2 x cos x — cos x = —cos x (—sin2 x + 1 )

S upport G raphically The graphs below appear to be identical.

y = si n 2 x c o s x — s in — x j

s i z s ; .. A , . A ,

i— ^y = —cos 3r

/ '

[ — 2 t t , 2 -7T] s c l: y b y [ - 2 , 2 ] s c l: 0 .5 [ — 2 tt, 2 x ] s c l: y b y [ — 2 , 2 ] s c l: 0 .5

p GuidedPractice

5. Simplify —csc(y — xj — tan2 x sec X.

&connectED.m cgraw-hili.com ] 315 Q §

You can simplify some trigonometric expressions by combining fractions.

H E E E E E Z E Simplify by Combining Fractions

Simplify sin v r J 1 — sin x1 + sin x

s i n x c o s x _ 1 + s i n x _ s i n x c o s x ( c o s x) _ (1 + s i n x ) ( l - s i n x) 1 — s i n x c o s x (1 — s i n x ) ( c o s x) ( c o s x ) ( l — s i n x)

s i n x c o s xc o s x — s i n x c o s x c o s x — s i n x c o s x

s i n x c o s xc o s x — s i n x c o s x c o s x — s i n x c o s x

c o s x — s i n x c o s x

( c o s 2 x ) ( s i n x — 1 )

( — c o s x ) ( s i n x — 1)

= —cos x

)► GuidedPracticeSimplify each expression.5 A c o s x 1 + s i n x

1 + s i n x c o s x

Common denom inator

M ultip ly.

Pythagorean iden tity

Subtract.

Factor the num erator and denominator.

D ivide out com m on factors,

6 B e s c x c s c x

1 + s e c x 1 — s e c x

ReviewVocabularyconjugate a binomial factor which when multiplied by the original binomial factor has a product that is the difference of two squares (Lesson 0-3)

In calculus, you will sometimes need to rewrite a trigonometric expression so it does not involve a fraction. When the denominator is of the form 1 ± u or u ± 1, you can sometimes do so by multiplying the numerator and denominator by the conjugate of the denominator and applying a Pythagorean identity.

Rewrite to Eliminate Fractions

Rewrite

1

\— as an expression that does not involve a fraction.1 + COS X r

1 — c o s X

1 + c o s X 1 + c o s x 1 — c o s X

_ 1 — COS X

1 — C O S2 X

_ 1 — COS X

COS X

s i n x s i n x

= C S C 2 X — cot X C SC X

M ultip ly num era to r and denom ina to r by the conjugate o f 1 + cos x, w h ich is 1 — cos x.

M ultip ly.


W rite as the d iffe rence o f tw o fractions.

Factor.

R eciprocal and Quotient Identities

p GuidedPracticeRewrite as an expression that does not involve a fraction.

7A. 7B. sec x + tan x

316 | Lesson 5-1 | T rig o n o m e tric Id en tities

Exercises = Step-by-Step Solutions begin on page R29.

Find the value of each expression using the given information. (Example 1)

1. If cot 9 = y, find tan 9.

22. If cos x = —, find sec x.

3. If tan a = find cot a.5

4. If sin (3 = —1> find csc (3.

5. If cos x = \ and sin x = —r —, find cot x.o 6

6 . If sec ip = 2 and tan ip = y/3, find sin (p.

7. If csc a = | and cot a = , find sec a .

8 . If sec 9 = 8 and tan 9 = 3 \[7, find csc 9.


9. sec 9 and cos 0; tan 0 == —5, cos 9 > 0

10. cot 9 and sec 9; sin 9 == j , tan 9 < 0

11. tan 9 and sin 9; sec 9 =- 4, sin 9 > 0

12. sin 9 and cot 9; cos 9 == ■§■, sin 9 <D 0

13. cos 9 and tan 9; csc 0 == |, tan 9 >• 0

14. sin 9 and cos 9; cot 9 == 8 , csc 9 < 0

15. cot 9 and sin 9; sec 9 =; —j , sm 9 > 0

16. tan 9 and CSC 0; cos 9 = 1 • a = —7 , sm 9 4 < 0


17. If csc 9 = —1.24, find sec - fi18. If cos x = 0.61, find sin |x —ll-19. If tan 9 = —1.52, find cot 9-f)-20. If sin 9 = 0.18, find cos ^0 —fl-21. If cot x —1.35, find tan |x —?)-Simplify each expression.

2 2 . csc x sec x — tan x

24. sec x cot x — sin x

2 g 1 — sin2 x csc2 x — 1

2 g sec x csc x — tan x sec x csc x

30. cot x — csc2 x cot x

(Examples 4 and 5)

23. CSC x — cos X cot X2 g tan x + sin x sec x

csc x tan x2 7 csc x cos x + cot x

sec x cot x

2 0 sec2 x cot2 X + 1

31. cot x — COS3 X CSC X

Simplify each expression. (Example 6)

c o s X . COS X32.

33.

34.

35.

36.

sec x + 1

1 — cos x tan x

1

sec x — 1

sin x 1 + cos x

1sec x + 1 sec x — 1

+ -COS X cot X

sec x + tan x sec x •sin x . sin x

tan x

csc x + 1 csc x — 1

37) SUNGLASSES Many sunglasses are made with polarized lenses, which reduce the intensity of light. The intensity of light emerging from a system of two polarizing lenses

I can be calculated by I = 70 ------^—■, where I0 is thecsc2 9

intensity of light entering the system of lenses and 9 is the angle of the axis of the second lens in relation to that of the first lens. (Example 6}

Axis 1

Axis 2

%>Unpolarized

light Lensl

a. Simplify the formula for the intensity of light emerging from a system of two polarized lenses.

b. If a pair of sunglasses contains a system of two polarizing lenses with axes at 30° to one another, what proportion of the intensity of light entering the sunglasses emerges?

Rewrite as an expression that does not involve a fraction.(Example 7)

38.

40.

42.

44.

46.

sin xcsc x — cot x

cot x sec x — tanx

3 tan x 1 — cos x

sin x 1 — sec x

5sec x + 1

39.

41.

43.

45.

47.

CSC X

1 — sin xcot x

1 + sin x2 sin x

cot x + csc xcot2 x cos x

csc x — 1

sin x tan x cos x + 1

Determine whether each parent trigonometric function shown is odd or even. Explain your reasoning.

49.

J ^l i V

n I n[—2 tv, 2 t t I scl:-^- by [—4 ,4 ] scl: 1

^||p)nnecO 3 1 7

50. SOCCER When a soccer ball is kicked from the ground, its height y and horizontal displacement x are related by

V =Z 2 £ _

2 u 0 c o sx sm ^ where i?0 is the initial velocity of cos 6

the ball, 9 is the angle at which it was kicked, and g is the acceleration due to gravity. Rewrite this equation so that tan 9 is the only trigonometric function that appears in the equation.

Write each expression in terms of a single trigonometric function.

51. tan x — csc x sec x

52. cos x + tan x sin x

53. csc x tan2 x — sec2 x csc x

54. sec x csc x — cos x csc x

55. # MULTIPLE REPRESENTATIONS In this problem, you will investigate the verification of trigonometric identities. Consider the functions shown.i. t/j = tan x + 1

y2 = sec x cos x — sin x sec xii. y3 = tan x sec x — sin x

y4 = sin x tan2 x

a. TABULAR Copy and complete the table below, without graphing the functions.

— 27V — 7T 0 7T 27V

b. GRAPHICAL Graph each function on a graphing calculator.

C. VERBAL Make a conjecture about the relationship between y x and y2. Repeat for y3 and y4.

d. ANALYTICAL Are the conjectures that you made in part c valid for the entire domain of each function? Explain your reasoning.

Rewrite each expression as a single logarithm and simplifythe answer.

56. In |sin x| — In |cos x|

57. In | sec x| — In |cos x|

58. In (cot2 x + 1) + In |sec x|

59. In (sec2 x — tan2 x) — In (1 — cos2 x)

60. ELECTRICITY A current in a wire in a magnetic field causes a force to act on the wire. The strength of the magnetic

F csc tufield can be determined using the formula B ■

where F is the force on the wire, I is the current in the wire, £ is the length of the wire, and 9 is the angle the wire makes with the magnetic field. Some physics books give the formula as F = IiBsin 9. Show that the two formulas are equivalent.

61. LIGHTWAVES When light shines through two narrow slits, a series of light and dark fringes appear. The angle 9, in radians, locating the with fringe can be calculated by

sin 9 = -A , where d is the distance between the two slits,a

and A is the wavelength of light.

a. Rewrite the formula in terms of csc 9.b. Determine the angle locating the 100th fringe when

light having a wavelength of 550 nanometers is shined through double slits spaced 0.5 millimeters apart.

H.O.T. Problem s Use Higher-Order Thinking Skills

62. PROOF Prove that the area of the triangle is

A = \Js(s — a)(s — b)(s — c) where s = j ( a + b + c).

(Hint: The area of an oblique triangle is A = be sin A.)

B

63. ERROR ANALYSIS Jenelle and Chloe are simplifying1 sm2 x— ---------— . Jenelle thinks that the expression

sinz x — cos x2

simplifies to — cos x . , and Chloe thinks that it1 - 2 c o s 2 x

simplifies to csc2 x — tan2 x. Is either of them correct? Explain your reasoning.

CHALLENGE Write each of the basic trigonometric functions in terms of the following functions.

64. sin x 65. cos x 66. tan x

REASONING Determine whether each statement is true or fa lse . Explain your reasoning.

67. csc2 x tan x = csc x sec x is true for all real numbers.

6 8 . The odd-even identities can be used to prove that the graphs of y = cos x and y = sec x are symmetric with respect to the y-axis.

PROOF Prove each Pythagorean identity.

@tan2 9 + 1 = sec2 9 70. cot2 9 + 1 = csc2 9

71. PREWRITE Use a chart or a table to help you organize the major trigonometric identities found in Lesson 5-1.

318 | Lesson 5-1 | T rig o n o m etric Id en tities

Spiral Review

Solve each triangle. Round to the nearest tenth, if necessary. Lesson 4-7)

73.

75. 76.

Find the exact value of each expression, if it exists. (Lesson 4-6)

78. cot ^sin^ 1 |-j 79. tan (arctan 3)

81. cos cos 1y - J 82. cos "K'f)84. ANTHROPOLOGY Allometry is the study of the relationship between the size

of an organism and the size of any of its parts. A researcher decided to test for an allometry between the size of the human head compared to the human body as a person ages. The data in the table represent the average American male. (Lesson 3-5)

a. Find a quadratic model relating these data by linearizing the data and finding the linear regression equation.

b . Use the model for the linearized data to find a model for the original data.

C. Use your model to predict the height of an American male whose head circumference is 24 inches.

Let U = {0 ,1 , 2, 3, 4, 5}, A = {6, 9 } , B = {6, 9 , 10}, C = {0 ,1 , 6, 9 , 11},D = {2, 5 ,11}. Determine whether each statement is true or fa lse .Explain your reasoning. (Lesson 0-1)

80. cos

83. sin

arccos |H)(cos“' l )

G row th o f th e A verage A m erican M ale ( 0 - 3 years o f age)

Head C ircum ference (in .) H eight (in.)

14.1 19.518.0 26.4

18.3 29.7

18.7 32.319.1 34.419.4 36.2

19.6 37.7Source: National Center for Health Statistics

85. A C B D C U

Skills R eview fo r S tandard ized Tests

87. SAT/ACT lfx > 0 ,th e nv2 t rv -i- 112X * - 1 + (X + l ) 2 1 + (x + 2 )2 -

x + 3x + 1 x + 2

A (x + 1) 2

B ( x - 1 ) 2 C 3 i - l D 3xE 3(x - l ) 2

88. REVIEW If sin x = m and 0 < x < 90°, then tan x ■

H

1 - m 2

F \raz

j

fzVl — m2 1 - m 2 m

1 - m 2

89. Which of the following is equivalent to1 - s i n " I 1 — c o s 2 (

A tan 9 B cot 9

■ tan 9?

C sin i D cos i

90. REVIEW Refer to the figure. If cos D = 0.8, what is the length of DF?

F 5 FG 4H 3.2

J t

&connectED.mcgraw-hill.com 319

NewVocabularyverify an identity 1

Verify Trigonometric identities In Lesson 5-1, you used trigonometric identities to rewrite expressions in equivalent and sometimes more useful forms. Once verified, these new

identities can also be used to solve problems or to rewrite other trigonometric expressions.

To verify an identity means to prove that both sides of the equation are equal for all values of the variable for which both sides are defined. This is done by transforming the expression on one side of the identity into the expression on the other side through a sequence of intermediate expressions that are each equivalent to the first. As with other types of proofs, each step is justified by a reason, usually another verified trigonometric identity or an algebraic operation.

You will find that it is often easier to start the verification of a trigonometric identity by beginning on the side with the more complicated expression and working toward the less complicated expression.

Verify a Trigonometric Identity

Verify that csc x—— = cos2 x.CSC X

The left-hand side of this identity is more complicated, so start with that expression first.

csc x — 1 _ cot x

= cot x sin x

C0S *siftin'

= cos2 X ✓


Reciprocal Identity

Quotient Identity


Notice that the verification ends with the expression on the other side of the identity.

^ GuidedPracticeVerify each identity.

1 A. sec 9 cot 8 — 1 = cot 1 B. tan' a = sec a csc a tan a — 1

There is usually more than one way to verify an identity. For example, the identity in Example 1 can also be verified as follows.

2 _ 1 2 -1^sc * --------= csc„ x ---------- — Write as the difference of two fractions.

= 1 — sin x Simplify and apply a Reciprocal identity.


ly ll 320 | Lesson 5-2

Stev

e A

llen/

Get

ty

Imag

es

When there are multiple fractions with different denominators in an expression, you can find a common denominator to reduce the expression to one fraction.

feiMijit- i Verify a Trigonometric Identity by Combining Fractions

Verify that 2 csc x =CSC X + cot X CSC x — cot X

The right-hand side of this identity is more complicated, so start there, rewriting each fraction using the common denominator (csc x + cot x ) ( c s c x — cot x ) .

1 ■ + ■ 1CSC X + c o t X CSC x — c o t x

CSC x — c o t X ■ + ■ CSC X + c o t X

( c s c X + c o t x ) ( c s c x - c o t x ) ( c s c X + c o t x ) ( c s c x — c o t x )

_______________ 2 c s c x( c s c z + c o t x ) ( c s c x — c o t x )

2 c s c x

CSC2 x — c o t 2 x

= 2 csc x

y GuidedPractice

2. Verify that T cosa - C0SQ+ s i n a 1 — s i n a = 2 sec a.

Start with the right-hand side of the identity.

Common denominator

Add.

Multiply,


sin a1 + cos a 1 - cos a

sin a 1 - cos a sin a y

1 - cos a

To eliminate a fraction in which the denominator is of the form 1 ± u or u ± 1, remember to try multiplying the numerator and denominator by the conjugate of the denominator. Then you can potentially apply a Pythagorean Identity.

StudyTipAlternate Method You do not always have to start with the more complicated side of the equation. If you start with the right-hand side in Example 3, you can still prove the identity.

csc a + cot a _ _ 1 cos a

sin a sin a1 + cos a

1 0 9 1 ' Verify a Trigonometric Identity by Multiplying

> ■ = csc a + cot a .Verify that ----J 1 — <

Because the left-hand side of this identity involves a fraction, it is slightly more complicated than the right side. So, start with the left side.

s i n a 1 + c o s a1 — c o s a 1 — c o s a 1 + c o s a

_ s i n a (1 + c o s a)1 — c o s 2 a

_ s i n a (1 + c o s a) s i n 2 a

1 + c o s a s i n a

1 , c o s CL

sm a sm a

= csc a + cot a ✓

Multiply numerator and denominator by the conjugate of 1 — cos a , which is 1 + cos a.

Multiply.


Divide out the common factor of sin a .

Write as the sum of two fractions.

Reciprocal and Quotient Identities

y GuidedPractice

3. Verify that t a n x s e c x + 1

= CSC x — cot X.

Until an identity has been verified, you cannot assume that both sides of the equation are equal. Therefore, you cannot use the properties of equality to perform algebraic operations on each side of an identity, such as adding the same quantity to each side of the equation.

Bc o n n ec t^m cg ra w -h iil.c o m frj 321

When the more complicated expression in an identity involves powers, try factoring.

StudyTipAdditional Steps When verifying an identity, the number of steps that are needed to justify the verification may be obvious. However, if it is unclear, it is usually safer to include too many steps, rather than too few.

Verify a Trigonometric identity by Factoring

Verify that cot 9 sec 0 csc2 9 — cot3 0 sec 9 = csc 9.

cot 6 sec 9 csc2 9 — cot3 9 sec 9 Start with the left-hand side of the identity.

= cot 9 sec 9 (csc2 9 — cot2 9) Factor.

= cot 9 sec 9 Pythagorean Identity_ cos 8 1

sin a cos (

1

Reciprocal and Quotient Identities

Multiply, sm tt

= csc 9 ✓ Reciprocal Identity

GuidedPractice4. Verify that sin2 x tan2 x csc2 x + cos2 x tan2 x csc2 x = sec2 x.

It is sometimes helpful to work each side of an identity separately to obtain a common intermediate expression.

Verify an Identity by Working Each Side Separately

> Verify that v =1 + sec x cos x

Both sides look complicated, but the left-hand side is slightly more complicated since its denominator involves two terms. So, start with the expression on the left.

tan2* _ sec2 x — 1 Pythagorean Identity1 + sec x 1 + sec x

(sec x — l)(sec x + 1)Factor.

1 + sec x= sec x — 1 Divide out common factor of sec x + 1.

From here, it is unclear how to transform sec x — 1 into 1 COg° * / so start with the right-hand side and work to transform it into the intermediate form sec x — 1.

-•-" d o s f * = cikx ~ £ t f Write as the difference of two fractions.

= sec x — 1 Use the Quotient Identity and simplify.

To complete the proof, work backward to connect the two parts of the proof.

tan2 x _ sec2 x — 1 1 + sec x 1 + sec x

_ (sec x — l)(sec x + 1 )


. , Factor.1 + sec x

: sec x — 1 Divide out common factor of sec x + 1.

: cckrx ~ - i f f Use the Quotient ldentitvand write 1 as U p

1 — cos X ,cos X Combine fractions.

p GuidedPractice5. Verify that sec4 x — sec2 x = tan4 x + tan2 x.

322 | Lesson 5-2 | V e r ify in g T rig o n o m etric Id en tities

ConceptSummary Strategies for Verifying Trigonometric IdentitiesStart with the more complicated side of the identity and work to transform it into the simpler side, keeping the other side of the identity in mind as your goal.

Use reciprocal, quotient, Pythagorean, and other basic trigonometric identities.

Use algebraic operations such as combining fractions, rewriting fractions as sums or differences, multiplying expressions, or factoring expressions.

Convert a denominator of the form 1 ± u or u ± 1 to a single term using its conjugate and a Pythagorean Identity.

Work each side separately to reach a common intermediate expression.

If no other strategy presents itself, try converting the entire expression to one involving only sines and cosines.

2 Identifying Identities and Nonidentities You can use a graphing calculator toinvestigate whether an equation may be an identity by graphing the functions related to each

side of the equation.

WatchOut!Using a Graph You can use a graphing calculator to help confirm a nonidentity, but you cannot use a graphing calculator to p /w e that an equation is an identity. You must provide algebraic verification of an identity.

>^ J 2 S ® 0 l Determirie Whether an Equation is an Identity

Use a graphing calculator to test whether each equation is an identity. If it appears to be an identity, verify it. If not, find a value for which both sides are defined but not equal.

cos 0 + 1 _ cos 0 tan2 0 sec /3 + 1

The graphs of the related functions do not coincide for all values of x for which the both functions are defined. When x = ^ , Y i ~ 1.7 but Y2 ~ 0.3. The equation is not an identity.

12=(«S(X )A (i

i n i n I

K=.7B539Hifi Y=.29£B9322

1—2tt, 2 id scl: i r by [—1,3] scl: 1 [—2tt, 2 id scl: 7r by 1—1,3l scl: 1

jj cos 0 + 1 _ cos 0 tan2 0 sec 0 — 1

The equation appears to be an identity because the graphs of the related functions coincide. Verify this algebraically.

c o s 0 _ c o s 0 s e c 0 + 1s e c 0 — 1 s e c 0 — 1 s e c 0 + 1

_ c o s 0 s e c 0 + c o s 0 s e c 2 0 — 1

c o s 0 ( — j ) + c o s 0 \ c o s p J

s e c 2 0 — 1

_ 1 + c o s 0s e c 2 0 — 1

c o s 0 + 1------ o *

t a n - 0

Multiply numerator and denominator by the conjugate of sec 0 — 1.

Multiply.

Reciprocal Identity

Simplify.

Commutative Property and Pythagorean Identity

[—2it, 2 id scl: tv by [—1,3l scl: 1

I1- GuidedPractice6A. c s c g = c o t f l t a n 2 9 + c o t e 6B. c o s x + 1

&connectED.m cgraw-hill.com 323


1. (sec2 8 —1) cos2 9 = sin2 9

2. sec2 6(1 — cos2 9) = tan2 9

3. sin 9 — sin 6 cos2 9 = sin3 9

4. csc 8 — cos 6 cot 9 = sin 9

5. cot2 9 csc2 9 — cot2 9 = cot4 9

6 . tan 9 csc2 9 — tan 9 = cot 9

sec 9 sin

V e r ify each id en tity . Examples 1-3)

7.

8 .

9.

1 0 .

1 1 .

1 2 .

1 — cos (

cos 9 1 + sin 6

sin 6 1 — cot 8

11 — tan2 6

1

= cot 9

+ 1 V V 1 3 O :— -— = 2 CSC I

sin 6 cos 9

sin 9 , 1 — cossin 9

+ tan 9 = sec

cos (

+

1 — tan

1

+

1 - cot"

1

- = sin 9 + cos i

-= 1

= 2 sec 9 sin Icsc 9 + 1 csc 8 —1

13. (csc 9 — cot 0)(csc 9 + cot 9) = 1

14. cos4 9 — sin4 9 = cos2 9 — sin2 9

1 , 1 _ . _ _ „ 215.

16.

1 — sin

cos 81 + sin - +

1 + s i n I

cos 9 1 — sin i

: 2 sec

- = 2 sec 9

17. csc4 cot4 9 = 2 cot2 9 + 1

18. csc 8 + 7. csc 8 — 3 csc 9 + 3csc 8 —1 csc 8+ 1

19) FIREWORKS If a rocket is launched from groundlevel, the maximum height that it reaches is given by

h = v- 8, where 9 is the angle between the ground

and the initial path of the rocket, v is the rocket's initial speed, and g is the acceleration due to gravity,9.8 meters per second squared. (Example 3)

t

b. Suppose a second rocket is fired at an angle of 80° from the ground with an initial speed of 1 1 0 meters per second. Find the maximum height of the rocket.

Verify each identity. (Examples 4 and 5)

20. (csc 9 + cot 9)( 1 — cos 9) = sin 9

21. sin2 9 tan2 9 = tan2 9 — sin2 9

1 — tan2 8 _ cos2 9 —12 2 .

23.

1 — cot

1 + csc ( sec 8

cos 8

= cos 9 + cot I

24. (csc 9 - cot 9)2 = } cos 1 v ' 1 + cos (

25. 1 + tan2

1 — tan2

1

2 cos

26. tan2 9 cos2 9 = 1 — cos2 9

27. sec 6 — cos 9 = tan 9 sin 9

28. 1 — tan4 9 = 2 sec2 9 — sec4

29. (csc 9 - cot 9)2 = 1 ~ cos e

1 + tan 9

1 + cos I

30.

31.

sin 8 + cos

2 + csc 9 sec 9 csc 8 sec 8

= sec i

= (sin 9 + cos 9)2

32. OPTICS If two prisms of the same power are placed next to each other, their total power can be determined using z = 2p cos 9, where z is the combined power of the prisms, p is the power of the individual prisms, and 9 is the angle between the two prisms. Verify that 2p cos 9 = 2p(l — sin2 9) sec 9. (Example 4)

33. PHOTOGRAPHY The amount of light passing through apolarization filter can be modeled using I = Im cos2 9, where I is the amount of light passing through the filter, Im is the amount of light shined on the filter, and 9 is the angle of rotation between the light source and the

filter. Verify that Im cos 9 = Im —cot2 9+ 1

■. (Example 4)

GRAPHING CALCULATOR Test whether each equation is an identity by graphing. If it appears to be an identity, verify it. If not, find a value for which both sides are defined but not equal. (Example 6)

34. tan x + 1 _ 1 + cot x tan x — 1 1 — cot x

35. sec x + tan x = 1sec x — tan x

36. sec2 x — 2 sec x tan x + tan2 x = —

37.

38.

c2 '

cot2 x — 1 _ j 1 + cot2 x

tan x — sec x

■ cos X1 + cos X

■ 2 sin2 x

tan x — 1tan x + sec x

39. cos2 x — sin2 x ■ cot x — tan x tan x + cot x

324 | Lesson 5-2 | V e rify in g T rig o n o m e tric Id en tities

«■

41 . sec x —T _ I secx — 1 IV sec x + 1 I tan x |

42. In |csc x + cot x\ + In |csc x — cot x| = 0

43. In |cot x| + In |tan x cos x| = In |cos x\

Verify each identity.

44. sec2 8 + tan2 8 = sec4 8 — tan4 6

45. —2 cos2 8 = sin4 8 — cos4 8 —1

46. sec2 8 sin2 8 = sec4 8 — (tan4 8 + sec2 8)

47. 3 sec2 8 tan2 9 + 1 = sec6 8 — tan6 8

48. sec4 x = 1 + 2 tan2 x + tan4 x

49. sec2 x csc2 x = sec2 x + csc2 x

50. ENVIRONMENT A biologist studying pollution situates a net across a river and positions instruments at two different stations on the river bank to collect samples. In the diagram shown, d is the distance between the stations and w is width of the river.


a. Determine an equation in terms of tangent a that can be used to find the distance between the stations.T . , , w c o s ( 9 0 ° — a)

b. Verify that d = ------------- .

C. Complete the table shown for d = 40 feet.

w 20 4 0 6 0 8 0 100 120

a

d. If a > 60° or a < 20°, the instruments will not function properly. Use the table from part c to determine whether sites in which the width of the river is 5, 35, or 140 feet could be used for the experiment.

HYPERBOLIC FUNCTIONS The hyperbolic trigonom etricfunctions are defined in the following ways.

sinh x = i(e * — e~x) csch x = ——— ,2 sinh x

cosh x = \\(ex + e~x) sech x = — \—2 cosh x

tanh x = x coth x - — i-—, x=f= 0cosh ,r tanh x

Verify each identity using the functions shown above.

51. cosh2 x — sinh2 x = 1 52. sinh(—x) = —sinh x

53. sech2 x = 1 — tanh2 x 54. cosh(—x) = cosh x

GRAPHING CALCULATOR Graph each side of each equation. Ifthe equation appears to be an identity, verify it algebraically.

c c s e c x _ ta n x s e c x _ , c o s x c s c x ~

56. sec x — cos2 x csc x = tan x sec x

57. (tan x + sec x)(l — sin x) = cos x

g g s e c x c o s x _______________1______________

c o t 2 x t a n 2 x — s in 2 x t a n 2 x

59. d p MULTIPLE REPRESENTATIONS In this problem, you will investigate methods used to solve trigonometric equations. Consider 1 = 2 sin x.

a. NUMERICAL Isolate the trigonometric function in the equation so that sin x is the only expression on one side of the equation.

b. GRAPHICAL Graph the left and right sides of the equation you found in part a on the same graph over [0, 27t). Locate any points of intersection and express the values in terms of radians.

c. GEOMETRIC Use the unit circle to verify the answers you found in part b.

d. GRAPHICAL Graph the left and right sides of the equation you found in part a on the same graph over —2tt < x < 2tt. Locate any points of intersection and express the values in terms of radians.

e. VERBAL Make a conjecture as to the solutions of 1 = 2 sin x. Explain your reasoning.


60. REASONING Can substitution be used to determine whether an equation is an identity? Explain your reasoning.

(6?) CHALLENGE Verify that the area A of a triangle is given by_ a 2 s i n f3 s i n 7

2 s i n (/3 + 7 )

where a, b, and c represent the sides of the triangle and a , (3, and 7 are the respective opposite angles.

62. WRITING IN MATH Use the properties of logarithms to explain why the sum of the natural logarithms of the six basic trigonometric functions for any angle 8 is 0 .

63. OPEN ENDED Create identities for sec x and csc x in terms of two or more of the other basic trigonometric functions.

64. REASONING If two angles a and (3 are complementary, is cos2 a + cos2 (3 = 1? Explain your reasoning. Justify your answers.

65. WRITING IN MATH Explain how you would verify a trigonometric identity in which both sides of the equation are equally complex.

connectED.mcgraw-hill.com | 325

Spiral Review

Simplify each expression. Lesson 5-1)

66. cos 9 csc 9 67. tan 9 cot 9 68. sin 9 cot 0

g g c o s 8 c s c 8 7 q s in 8 c s c 8 1 — c o s 2 0ta n 8 ' c o t 8 ' sjn 2 q

72. BALLOONING As a hot-air balloon crosses over a straight portion of interstate highway, its pilot eyes two consecutive mileposts on the same side of the balloon. When viewing the mileposts, the angles of depression are 64° and 7°. How high is the balloon to the nearest foot? _esson 4-7)

Locate the vertical asymptotes, and sketch the graph of each function. (Lesson 4-5)1 173. y = - tan x 74. y — csc 2x 75. y = - sec 3x

Write each degree measure in radians as a multiple of it and each radian measure indegrees. (Lesson 4-2)

76. 660° 77. 570° 78. 158°

79. ^ 2 1 80. 1Z2L 81. 94 6

Solve each inequality. Lesson 2-6)

82. x2 — 3x — 18 > 0 83. x2 + 3x - 28 < 0 84. x2 - 4x < 5

85. x2 + 2x > 24 86. - x 2 - x + 12 > 0 87. - x 2 - 6 x + 7 < 0

88. FOOD The manager of a bakery is randomly checking slices of cake prepared by employees to ensure that the correct amount of flavor is in each slice. Each 12-ounce slice should contain half chocolate and half vanilla flavored cream. The amount of chocolate by which each slice varies can be represented by g(x) = h x - 12|. Describe the transformations in the function. Then graph the function. (Lesson 1 -5)

Skills Review fo r S tandard ized Tests

89. SAT/ACTa, b, a, b, b, a, b, b, b, a, b, b, b ,b , a , ...

If the sequence continues in this manner, how many bs are there between the 44th and 47th appearances of the letter a?

A 91 C 138 E 230B 135 D 182

90. Which expression can be used to form an identity

with sec e + csc e when tan 9 * - 1 ?1 + tan 8

F sin 9G cos 9H tan 9J csc 9

91. REVIEW Which of the following is not equivalent to cos 9 when 0 < 9 < y ?

A ---- , LOS 6—-— C cot 9 sin 9c o s 8 + s in 8

„ 1 — s in 2 9 r-» i n r .B ---------- — D tan 9 csc 9c o s 8

92. REVIEW Which of the following is equivalent to sin 9 + cot 9 cos 9 ?

F 2 sin 9

G - r 3— s m 8

H cos2 9

I s in 8 + c o s 8

sm ^ 8

326 [ Lesson 5-2 | V erify in g T rig o n o m e tric Id en tities

Tom

Car

ter/

Ph

otoE

dit

Solving Trigonometric Equations

You verifiedtrigonometricidentities.(Lesson 5-2)

t Solve trigonometric equations using algebraic techniques.

Solve trigonometric equations using basic identities.

: Why?

A baseball leaves a bat at a launch angle 6 and returns to its initial batted height after a distance of d meters. To find the velocity v0 of the ball as it leaves the bat, you can solve the

trigonometric equation d -2v02 sin 0cos 9

< 0 '

IUse Algebraic Techniques to Solve In Lesson 5-2, you verified trigonometric equations called identities that are true for all values of the variable for which both sides ar e defined. In this lesson we will consider conditional trigonometric equations, which may be true for certain

values of the variable but false for others.

Consider the graphs of both sides of the conditional trigonometric equation cos x ■ 12 '

The graph shows that cos x = j has two solutions on the interval [0, 2ir), x = y and x = -y-.

Since y = cos x has a period of 2tt, cos x = -i has infinitely many solutions on the interval (—oo, oo) Additional solutions are found by adding integer multiples of the period, so we express all solutions by writing

x = — + 2 mr and x = 4^ + 2 «7V, where n is an integer.

To solve a trigonometric equation that involves only one trigonometric expression, begin by isolating this expression.

So,ve by Isolating Trigonometric Expressions

Solve 2 tan x — V3 = tan x.

2 tan x — \/3 = tan x

tan x — V3 = 0

tan x = V3

Original equation

Subtract tan x from each side to isolate the trigonometric expression.

Add V 3 to each side.

The period of tangent is i t , s o you only need to find solutions on the interval [ 0 , tt ) . The only solution on this interval is x = y . The solutions on the interval (—oo, oo) are then found by adding integer multiples tv. Therefore, the general form of the solutions is

where n is an integer.

^ GuidedPractice1. Solve 4 sin x — 2 sin x + \[2.

StudyTipFind Solutions Using the Unit Circle Since sine corresponds to the y-coordinate on the unit circle, you can find the solutions of sin x= on the interval [0 ,2tt] using the unit circle, as shown.

Any angle coterminal with these angles will also be a solution of the equation.

>So,ve by Taking the Square Root of Each Side

Solve 4 sin2 x + 1 = 4.

4 sin2 x + 1 = 4

4 sin2 x = 3

sin2 x = 4 4

sin x = ± V 3

Original equation

Subtract 1 from each side.

Divide each side by 4.

Take the square root of each side.

On the interval [0, 2tt), sin x = when x = - j and x = -=y- and sin x = ~ w h e n x = ~

and x Since sine has a period of 27V, the solutions on the interval (—oo, oo) have the general

form x = j + 2m r, x = ~ + 2n-n, x ■■ 4*tt 5tc— + 2 n77, and x = — + 2 mr, where n is an integer.

f- GuidedPractice2. Solve 3 cot2 x + 4 = 7.

When trigonometric functions cannot be combined on one side of an equation, try factoring and applying the Zero Product Property. If the equation has quadratic form, factor if possible. If not possible, apply the Quadratic Formula.

^.PfufflH^lSolve by Factoring

Watch Out!D ivid ing by Trigonom etric Factors Do not divide out the cos x in Example 3a. If you were to do this, notice that you might conclude that the equation had no solutions, when in fact, it has two on the interval [0, 2-k).

Find all solutions of each equation on the interval [0, 2-k ).

a. cos x sin x = 3 cos x

cos x sin 9 = 3 cos x Original equation

cos x sin x — 3 cos x = 0 Isolate the trigonometric expression.

cos x(sin x — 3) = 0 Factor,

cos x = 0 or sin x — 3 = 0 Zero Product Property

sin x = 3 Solve for xon [0, 2 - k ] ,x = £ and

The equation sin x = 3 has no solution since the maximum value the sine function can attain is 1. Therefore, on the interval [0, 27t), the solutions of the original equation are j and

b. cos4 x + cos2 x — 2 = 0

cos4 x + cos2 x — 2 = 0

(cos2 x)2 + cos2 x — 2 = 0

(cos2 x + 2 )(cos2 x — 1 ) = 0

cos2 x + 2 = 0 or

cos2 x = — 2

cos x = ± V — 2

cos2 x — 1 = 0

cos2 x = 1

cos x = ± V T or +1

Original equation

Write in quadratic form.

Factor.

Zero Product Property

Solve for cos2 x.Take the square root of each side.

The equation cos x = ± V —2 has no real solutions. On the interval [0, 27t), the equation cos x = ± 1 has solutions 0 and tt.

^ GuidedPractice3A. 2 sin x cos x = \[2. cos x 3B. 4 cos2 x + 2 cos j — 2\[2 cos x = \[2

328 | Lesson 5-3 I Solving T rig o n o m e tric Equations

StudyTipExact Versus Approximate Solutions When solving trigonometric equations that are not in a real-world context, write your answers using exact values rather than decimal approximations. For example, the general solutions of the equation tan x = 2 should be expressed as A-=tan_1 2 + m ror x = arctan 2 + mr.

v

StudyTipOptimal Angle Ignoring wind resistance and other factors, a baseball will travel the greatest distance when it is hit at a 45° angle. This is because sin 2(45) = 1, which maximizes the distance formula in the example.

Some trigonometric equations involve functions of multiple angles, such as cos 2x = To solve these equations, first solve for the multiple angle.

> i f f lS V ff lf i j j f Q . ; Trigonometric Functions of Multiple Angles

BASEBALL A baseball leaves a bat with an initial speed of 30 meters per second and clears a fence 90.5 meters away. The height of the fence is the same height as the initial height of the

v g 2 s i n 2 8batted ball. If the distance the ball traveled is given by a = ---- —---- , where 9.8 is in meters9.oper second squared, find the interval of possible launch angles of the ball.

---\V

v0 s i n 28 9 8

3 0 2 s i n 289 . 8

9 0 0 s i n 289 . 8

886.9 = 900 sin 29

8 8 6 . 9

d =

90.5 =

90.5 =

9 0 0sin 26

sm - l 8 8 6 . 9 _ r,„

Original formula

d= 90.5

Simplify.

Multiply each side by 9.8.


Definition of inverse sine

Recall from Lesson 4-6 that the range of the inverse sine function is restricted to acute angles of < in the interval [—90°, 90°]. Since we are finding the inverse sine of 29 instead of 9, we need to consider angles in the interval [-2(90°), 2(90°)] or [-180°, 180°]. Use your calculator to find the acute angle and the reference angle relationship sin (180° — 9) = sin 9 to find the obtuse angle.

• - 1 8 8 6 . 9 _

9 0 0

sin-1 « 80.2° and sin (180° - 80.2°) = sin 99.8°

Definition of inverse sine

80.2° or 99.8° = 29

40.1° or 49.9° = 9 Divide by 2.

The interval is [40.1°, 49.9°]. The ball will clear the fence if the angle is between 40.1° and 49.9°.

CHECK Substitute the angle measures into the original equation to confirm the solution.

v02 s i n 28v02 s i n 28 d = -----

90.5 =

9 . 8

3 0 2 s i n ( 2 • 4 0 . 1 ° )

>►

9 . 8

90.5 ~ 90.497 ✓

Original formula

0 = 40.1° or 0 = 49.9°

Use a calculator.

d = -9 . 8

90.5 =3 0 2 s i n ( 2 • 4 9 . 9 ° )

90.5 ~ 90.497 ✓

GuidedPractice

4. BASEBALL Find the interval of possible launch angles required to clear the fence if:

A. the initial speed was increased to 35 meters per second.

B. the initial speed remained the same, but the distance to the fence was 80 meters.

329

2 Use Trigonometric Identities to Solve You can use trigonometric identities along withalgebraic methods to solve trigonometric equations.

StudyTipAlternate Method An alternate way to check Example 5 is to graph y = 2 cos2 x - sin x — 1. It has zeros and ^ onb o 2

[0 , 2 it] scl: y by [ —2,4] scl: 1

Solve by Rewriting Using a Single Trigonometric Function

>

Find all solutions of 2 cos2 x — sin x — 1 = 0 on the interval [0, 2n).

2 cos2 x — sin x — 1 = 0

2 ( 1 — sin2 x) — sin x — 1 = 0

2 — 2 sin2 x — sin x — 1 = 0

— 2 sin2 x — sin x + 1 = 0

—1 ( 2 sin x — l)(sin x + 1 ) = 0

2 sin x — 1 = 0 or sin x + 1 = 0

sin x = \ sin x = — 1

6 63tt2

Original equation


Multiply.

Simplify.

Factor.


Solve for sin x.

Solve for x on [0 ,2 it].

CHECK The graphs of Yi = 2 cos2 x — sin x and Y2 = 1 intersect

at y , and ~ on the interval [0 , 2 ir] as shown ✓

TInteraction K=.S2559H7B V=1

10,2 tt] scl: y by [ —2,4] scl: 1p GuidedPracticeFind all solutions of each equation on the interval [0, 2tz).

5A. 1 — cos x = 2 sin2 x 5B. cot2 x csc2 x + 2 csc2 x — cot2 x = 2

Sometimes you can obtain an equation in one trigonometric function by squaring each side, but this technique may produce extraneous solutions.

H 2 2 3 1 jl l ! ! 3 Solve by Squaring

Find all solutions of csc x — cot x = 1 on the interval [0, 2rt].

csc x — cot X = 1

CSC X = 1 + cot X

(csc x) 2 = (1 + cot x) 2

CSC2 X = 1 + 2 cot X + cot2 X

1 + cot2 x = 1 + 2 cot x + cot2 x

0 = 2 cot x

0 = cot x

CHECK

x - f o r f

CSC x — cot X = 1_ 7V cot ■£ = 1

2

1 - 0 = ! ✓

Original equation

Add cot x to each side.

Square each side.

Multiply,


Subtract 1 + cot2 xfrom each side.


Solve for x on [0 ,2-ir],

Original equation

Substitute.

Simplify.

csc x — cot x = 1

3*77 , 37T ? ^CSC —— cot —- = 1 2 2

- 1 - 0 / I X

Therefore, the only valid solution is y on the interval [0, 2tt].

p GuidedPractice6A. sec x + 1 = tan x 6B. cos x = sin x — 1

330 | Lesson 5-3 1 Solving T rig o n o m etric Equations


Solve each equation for all values of x. (Examples 1 and 2)

/T7)5 sin x + 2 = sin x 2. 5 = sec2 x + 3

(3. 2 = 4 cos2 x + 1 4. 4 tan x — 7 = 3 tan x — 6

5. 9 + cot2 x = 12 6. 2 — 10 sec x = 4 — 9 sec x

7. 3 csc x = 2 csc x + \[2 8. 11 = 3 csc2 x + 7

,9. 6 tan2 x — 2 = 4 10. 9 + sin2 x = 10

11. 7 cot x — \/3 = 4 cot x 12. 7 cos x = 5 cos x + \[3

Find all solutions of each equation on [0, 2tt]. (Example 3)

sin4 x + 2 sin2 x — 3 = 0

14. — 2 sin x = — sin x cos x

( K ) 4 cot x = cot x sin2 x

16. csc2 x — csc x + 9 = 11

17. cos3 x + cos2 x — cos x = 1

18. 2 sin2 x = sin x + 1

19) TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. (Example 4)

a. If the ball is hit at 50 feet per second, neglecting air resistance, use d = v02 sin 28 to find the interval of

possible angles of the ball needed to clear the net.

b . Find 6 if the initial velocity remained the same but the distance to the net was 50 feet.

20. SKIING In the Olympic aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can reach is given by

vn2 sin2 8p e a k = 2g ' w ere $ ls acceleration due to

gravity or 9.8 meters per second squared. (Example 4)

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier's initial speed?

b . Use your answer from part a to determine how long it took the skier to reach the maximum height

vn sin 6 if f = u 11 p e a k g

Find all solutions of each equation on the interval [0, 2it].(Examples 5 and 6)

21. 1 = cot2 x + csc x

2 2 . sec x = tan x + 1

an2 x = 1 — sec x

24. csc x + cot x = 1

/25^)2 — 2 cos2 x = sin x + 1

26. cos x — 4 = sin x — 4

27. 3 sin x = 3 — 3 cos x

28. cot2 x csc2 x — cot2 x = 9

29. sec2 x — 1 + tan x — V3 tan x = \p3

30. sec2 x tan2 x + 3 sec2 x — 2 tan2 x = 3

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power P R of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and Pv.

O p h th a lm ic P rism P o s itio n o f Base

Base

R eso lv ing Prism P o w er

Pv =PRs\n9

Ph = Pr c° s 8

Using the equations above, determine the values of 9 for which P v and PH are equivalent.

Find all solutions of each equation on the interval [0, 2tc].

32. tan x sec x

cos x

+ cos x = 2

1 + sin x1 + sin x + ■ cos X

= - 4

34. sin x + cos x 1 — sin x _ CQg xtan x sin x

35. cot x cos x + 1 = - | O il L A.

sec x — 1 tan2 x

GRAPHING CALCULATOR Solve each equation on the interval [0, 2-rc] by graphing. Round to the nearest hundredth.

37. sin *kx + cos to = 3x

39. x log x + 5x cos x = — 2

36. 3 cos 2x = ex + 1

38. x2 = 2 cos x + x

jflFcconnectED.mcgraw-hill.com ] 3 3 1 Q j

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t = 8.05 cos (^x — tt j + 66.95, where x is a function oftime, x = 1 represents January 15, x = 2 represents February 15, and so on.a. Use a graphing calculator to estimate the temperature

on January 31.b. Approximate the number of months that the average

daily temperature is greater than 70° throughout the entire month.

c. Estimate the highest temperature of the year and the month in which it occurs.

Find the x-intercepts of each graph on the interval [0, 2ttI.

41.y = s in 2 x cos x — cos x

7 K .1 7 * X

Find all solutions of each equation on the interval [0, 4-ir].

45. 4 tan x = 2 sec2 x 46. 2 sin2 x + 1 = — 3 sin x47. csc x cot2 x = csc x 48. sec x + 5 = 2 sec x + 3

49. GEOMETRY Consider the circle below.

a. The length s of arc AB is given by s = r(20) where 0 < 8 < tv. When s = 18 and AB = 14, the radius isr = • -. Use a graphing calculator to find the

measure of 29 in radians.The area of the shaded region is given by

2^ smA = . Use a graphing calculator to find theradian measure of 0 if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

Solve each inequality on the interval [0, 2ir).

50. 1 > 2 sin x 51. 0 < 2 cos x — \[2

52. Co s ( f - x ) > ^

V354. cos x < — r -

53. sin {x — y j < tan x cot x

55. \/2 sin x — 1 < 0

332 | Lesson 5-3 | Solving T rig o n o m etric Equations

56. REFRACTION When light travels from one transparent medium to another it bends or refracts, as shown.

Refraction is described by n2 sin 01 = n x sin 92, where n x is the index of refraction of the medium the light is entering, n2 is the index of refraction of the medium the light is exiting, 9l is the angle of incidence, and 92 is the angle of refraction.a. Find 92 for each material

shown if the angle of incidence is 40° and the index of refraction for air is 1 .0 0 .

b. If the angle of incidence is doubled to 80°, will the resulting angles of refraction be twice as large as those found in part a?


57. ERROR ANALYSIS Vijay and Alicia are solvingtan2 x — tan x -I- V3 = V3 tan x. Vijay thinks that

the solutions are x = ^ + nix, x = ^ - + n-K. x = ^ + «tt, 4 4 3and x = + )!7T. Alicia thinks that the solutions arex = ^ + wn and x = - j + nix. Is either of them correct?

Explain your reasoning.

CHALLENGE Solve each equation on the interval [0, 2tv].

58. 16 sin5 x + 2 sin x = 12 sin3 x

60. REASONING Are the solutions of csc x = \[2 and cot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

63. WRITING IN MATH Explain the difference in the techniques that are used when solving equations and verifying identities.

M ateria lIndex of

R efraction

glass 1.52

ice 1.31

plastic 1.50

water 1.33

Spiral Review

Verify each identity. (Lesson 5-2)1 + sinfl _ cot2 8 gg 1 + tan 8 _ sin 9

sin 8 csc 6 —1 1 + cot 8 cos 9

Find the value of each expression using the given information. (Lesson 5-1)

67. tan 8; sin 8 = tan 8 > 0 68. csc 9, cos 8 = —f-, csc 8 < 0 69. sec 8; tan 8 = — 1, sin 8 < 02 3

6 6 . ■ = i

70. POPULATION The population of a certain species of deer can be modeled by p = 30,000 + 20,000 cos (yjyf)/ where p is the population and t is the time in years. (Lesson 4-4)

a. What is the amplitude of the function? What does it represent?b. What is the period of the function? What does it represent?c. Graph the function.

Given f(x ) = 2x2 — 5x + 3 and g(x) = 6x + 4, find each of the following. (Lesson 1 -6)

71- ( f ~ g ) ( x ) 72. ( f . g ) ( x ) 73. ( | W )

74. BUSINESS A small business owner must hire seasonal workers as the need arises. The following list shows the number of employees hired monthly for a 5-month period.

If the mean of these data is 9, what is the population standard deviation for these data? Round to the nearest tenth. (Lesson 0-8)

Montfi Number of Additional Employees Needed

August 5! September 14

October 6NovemberDecember

8

I


75. SAT/ACT For all positive values of m and n, if 3x - 2 , then x =m — nx

\ 2m — 2n

B 3 + 2 n2m

c 2m - 32 n

D 2m3 + 2 n

E 32 m - 2 n

76. If cos x = —0.45, what is sinF -0 .55

G -0 .45H 0.45

J 0.55

77. Which of the following is not a solution of 0 = sin 8 + cos 8 tan2 81

3iT 4

7tt4

A

B

C 2tt

D 5tt2

78. REVIEW The graph of y = 2 cos 8 is shown. Which is a solution for 2 cos 8 = 1?

F

G

8 tt3

IOtt3

H 13tv

T 1521 J 3

333

Graphing Technology Lab I I I. k Solving Trigonometric OOOO

o o o oo o o o

Inequalities C D O OV J

Objective

Use a graphing calculator to solve trigonometric inequalities.

334 I Lesson 5-3

You can use a graphing calculator to solve trigonometric inequalities. Graph each inequality. Then locate the end points of each intersection in the graph to find the intervals within which the inequality is true.

Activity 1 Graph a Trigonometric Inequality

Graph and solve sin 2x > cos x.

Replace each side of this inequality with y to form the new inequalities.

sin 2x > Yi and Y2 > cos x

Graph each inequality. Make each inequality symbol by scrolling to the left of the equal sign and selecting enter [ until the shaded triangles are flashing. The triangle above represents greater than, and the triangle below represents less than (Figure 5.3.1). In the MODE I menu, select RADIAN.

RTSffB Graph the equations in the appropriate window. Use the domain and range of each trigonometric function as a guide (Figure 5.3.2).

P lo t i PlotZ P lo tsk V iB s in C 2 H> ’'V 2 B c o s < > 0 W j =W h= sV e = \V s =

- 2 tt , 2 tt ] sd: - j - by [— 1, 1] sd: 0.1

Figure 5.3.1 Figure 5.3.2

PTTm The darkly shaded area indicates the intersection of the graphs and the solution of thesystem of inequalities. Use CALC: intersect to locate these intersections. Move the cursor over the intersection and select enter 13 times.

[—2 t t , 2 -jt] scl: y by [— 1, 1] scl: 0.1

The first intersection is when 1 0.866 or Since cos ~ 2 6

V3> the

intersection is at x = The next intersection is at x = Therefore, one of theo 2

solution intervals is TV 7T . Another interval is 5tt 3tt6 ' 2 . 6 ' 2 . . There are an infinite

number of intervals, so the solutions for all values of x are

and + 2 mtt, 4 r- + 2 m7VD 2

2/ZTT, y + 2MIT

ExercisesGraph and solve each inequality.

1. sin 3x < 2 cos x 2. 3 cos > 0.5 sin 2x

4. csc 2x > sin 8x 5. 2 tan 2x < 3 sin 2x

3. sec x < 2 cos x

6 . tan x > cos x

Mid-Chapter QuizLessons 5-1 through 5-3

Find the value of each expression using the given information.(Lesson 5-1)

1. sin e and cos 0, cot e = 4, cos e > 0

2. sec e and sin 0, tan e = —| , csc 9 > 0

3. tan e and csc 9, cos 9 = j , sin 9 > 0

Simplify each expression. (Lesson 5-1)

sin (—x)

6 .

tan ( - * ) sin (90° - x)

cot2 (90° - x) + 1

5.

7.

cot2 X + 1 sin x

1 + sec x

8. ANGLE OF DEPRESSION From his apartment window, Tim can see the top of the bank building across the street at an angle of elevation of 0, as shown below. (Lesson 5-1)

employee

a. If a bank employee looks down at Tim’s apartment from the top of the bank, what identity could be used to conclude thatsin 9 = cos e’l

b. If Tim looks down at a lower window of the bank with an angle of depression of 35°, how far below his apartment is the bank window?

9. MULTIPLE CHOICE Which of the following is not equal to csc 0?(Lesson 5-1)

A sec (90° - 0)

B V c o t2 0 + 1

C —sin en 1

sin (90° - 6)

Verify each identity. ;Lesson 5-2)

cos e cos e1 0 . = - 2 tan i1 + sin 6» 1 - sin t

11. csc2 9 - sin2 e - cos2 9 - cot2 0 = 0

cos (12. sin 0 + tan e csc 0

13. cos e 1 + sin e ■■ sec 0 - tan i

14. = cot2 0 + csc 0 + 1sin e cos e

15. 1 + sin 6> sin 9 + sin 9

1 - sin icsc e

1 — sin i

Find all solutions of each equation on the interval [0 ,2 ir],(Lesson 5-3)

16. 4 sec 0 + 2 \ /3 = sec 0

17. 2 tan 0 + 4 = tan 0 + 5

18. 4 co s2 0 + 2 = 3

19. cos 0 - 1 = sin 0

20. MULTIPLE CHOICE Which of the following is the solution set for cos 0 tan 0 - sin2 0 = 0 ? i Lesson 5-3)

F y / j , where n is an integer

G y + m r, where n is an integer

H 77 + 2mx, where n is an integer

J n-k , where n is an integer

Solve each equation for all values of 0. Lesson 5-3)

21. 3 sin2 0 + 6 = 2 sin2 0 + 7

22. sin 0 + cos 0 = 0

23. sec 0 + tan 0 = 0

24. 3 - 3 cos2 0 = 1 + sin2 0

25. PROJECTILE MOTION The distance tf that a kickball travels in feet with an acceleration of 32 feet per second squared is given by d

v 2 sin 20- 0 , where v0 is the object’s initial speed and 9 is the angle32 ......at which the object is launched. If the ball is kicked with an initial speed of 82 feet per second, and the ball travels 185 feet, find the possible launch angle(s) of the ball. (Lesson 5-3)

26. FERRIS WHEEL The height h of a rider in feet on a Ferris wheel after f seconds is shown below, le s so n 5-3)

/7(f) = 75 + 70 sin ( § f - y )

a. If the Ferris wheel begins at t = 0, what is the initial height of a rider?

b. When will the rider first reach the maximum height of 145 feet?

335

• You found values of trigonometric functions by using the unit circle.(Lesson 4-3)

• « | Use sum and difference I identities to evaluate

trigonometric functions.

2 Use sum and difference identities to solve trigonometric equations.

• When the picture on a television screen is blurry or a radio station w ill not tune in properly, the problem is too often due to interference. Interference results when waves pass through the same space at the same time. You can use trigonometric identities to determine the type of interference that is taking place.

NewVocabularyreduction identity 1

Evaluate Itigonometric Functions In Lesson 5-1, you used basic identities involving only one variable. In this lesson, we will consider identities involving two variables. One of

these is the cosine o f a difference identity.

Let points A, B, C, and D be located on the unit circle, a and f3 be angles on the interval [0, 2ttJ, and a > (3 as shown in Figure 5.4.1. Because each point is located on the unit circle, x 2 + y 2 = 1 , x 22 + Vi ~ 1' and x32 + ] j 2 = 1- Notice also that the measure of arc CD = a — (a — 13) or /3 and the measure of arc AB = p.

336 I Lesson 5-4

Figure 5.4.1 Figure 5.4.2

Since arcs AB and CD have the same measure, chords AC and BD shown in Figure 5.4.2 are congruent.

AC = BD

\ J(x 2 - l ) 2 + (y 2 - 0 )2 = y J (x 3 - x { ) 2 + ( y 3 - y x)2

(x 2 - l ) 2 + (y2 - 0)2 = (x 3 - x j 2 + ( y 3 - y x) 2

x 22 - l x 2 + l + y 2 = x 32 - 2* 3* ! + x 2 + y 2 - 2 y 3y + y 2

(x 2 + y 22) ~ 2 x 2 + 1 = Otj2 + j/a2) + (x 32 + y 32) - 2 x 3x l — 2 y 3y 1

1 — 2 x 2 + 1 = 1 + 1 — 2 x 3x 1 — 2 y 3y 1

2 — 2x2 = 2 — 2x3x l - 2y3y x

- 2 x 2 = - 2 x 3Xl - 2 y3y

x2 = X3X1 + y3y x

Chords 4Cand BD are congruent.

Distance Formula

Square each side.

Square each binomial.

Group similar squared terms.

Substitution

Add.


Divide each side by —2.

In Figure 5.4.1, notice that by the unit circle definitions for cosine and sine, x l = cos /3,x2 = cos (a — f3), x3 = cos a , y 1 = sin f3, and y3 = sin a . Substituting, x2 = x3x1 + y3y x becomes

cos (a — (3) = cos a cos (3 + sin a sin f3. Cosine Difference Identity

We can now obtain the identity for the cosine of a sum.cos [a + (— f3)] = cos a cos (3 + sin a sin (3 a — /3 = a + (—())

cos [a + (—/3)] = cos a cos (—/3) — sin a sin (—(3) cos(— 0) = cos (/3), sin /3 = —sin (-0)cos [a + 9] = cos a cos 8 — sin a sin 9 Let (—fi) = 0 Ju

pit

erim

ages

/Th

inks

tock

/Ala

my

These two cosine identities can be used to establish each of the other sum and difference identities listed below.

KeyConcept Sum and Difference Identities

Sum Identities D ifference Identities

cos (a + j9) = cos a cos /3 — sin a sin /3 cos (a - jS) = cos a cos /3 + sin a sin

sin (a + /3) = sin a cos /3 + cos a sin /3 sin (a — /3) = sin a cos 13- cos a sin /3. , , tan a + tan /3tan a + /3 = — ------—^1 — tan a tan p , , tan a - tan 0tan (a —)3) = ——-----~ ~1 + tan a tan 0

V ................ JYou will prove the sine and tangent sum and difference identities in Exercises 57-60.

By writing angle measures as the sums or differences of special angle measures, you can use these sum and difference identities to find exact values of trigonometric functions of angles that are less common.

Evaluate a Trigonometric Expression

StudyTipCheck Your Answer You can check your answers by using a graphing calculator. In Example 1a, sin 15° « 0.259 and

« 0.259. Make sure4that the calculator is in the correct mode.

Find the exact value of each trigonometric expression,

a. sin 15°

Write 15° as the sum or difference of angle measures with sines that you know.

sin 15° = sin (45° - 30°)

= sin 45° cos 30° — sin 30° cos 45°

_ \fl V3 1 V22 ' 2 2 * 2

_ V6 V ?4 4

V6 - V2

b. ta n ^ f

ta n i f = ta n ( f + f )

ta n y + ta n ^

V 3 + 1 1 - V3(l)

_ V 3 + 1 1 - V3

_ V3 + 1 1 + V3 1 - V3 1 + V3

_ V 3 + 1 + 3 + V 3 1 - V 3 + V 3 - 3

4 + 2 V 3 - 2

= - 2 - V 3

f GuidedPractice

1A. cos 15°

: cos 45°

4 5 ° -3 0 ° = 15°

Sine Difference Identity

sin 30° = •!•, sin 45°

Multiply.

Combine the fractions.

it . i t _ 7it3 4 12

Tangent Sum Identity

tan y = V 3 and tan - j = 1

Simplify.

Rationalize the denominator.

Multiply.

Simplify.

Simplify.

IB . s i n f

^ , C O ! 3 0 '- ^

337

be found after

^ GuidedPractice

2. ELECTRICITY An alternating current i in amperes in another circuit can be found after t seconds using i = 2 (sin 285)f, where 285 is a degree measure.A. Rewrite the formula in terms of the difference of two angle measures.B. Use a sine difference identity to find the exact current after 1 second.

If a trigonometric expression has the form of a sum or difference identity, you can use the identity to find an exact value or to simplify an expression by rewriting the expression as a function of a single angle.

EE3MI Rewrite as a Single Trigonometric Expression

a. Find the exact value of tan 32° + tan 13° 1 - t a n 32° tan 13°'

t a n 3 2 ° + t a n 1 3 °

1 — t a n 3 2 ° t a n 1 3 °= tan (32° + 13°)

= tan 45° or 1

b. Simplify sin x sin 3x — cos x cos 3x.

sin x sin 3x — cos x cos 3x = —(cos x cos 3x — sin x sin 3x)

- —cos (x + 3x) or —cos 4x

p GuidedPractice

3A. Find the exact value of cos ^ cos + sin ^ sin8 2 4 8 2 4

3B. Simplify J f f l f a - ta n T s v 1 + t a n 6x t a n 7 x

Tangent Sum Identity

Simplify.

Distributive and Commutative PropertiesTangent Sum Identity

Use a Sum or Difference Identity

ELECTRICITY An alternating current i in amperes in a certain circuit can t seconds using i = 3 (sin 165)f, where 165 is a degree measure.

a. Rewrite the formula in terms of the sum of two angle measures.

/ = 3 (sin 165)f Original equation

= 3 [sin (120 + 45)]f 120 + 45 = 165

Sum and difference identities are often used to solve real-world problems.

b. Use a sine sum identity to find the exact current after 1 second.

i = 3 [sin (120 + 45)]f Rewritten equation

= 3 sin (120 + 45) f = 1

= 3[(sin 120)(cos 45) + (cos 120)(sin 45)] Sine Sum Identity

_ 3V6 - 3V2

Substitute.

Multiply.

Simplify.

The exact current after 1 second is amperes.

Real-WorldCareerLineworker Lineworkers are responsible for the building and upkeep of electric power transmission and distribution facilities. The term also applies to tradeworkers who install and maintain telephone, cable TV, and fiber optic lines.

338 | Lesson 5-4 j Sum and D iffe re n c e Id en tities

Sum and difference identities can be used to rewrite trigonometric expressions as algebraic expressions.

ReadingMathCofunction Identities The “co” in cofunction stands for “complement.” Therefore, sine and cosine, tangent and cotangent, and secant and cosecant are all complementary functions or cofunctions.

Write as an Algebraic Expression

Write sin (arctan a/3 + arcsin x) as an algebraic expression of x that does not involve trigonometric functions.

Applying the Sine of a Sum Identity, we find that

sin (arctan \f?> + arcsin x) = sin (arctan V3) cos (arcsin x) + cos (arctan \/3) sin (arcsin x).

If we let a = arctan \[?> and (3 = arcsin x, then tan a = \/3 and sin (3 = x. Sketch one right triangle with an acute angle a and another with an acute angle (3. Label the sides such that tan a = V3 and the sin (3 = x. Then use the Pythagorean Theorem to express the length of each third side.

Using these triangles, we find that sin (arctan V3) = sin a or -y -, cos (arctan V3) =

cos (arcsin x) = cos (3 or V l — x2, and sin (arcsin x) = sin [3 or x.

cos a or —,

Now apply substitution and simplify.

sin (arctan V3 + arcsin x) = sin (arctan \[?>) cos (arcsin x) + cos (arctan V3) sin (arcsin x)

= ^ - { \ l l - x 2) + \ x

V 3 — 3x2 x x + V3 — 3x2— T — + 2 ° r ----------- 2---------

► GuidedPractice

Write each trigonometric expression as an algebraic expression.

4A. cos (arcsin 2x + arccos x) 4B. sin ^arctan x - arccos

Sum and difference identities can be used to verify other identities.

B R fffiffT S iH Verify Cofunction Identities

>Verify sin ( y — = cos x-

■ / I T \ • I t ITsin (y ~ *) = Sln cos x ~ cos sm

= l(cos x) — 0 (sin x)

= cos x S

Sine Difference Identity

sin j - = 1 and cos — =

Multiply.

GuidedPractice

Verify each cofunction identity using a difference identity.

5A. cos - x'j = sin x 5B. csc |y — : sec i

339

Sum and difference identities can be used to rewrite trigonometric expressions in which one of the angles is a multiple of 90° or ^ radians. The resulting identity is called a reduction identity because

it reduces the complexity of the expression.

TechnologyTipViewing Window When checking your answer on a graphing calculator, remember that one period for y = sin x or y = cos x is 2tv and the amplitude is 1. This will help you define the viewing window.

B 2 0 W Verify Reduction Identities

Verify each reduction identity,

a. sin = —cos 0

sin = sin 8 cos + cos 9 sin $2. Sine Sum Formula

= sin 0 (0 ) + cos 9(—1 )

= —cos 9 ✓

■ ■ ■ ■ ■ ■ ■ ■ I

b. tan (x — 180°) = tan x

tan (x - 180)° = yta n x — tan 1 8 0 ° + ta n x tan 1 8 0 °

ta n x — 01 + tan x(0)

= tan x ✓

p GuidedPractice

Verify each cofunction identity.

6A. cos(360° — 9) = cos 9

cos -y - = 0 and sin -y - = - 1

Simplify.

Tangent Sum Formula

tan 180° = 0

Simplify.

6 Bi. sin + xj = cos;

) Solve Trigonometric Equations You can solve trigonometric equations using the sum ■ and difference identities and the same techniques that you used in Lesson 5-3.

B 2 B E E 3 3 B Solve a Trigonometric Equation

Find the solutions of cos |y + x) + cos |y — x) = j on the interval [0, 2iv].

>

cos - j cos x — sin y sin x + cos y cos x + sin y sin x = ~

■kcos x) - ^ -(s in x) + -kcos x) + ~ ( s i n x) •

cos x = 2

CHECK The graph of y = cos |y + xj + cos ^y - xj — j

has zeros at y and y^ on the interval [0 , 2 tt]. ✓

► GuidedPractice

7. Find the solutions of cos (x + tv) — sin (x — 7v) = 0 on the interval [0 , 2 tv],

Original equation

Cosine Sum Identities

Substitute.

Simplify.

On the interval [0, 2tv], cos x = — when x = ^ and x =

340 | Lesson 5 -4 j Sum and D iffe re n c e Iden tities

Exercises = step -by-step Solutions begin on page R29. ®'-aH

Find the exact value of each trigonometric expression.(Example 1)

cos 75°

0 . , n ^

7. VOLTAGE Analysis of the voltage in a hairdryer involves terms of the form sin (nwt — 90°), where n is a positive integer, w is the frequency of the voltage, and t is time. Use an identity to simplify this expression. (Example 2)

8. BROADCASTING When the sum of the amplitudes of two waves is greater than that of the component waves, the result is constructive interference. When the component waves combine to have a smaller amplitude, destructive interference occurs.

Consider two signals modeled by y = 10 sin (2f + 30°) and y = 10 sin (2f + 210°). (Example 2)

a. Find the sum of the two functions.b. What type of interference results when the signals

modeled by the two equations are combined?

9. WEATHER The monthly high temperatures for Minneapolis can be modeled by/(x) = 31.65 sin (^x — 2.09j + 52.35,

where x represents the months in which January = 1, February = 2, and so on. The monthly low temperatures for Minneapolis can be modeled by

g(x) = 31.65 sin ( j x - 2.09) + 32.95. (Example 2)

a. Write a new function h(x) by adding the two functions and dividing the result by 2 .

b. What does the function you wrote in part a represent?

10. TECHNOLOGY A blind mobility aid uses the same idea as a bat's sonar to enable people who are visually impaired to detect objects around them. The sound wave emitted by the device for a certain patient can be modeled by b = 30 (sin 195°)f, where t is time in seconds and b is air pressure in pascals. (Example 2)

a. Rewrite the formula in terms of the difference of two angle measures.

b. What is the pressure after 1 second?

2 . s i n ( - 2 1 0 ° )

1 7 t t4. cos

6 . tanrrr

12

12

Find the exact value of each expression. (Example 3)

£ j \ t a n 4 3 ° — t a n 1 3 °

1 + t a n 4 3 ° t a n 1 3 °

1 2 . cos -|y cos -j- + sin -|y sin y

sin 15° cos 75° + cos 15° sin 75°

14. sin j cos y j - cos y sin

((p . cos 40° cos 20° - sin 40° sin 20°

t a n 4 8 ° + t a n 1 2 °16.1 - t a n 4 8 ° t a n 1 2 °

Simplify each expression. (Example 3)

t a n 26 — t a n 917.1 + t a n 29 t a n (

18. cos y cos x + sin y sin x

19. sin 3y cos y + cos 3y sin y

2 0 . cos 2x sin x — sin 2x cos x

21 . cos x cos 2x + sin x sin 2x

2 2 t a n 59 + t a n 9 t a n 59t a n 9—1

(2 3 ) SCIENCE An electric circuit contains a capacitor, an inductor, and a resistor. The voltage drop across the inductor is given by VL = IwL cos (ivt + y j , where I is

the peak current, w is the frequency, L is the inductance,and t is time. Use the cosine sum identity to express VL as a function of sin wt. (Example 3)

Write each trigonometric expression as an algebraic expression. (Example 4)

24. sin (arcsin x + arccos x)

25. cos (sin 1 x + cos 1 2 x)

26. cos jsin - 1 x — tan - 1 ~ -

sin |sin_ 1 ^y- — tan - 1 x

)&

28. cos (arctan V 3 — arccos x)

29. tan (cos- 1 x + tan - 1 x)

30. tan ^sin- 1 y — cos- 1 xj

31. tan |sin- 1 x + y j

Verify each cofunction identity using one or more difference identities. (Example 5)

32. tan ^y — xj = cot x

33. sec ( ? - * )

34. cot ^y — x) = tan x

341& ■

35. cos (u — 9) = —cos 9

36. cos (2tt + 9) = cos 9

37. sin (7V — 9) = sin 9

38. sin (90° + 9) = cos 9

39. cos (270° — 8) = —sin 9

Find the solution to each expression on the interval [0, 2 tv].

(Example 7)

Verify each reduction identity. (Example 6)

40. COS tf+ ')— sin (f+ , ) = 0

41. cos (7T + X) + cos (IT + x) = 1

42. COS (?+ *)+ sin (f ♦ , ) = 0

g ) sin( ?

+' )

+ sin (I - ■ )_ 1

2

44. sin + xj + sin + xj = — 2

45. tan (tx + x) + tan (7T + x) = 2


46. tan x — tan y =

47. cot a — tan /3 =

s i n ( x - y )

c o s x c o s yc o s ( a + P)

48.

s i n a c o s f3 ( t a n u — t a n v) _ s i n ( u — v)( t a n u + t a n v) s i n (u + v)

49. 2 sin a cos b = sin (a + b) + sin (a — b)

GRAPHING CALCULATOR Graph each function and make a conjecture based on the graph. Verify your conjecture algebraically.

50. y = y[sin (* + 27v) + sin (x — 27v)]

51. y = cos2 |x + y j + cos2 x — y j

PROOF Consider A XYZ. Prove each identity.(Hint: x + y + z = it)

52. cos(x + y) = —cos z

53. sin z = sin x cos y + cos x sin y

54. tan x + tan y + tan 2 = tan x tan y tan z

55. CALCULUS The difference quotient is given by — + ^

a. Let f(x ) = sin x. Write and expand an expression for the difference quotient.

b. Set your answer from part a equal to y. Use a graphing calculator to graph the function for the following values of h: 2 ,1 ,0 .1 , and 0 .0 1 .

C. What function does the graph in part b resemble as h approaches zero?

56. ANGLE OF INCLINATION The angle o f inclination 9 of a line is the angle formed between the positive x-axis and the line, where 0 ° < 9 < 180°.a. Prove that the slope m

of line £ shown at the right is given by m = tan 9.

b. Consider lines £1 and t 2 below with slopes m1 and m2, respectively. Derive a formula for the angle 7 formed by the two lines.

c. Use the formula you found in part b to find the angle

formed by y = -^-x and y = x.


PROOF Verify each identity.

t a n a + t a n /357. tan (a + f3) ■1 — t a n a t a n (3

t a n a - t a n /358. tan ( a - B ) = ---------------- -1 + t a n a t a n /3

59. sin (a + f3) = sin a cos /3 + cos a sin (i

60. sin (a — (3) = sin a cos (3 — cos a sin (3

61. REASONING Use the sum identity for sine to derive an identity for sin (x + y + 2 ) in terms of sines and cosines.

CHALLENGE If sin x = ~ and cos y = j , find each of the

following if x is in Quadrant IV and y is in Quadrant I.

62. cos (x + y) (63) sin (x — y) 64. tan (x + y)

65. REASONING Consider sin 3x cos 2x = cos 3x sin 2x.a. Find the solutions of the equation over [0, 2tt]

algebraically.b. Support your answer graphically.

PROOF Prove each difference quotient identity.

6 6 .

67.

s i n (x + h) — s i n x

c o s (x + h) — c o s x h : cos

I c o s h — 1 \c \ h I

I c o s h — 1 \* V h j

+ COS X

sin x -

s i n hI T

s i n hI T

68. WRITING IN MATH Can a tangent sum or difference identity be used to solve any tangent reduction formula? Explain your reasoning.

342 I Lesson 5 -4 I Sum and D iffe re n c e Id en tities

Spiral Review

69. PHYSICS According to Snell's law, the angle at which light enters water a is related to the angle at which light travels in water (3 by sin a = 1.33 sin /3. At what angle does a beam oflight enter the water if the beam travels at an angle of 23° through the water? (Lesson 5-3)

Find the exact value of each expression, if it exists. (Lesson 4-6)

72. sin - 1 (—1) 73. tan - 1 V3 74. tan |arcsin -|j

75. MONEY Suppose you deposit a principal amount of P dollars in a bank account that pays compound interest. If the annual interest rate is r (expressed as a decimal) and the bank makes interest payments n times every year, the amount of money A you would have after t years is given by A(t) = P (l + "f. (Lesson 3-1)a. If the principal, interest rate, and number of interest payments are known, what type of

function is A{t) = P (l + - -)”f? Explain your reasoning.

b. Write an equation giving the amount of money you would have after t years if you deposit $1000 into an account paying 4% annual interest compounded quarterly (four times per year).

c. Find the account balance after 20 years.

List all possible rational zeros of each function. Then determine which, if any, are zeros. Lesson 2-4)

76. p(x) = x4 + x3 — l lx — 5x + 30 77. d(x) = 2x4 — x3 — 6x2 + 5x — 1 78. f(x ) = x3 — Ix 1 — 5x — 6

Skills Review fo r S tandard ized Tests

79. SAT/ACT There are 16 green marbles, 2 red marbles, and 6 yellow marbles in a jar. How many yellow marbles need to be added to the jar in order to double the probability of selecting a yellow marble?A 4 C 8 E 16

B 6 D 12

80. REVIEW Refer to the figure below. Which equation could be used to find mZG?

H

F • ^ 3 sm G = — H cot G = F tan 94 4 G cot 8

G n 3 cos G = —4 I tan G = ■7 4 H sec 9

J csc 9

81. Find the exact value of sin <

A

B

C

D

4V2 - V6

42 + V3

42 - V3

82. REVIEW Which of the following is equivalent to cos 6 (cot2 8 + 1 )?

csc 8

343 g §

Another reduction identity involves the sum or difference of the measures of an angle and a quadrantal angle. This can be illustrated by comparing graphs of functions on the unit circle with Tl-Nspire technology.

Activity 1 Use the Unit Circle

Graphing Technology LabReduction Identities

______

Use Tl-Nspire technology and quadrantal angles to reduce identities.

StudyTipAngle Measure The Tl-Nspire only measures angles between 0 and 7r.

Use the unit circle to explore a reduction identity graphically.

ETEjffl Add a Graphs page. Select Zoom-Trig from the Window menu, and select Show Gridfrom the View menu. From the File menu under Tools, choose Document Settings, set the Display Digits to Float 2, and confirm that the angle measure is in radians.

ETETW Select Points & Lines and then Point from the menu. Place a point at (1, 0). Next, select Shapes, and then Circle from the menu. To construct a circle centered at the origin through (1, 0), click on the screen and define the center point at the origin. Move the cursor away from the center, and the circle will appear. Stop when you get to a radius of 1 and (1 , 0 ) lies on the circle.

E3330 Place a point to the right of the circle on the =Ai- AUT0 REALx-axis, and label it A. Choose Actions and then Coordinates and Equations from the menu, and then double-click the point to display its coordinates. From the Construction menu, choose Measurement transfer. Select the x-coordinate of A, the circle, and the point at (1, 0). Label the point created on the circle as B, and display its coordinates.

PTT7T1 With O as the origin, calculate and label the measure of /.AOB. Select Text from the Actions menu to write the expression a — 7r. Then select Calculate from the Actions menu to calculate the difference of the x-coordinate of A and 7r.

PTTfftfl Move A along the x-axis, and observe the effect on the measure of /A O B.

A -n0.858407*

"6 28. ,(-0.653644, -0.7)56802)'

£71239 rad AOB

a ( 4 .0 )

(4 . 6.2J3

Location of A m/AOB (radians)(4,0) 2.7124(3,0) 3.0708(2,0) 2.5708

(5,0) 2.2123

(-2 ,0 ) 0.5708

ETBT3 From the Construction menu, chooseMeasurement transfer. Select the x-axis and the value of a — tt. Label the point as C and display its coordinates. Using Measurement transfer again, select the x-coordinate of C, the circle, and the point at (1, 0). Label the point as D and display its coordinates.

1 ’ - I 1 1< |B 116 I au to rea l a " jJ . --------- /

~6

: :/4-n0.858407 ' i

OB 2.71239 rad

j 0 . 663644,0.756

% ■ ( 0.853407. 0}

302)

O28 , VB °A -(4^0 )

(-0.653644,-0.7 56802)

f s j x b . / _ „ • ~ 1 *

V ___________________________________________________________ _____

Analyze the Results1 A. In Step 5, how are the location of A and the measure of /.AOB related?IB. Consider the locations of points B and D. What reduction identity or identities does this

relationship suggest are true?

I C. MAKE A CONJECTURE If you change the expression a — Tt to a + 7T, what reduction identities do you think would result?

344 | Lesson 5-4

Use graphs to identify equal trigonometric functions.

HTSfln Open a new Graphs page. Select Zoom-Trig from the Window menu.

ETTiflW Graph/ ( x ) = cos x,f(x) = cos (x — tv) , and/ ( x ) = sin x . Using the Attributes feature from the Actions menu, make the line weight of/ ( x ) = cos (x — tv) medium and the line style of / ( x ) = sin x dotted.

ETCTfl Use translations, reflections, or dilations to transform/(x) = sin x so that the graphcoincides with/(x) = cos x. Select the graph and drag it over/(x) = cos x. As you move the graph, its function will change on the screen.

FTTffn Use translations, reflections, or dilations to transform/(x) = cos (x — tv) so that the graph coincides with the other two graphs. Again, as you move the graph, its function will change on the screen.

Analyze the Results2A. Write the identity that results from your transformation of/(x) = sin x in Step 3. Graph the

functions to confirm your identity.

2B. Write the identity that results from your alteration of/(x) = cos (x - tv) in Step 3. Graph the functions to confirm your identity.

2C. MAKE A CONJECTURE What does the reflection of a graph suggest for the purpose of developing an identity? a translation?

ExercisesUse the unit circle to write an identity relating the given expressions. Verify your identity by graphing.

1. cos (90° — x ) , sin x 2. cos — x j , sin x

Insert the trigonometric function that completes each identity.

3. cos x = _________ (x — 4. cot x = -----------------(x + 90°)

5. sec x = _________ (x — 180°) 6. cscx = --------------(x + f )

Use transformations to find the value of a for each expression.

7. sin ax = 2 sin x cos x 8. cos 4ax = cos2 x - sin2 x

9 . a sin2 x = 1 - cos 2 x 1 0 . 1 + cos 6ax = 2 cos2 x

345

The speed at which a plane travels can be described by a mach number, a ratio of the plane’s speed to the speed of sound.Exceeding the speed of sound produces a shock wave in the shape of a cone behind the plane. The angle 6 at the vertex of this cone is related to the mach number M describing the plane’s speed by the half-angle equation sin |2Use product-to-sum identities to

evaluate trigonometric expressions and solve trigonometric equations.

• You proved and used sum and difference identities.(Lesson 5-4)

• -4 Use double-angle, power-reducing, I and half-angle identities to evaluate

trigonometric expressions and solve trigonometric equations.

IUse Multiple-Angle Identities By letting a and (3 both equal 9 in each of the angle sum identities you learned in the previous lesson, you can derive the following double-angle identities.

KeyConcept Double-Angle Identities

sin 29 = 2 sin 0cos 9 cos 29 = cos2 9 - sin2 9

tan 29 — 2 tan^1 - tan2 9

cos 29 = 2 cos2 9 — 1

cos 29 = 1 - 2 sin2 9

Proof Double-Angle Identity for Sine

sin 29 = sin(0 +9) 29=9+9= sin 9 cos 9 + cos 9 sin 9 Sine Sum Identity where a = 0 = 0= 2 sin 9 cos 9 Simplify.

V JYou will prove the double-angle identities for cosine and tangent in Exercises 63-65.

■ m ] Evaluate Expressions Involving Double Angles

If sin 0 = on the interval |ir, -y-j, find sin 20, cos 20, and tan 20.

Since sin 9 = ^ on the interval (jc, ~ j , one point on the

terminal side of 9 has y-coordinate —7 and a distance of 25 units from the origin, as shown. The x-coordinate of this point is

therefore -V 2 5 2 — 72 or —24. Using this point, we find that

cos 9 = — orr24

'25 and tan . 1 7 I = 1 or — .x 25

Use these values and the double-angle identities for sine and cosine to find sin 29 and cos 29. Then find tan 29 using either the tangent double-angle identity or the definition of tangent.

cos 29 = 2 cos2 9 —1sin 29 = 2 sin 9 cos

= 2'1 7 , \ I 2 4 \ 336 nl 24V‘I 25) 1 2 5 j 625 = \ 25 / 1 or 527

625

Method 1 tan 29 = 2 tan 9 1 — tan2 f

2(i)Method 2 tan 29 =

1 -&)or

2 527

sin 29 cos 29336625 336527 527625

p GuidedPractice

1. If cos e = -|on the interval (o, y j , find sin 29, cos 29, and tan 29.

Igjjl 346 j Lesson 5-5

SPL/

Phot

o Re

sear

cher

s, I

nc.

StudyTipMore than One Identity Notice that there are three identities associated with cos 20. While there are other identities that could also be associated with sin 29 and tan 29, those associated with cos 29 are worth memorizing because they are more commonly used.

( ^ 2 3 ^ 1 5 Solve an Equation Using a Double-Angle IdentitySolve sin 20 — sin 0 = 0 on the interval [0, 2ir].

Use the sine double-angle identity to rewrite the equation as a function of a single angle.

sin 29 — sin 9 = 0

2 sin 9 cos 9 — sin 0 = 0

sin 9 (2 cos 9 — 1) = 0

sin 0 = 0 or 2 cos 0 — 1 = 0

0 = 0 or 7t 1

Original equation

Sine Double-Angle Identity

Factor.


Therefore, 0 = - j or

The solutions on the interval [0, 2ir] are 0 = 0, 7t, or

P' GuidedPractice Solve each equation on the interval [0, 2tv].

2A. cos 2a = —sin2 a 2B. tan 2/3 = 2 tan f3

The double angle identities can be used to derive the power-reducing identities below. These identities make calculus-related manipulations of functions like y = cos2 x much easier.

KeyConcept Power-Reducing Identities

sin2 $ = 1 - “ s 20 cos2 0 - 1 + c o s 20 tan 2 0 - 11 7 cos,2® 1 + cos 26

Proof Power-Reducing Identity for Sine

1 - cos 20 _ 1 - (1 - 2 Sin2 0) 2 2

Cosine Double-Angle Identity

2 sin2 0 2 Subtract.

= sin2 0 Simplify.V ......... J

You will prove the power-reducing identities for cosine and tangent in Exercises 82 and 83.

J g g J I S S B Use an lden% t0 Reduce a PowerRewrite sin4 x in terms with no power greater than 1.

sin4 x = (sin2 x)2 (sin2 x )2 = sin4 x

(1 — c o s 2x \2^ 1 — cos 2x j 2

1 — 2 cos 2x + cos2 2x

1 — 2 cos 2x +

41 + cos 4x

4

2 — 4 cos 2x + 1 + cos 4x

Sine Power-Reducing Identity

Multiply.

Cosine Power-Reducing Identity

Common denominator

= i ( 3 — 4 cos 2x + cos 4x) Factor.O

f- GuidedPracticeRewrite each expression in terms with no power greater than 1.

3A. cos4 x 3B. sin3 0

yj

F ranco is V iete (1540-1603)Born in a village in western France, Viete was called to Paris to decipher messages for King Henri III. Extremely skilled at manipulating equations, he used double-angle identities for sine and cosine to derive triple-, quadruple-, and quintuple-angle identities.

H23UE3U Solve an Equation Using a Power-Reducing Identity

Solve cos2 x — cos 2x = j .

Solve A lgebraicallyt) 1

cos x — cos 2x = —

1 + cos 2x _ 1 - cos 2x — —

1 + cos 2x — 2 cos 2x = 1

cos 2x — 2 cos 2 x = 0

—cos 2x = 0

cos 2 x = 0

TV

2X = f

4 4

Original equation


Multiply each side by 2.


Subtract like terms.

Multiply each side by —1.

Solutions for double angle in [0 ,2iv]

Divide each solution by 2.

The graph of y = cos 2x has a period of tv, so the solutions are x = ^ + mr or 4- «7V, n 6

Support GraphicallyThe graph of y = cos2 x — cos 2x — ^ has zeros at ~ and ^ on the interval [0 , tv]. ✓

GuidedPractice Solve each equation.

4A. cos4 a — sin4 a = i 4B. sin2 3/3 = sin2 [3

[0, ivl scl: y b y 1— 1 .5 ,1 .5 ] scl: 1

0By replacing 9 with — in each of the power-reducing identities, you can derive each of the following half-angle identities. The sign of each identity that involves the + symbol is determined by checking the quadrant in which the terminal side of j lies.

WatchOut!Determining Signs To determine which sign is appropriate when using a half-angle identity, check the quadrant in which - | lies, not the quadrant in which 9 lies.

>KeyConcept Half-Angle Identities

sin | 1 - cos ( tan | -Vi - cos 6

t a n | =

+ cos (

1 - cos 9 sin 9

sin 8

Proof Half-Angle Identity for Cosine

Rewrite 0 as 2 •

I + cos 2x Substitute x - 8

= ± V cos^ x

= cos |


Simplify.

Substitute.

You will prove the half-angle identities for sine and tangent in Exercises 66-68.

3 4 8 | Lesson 5-5 | M u l t ip le - A n g le a n d P ro d u c t- to - S u m Id e n t i t ie s

Evaluate an Expression Involving a Half Angle

Find the exact value of cos 112.5°.

Notice that 112.5° is half of 225°. Therefore, apply the half-angle identity for cosine, noting that since 112.5° lies in Quadrant II, its cosine is negative.

cos 112.5° = cos 225° 112.5° = 225°

+ cos 225°

1 - V2

2 - V2

V2 - V 2 V 2 - \/2 7= — o r r-----

V I 2

Cosine Half-Angle Identity (Quadrant II angle)

cos 225° = - ~

Subtract and then divide.

Quotient Property of Square Roots

V 2 - V ICHECK Use a calculator to support your assertion that cos 112.5° = -------- ------.

StudyTipTangent Half-Angle IdentitiesWhen evaluating the tangentfunction for half-angle values, it isusually easiest to use the form ofthe tangent half-angle identitytan - = 1 ~ c| --~ since its 2 sin 6 denominator has only one term.

cos 112.5° « -0.3826834324 and

^ GuidedPracticeFind the exact value of each expression.

5A. sin 75°

V 2 - V 2-0.3826834324 ✓

5B. tan ZlL1 2

Recall that you can use sum and difference identities to solve equations. Half-angle identities can also be used to solve equations.

5 Solve an Equation Using a Half-Angle Identity

Solve sin2 x = 2 cos2 ^ on the interval [0, 2ir].

sin2 x = 2 cos2 -|

sin x = 2 ± + COS X

sin2 x — '. = z j l+ c o s x j

sin2 x = 1 + cos x

1 — cos2 x = 1 + cos x

Original equation

Cosine Half-Angle Identity

Simplify.

Multiply.

—cos x — cos x = 0

cos x (—cos x — 1 ) = 0

cos x = 0 or7T 37T * = y o r -



Factor.

—cos x — 1 = 0

cos x = — 1; therefore, x = 7r.


Solutions in [0 ,2 ir]

The solutions on the interval [0, 2tx] are x = y , - y - , or i t .

p GuidedPracticeSolve each equation on the interval [0, 2-k],

6A. 2 sin2 + cos x = 1 + sin x 6B. > tan -J + 8 cos x tan 4 = 1

[connectED.m cgraw-hill com | 349

2 Use Product-to-Sum Identities To work with functions such as y = cos 5x sin 3x in calculus, you will need to apply one of the following product-to-sum identities.

StudyTipProofs Remember to work the more complicated side first when proving these identities.

KeyConcept Product-to-Sum Identities

sin a sin /3 = l[co s (a — /3) — cos (a + /3)] sin a cos 0 = l[s in (a + /3) + sin (a - /3)]

cos a cos /3 = -l-[cos (a - /3) + cos (a + /3)] cos a sin /3 = -l[sin (a + /3) - sin (a - /3)]

Proof Product-to-Sum Identity for sin a cos (3

j[s in (a + /9) + sin (a - /?)] More complicated side of identity

= -|(sin a cos /3 + cos a sin f) + sin a cos (3 - cos a sin /3) Sum and Difference Identities

= j(2 sin a cos 0) Combine like terms.

= sin a cos /3— _ ----------- _ _ _ — ,---------------------------------------------

Multiply.J

You will prove the remaining three product-to-sum identities in Exercises 84-86,

Use an Identity to Write a Product as a Sum or Difference

Rewrite cos 5x sin 3x as a sum or difference.

cos 5x sin 3x = j-fsin (5x + 3x) — sin (5x — 3x)] Product-to-Sum Identity

1= y(sin 8x — sin 2x) Simplify.

1 1= y sin 8 x — — sin 2x Distributive Property

p GuidedPractice Rewrite each product as a sum or difference.

7A. sin 40 cos 9 7B. sin 7x sin 6x

These product-to-sum identities have corresponding sum-to-product identities.

sin a + sin/3 = 2 sin | a ^ j cos (— cos a + cos /3 = 2 cos cos ^

sin a - sin /3 = 2 cos sin | a cos a - cos /3 = —2 sin j — sin ■ - - j

Proof Sum-to-Product Identity for sin a + sin /3

2Sm( 2 /3) COS( 2P)= 2 sin /c o s y a + P . a — 3 Substitute x = — y ~ and y = —

= 2 | j [ s in ( x + y ) + s in ( x - y ) ] | Product-to-Sum Identity

- s m ( a+/ + a ~ / ) + sm(a+/ a ~ / ) Substitute and simplify.

= s i n ( ^ ) + s i n ( f ) Combine fractions.

= sin a + sin /3 Simplify.------ -------- ------- -------------------------------- ,

You will prove the remaining three sum-to-product identities in Exercises 87-89.

350 | L e s s o n 5 -5 j M u l t ip le - A n g le a n d P ro d u c t- to -S u m Id e n t i t ie s

-• m m Use a Product-to-Sum or Sum-to-Product Identity

Find the exact value of sin 4?- + sin12 12

sin y y + sin ■— = 2 sin15H + 7T

12 12\ 2

IT TT— COS4 6

cos/ 57T tr 1

12 12\ 2

= #)(#)= V6

2

)► GuidedPracticeFind the exact value of each expression.

8 A. 3 cos 37.5° cos 187.5°

Sum-to-Product Identity

Simplify.

. -JT V2 . IV V3Sln _ = _ and cos — =

Simplify.

8 B. COS y ^ - COS y y

You can also use sum-to-product identities to solve some trigonometric equations.

•) Solve an Equation Using a Sum-to-Product Identity

WatchOutPeriods for M u ltip le A ngle Trigonom etric Functions Recall \from Lesson 4-4 that the periods /of y = sin /wand y =cos kx are not2 ir.k

Solve cos 4x + cos 2x = 0.

Solve Algebraically

cos 4x + cos Zx — 0

14x + 2x\ „ l 4tx — 2 x \ _ n2 cos I— - — I cos I— — j 0

(2 cos 3x)(cos x) = 0

Original equation

Cosine Sum-to-Product Identity

Simplify,

Set each factor equal to zero and find solutions on the interval [0, 27v].

2 cos 3x = 0

cos 3x = 0

3x = ^ or -

cos x = 0

T\ 2

* = £ < * §

First factor set equal to 0


Multiple angle solutions in [0 ,2 ir]

Divide each solution by 3.

The period of y = cos 3x is — , so the solutions are

Support GraphicallyTT TC 57TThe graph of y = cos 4x + cos 2x has zeros at —, y , — ,

, and on the interval [0, 27r]. ✓6 2 6

Second factor set equal to 0

Solutions in [0 ,2 ir]

3tt2

■ 27xn, n G !

p GuidedPracticeSolve each equation.

9A. sin x + sin 5x = 0

ZeroK=.S£3£9B?B Y=0

[0 , 2 tv1 scl: ~ r - b y I — 3 , 3 ] scl: 1

9B. cos 3x — cos 5x = 0


Find the values of sin 20, cos 20, and tan 20 for the given value and interval. (Example 1)

Q cos 9 = (270°, 360°)

2. tan 9 = (180°, 270°)

cos 0 = (90°, 180°)CD4. sin 9 = 27rj

'NlD tan 9 = ~ , 2 ttJ

6 . tan 0 = V 3, |o, —j

0 sin„ 5

8 . cos 9 =

> ( * * )

_513

Solve each equation on the interval [0, 2tt|. (Example 2)

9. sin 29 = cos 0

10. cos 29 = cos 9

11. cos 29 — sin 0 = 0

12. tan 29 — tan 2 0 tan2 9 = 2

13. sin 29 csc 9 = 1

14. cos 29 + 4 cos 9 = —3

15) GOLF A golf ball is hit with an initial velocity of 8 8 feet per second. The distance the ball travels is found by

v02 sin 26a = ---- —---- , where v0 is the initial velocity 9 is the anglethat the path of the ball makes with the ground, and 32 is in feet per second squared. (Example 2)

a. If the ball travels 242 feet, what is 9 to the nearest degree?

b. Use a double-angle identity to rewrite the equation for d.

Rewrite each expression in terms with no power greater than 1. (Example 3)

16. cos3 9 17. tan3 9

18. sec4 9 19. cot3 9

20. cos4 9 —sin4 9 21. sin2 9 cos3 9

22. sin2 9 - cos2 9 23. sin4 6 cos2 6

Solve each equation. (Example 4)

24. 1 — sin2 ■ cos 29 = —

25. cos2 9 — ^ cos 29 ■

26. sin2 9 — 1 = cos2 <

27. cos2 9 — sin 9 = 1

28. MACH NUMBER The angle 9 at the vertex of thecone-shaped shock wave produced by a plane breaking the sound barrier is related to the mach number M describing the plane's speed by the half-angle equation

6 1sin — = — . (Example 5)

a. Express the mach number of the plane in terms of cosine.

b. Use the expression found in part a to find the mach

number of a plane if cos 9 =

Find the exact value of each expression.

29. sin 67.5°

31. tan 157.5°

30. cos

32. sin

12

117712

Solve each equation on the interval [0, 2u], (Example 6)

33. sin — + cos 9 = 1

35. 2 sin = sin 9

34. tan — = sin —

36. 1 — sin2 — cos 4 = 4

Rewrite each product as a sum or difference. (Example 7)

37. cos 39 cos 9 38. cos 12x sin 5x

39. sin 3x cos 2x 40. sin 89 sin 9

Find the exact value of each expression. (Example 8)

41. 2 sin 135° sin 75°

43. I sin 172.5° sin 127.5°

45. sin 75° + sin 195°

47. 3 sin 17-rc12

■ 3 sin12

42. cos + cos

44. sin 142.5° cos 352.5°

46. 2 cos 105° + 2 cos 195°

48. cos ^ + cos

Solve each equation. (Example 9)

49. cos 9 — cos 39 = 0 50. 2 cos 49 + 2 cos 29 = 0

51. sin 39 + sin 50 = 0 52. sin 29 — sin 9 = 0

53. 3 cos 69 — 3 cos 49 = 0 54. 4 sin 9 + 4 sin 39 = 0

352 | Lesson 5-5 | M u ltip le -A n g le a n d P rod uct-to -S um Id en tities

5 5 . ^/l + cosfa 56 \j1 ~ C° S 166

Write each expression as a sum or difference.

57. cos (a + b) cos (a — b) 58. sin (9 — 7r) sin (9 + 7r)

59. sin (b + 9) cos (b + tt) 60. cos (a — b) sin (b — a)

61. MAPS A Mercator projection is a flat projection of the globe in which the distance between the lines of latitude increases with their distance from the equator.

Simplify each expression.

The calculation of a point on a Mercator projection

contains the expression tan |45° + y j, where £ is the latitude of the point.a. Write the expression in terms of sin i and cos Lb. Find the value of this expression if i = 60°.

62. BEAN BAG TOSS Ivan constructed a bean bag tossing game as shown in the figure below.

a. Exactly how far will the back edge of the board be from the ground?

b. Exactly how long is the entire setup?

PROOF Prove each identity.

63. cos 29 = cos2 9 — sin2 9

e c o n 2 t a n 665. tan 29 = —1 - t a n 2 6

, 6 . . Il — c o s 667. t a n ^ + y ^ ^ ^

Verify each identity by using the power-reducing identities and then again by using the product-to-sum identities.

69. 2 cos2 59 — 1 = cos 10070. cos2 29 — sin2 29 = cos 49

64. cos 29 = 2 cos2 0 —1

6 6 . sin-sin| = ± ^ -

6 8 . tan^ = s in 92 1 + c o s <

Rewrite each expression in terms of cosines of multiple angles with no power greater than 1.

71. sin6 9

73. cos

72. sin 9

74. sin4 9 cos4 i

75. MULTIPLE REPRESENTATIONS In this problem, you will investigate how graphs of functions can be used to find identities.a. GRAPHICAL Use a graphing calculator to graph

f(x ) = 4^sin 9 cos y ~ cos 9 sin - j j on the interval t—27V, 2 7 t].

b. ANALYTICAL Write a sine function h(x) that models the graph of f(x ). Then verify that/(x) = h(x) algebraically.

c. GRAPHICAL Use a graphing calculator to graph g(x) = cos2 — y j — sin2 — y j on the interval [ — 2 t t , 2 7 t].

d. ANALYTICAL Write a cosine function k(x) that models the graph of g(x). Then verify thatg(x) = k(x) algebraically.

H.O.T. Problem s Use H ig h e r-O rd e r T h in k in g S k ills

76. CHALLENGE Verify the following identity,

sin 29 cos 9 — cos 29 sin 9 = sin 9

REASONING Consider an angle in the unit circle. Determine what quadrant a double angle and half angle would lie in if the terminal side of the angle is in each quadrant.

77. I 78. II 79. Ill

CHALLENGE Verify each identity.

80. sin 49 = 4 sin 9 cos 9 — 8 sin3 9 cos t

8 l ) cos 49 = 1 — 8 sin2 9 cos2 9

PROOF Prove each identity.

82. cos2 9 = X+ c2os 16

83. tan2 9 =1 + c o s 2 6

84. cos a cos (3 = -|[cos (a — /3) + cos (a + (3)}

85. sin a cos f3 = -|[sin (a + (3) + sin (a - f3)]

8 6 . cos a sin f3 = -|[sin {a + (3) — sin (a — (3)]

13 = 2 cos ( " li j cos 2 ' j

io ^ = 2 co s(2 ± £ ) s i o ( ^ )

_ Ia + /3\ . l a — /3\a - cos (3 = — 2 sm I—-— I sm I—- — I

87. cos a + cos

8 8 . sin a — sin (3 =

89. cos

90. WRITING IN MATH Describe the steps that you would use to

find the exact value of cos 89 if cos 9 =

t yoi V2 5 '

(TonS 353

Spiral Review

91. cos ~ 92. cos 93. sin 12 1 2 o

94. sin 95. cos 96. sin

F in d the exact v a lu e o f each trig o n o m etric expression. (Lesson 5-4)

97. GARDENING Eliza is waiting for the first day of spring in which there will be 14 hours of daylight to start a flower garden. The number of hours of daylight H in her town can be modeled by H = 11.45 + 6.5 sin (0.0168d — 1.333), where d is the day of the year, d = 1 represents January 1, d = 2 represents January 2, and so on. On what day will Eliza begin gardening? (Lesson 5-3)

Find the exact value of each expression. If undefined, write undefined. [Lesson 4-3)

98. csc ( - y ) 99. tan 210° 100. sin — 101. cos (-3780°)

Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing. (Lesson 2-1)

102. f(x ) = ~ 1 0 3 ./(x ) = 4x4 104. f(x ) = —3x6 1 0 5 ./(x ) = 4 x5


106. REVIEW Identify the equation for the graph.

A y == 3 cos 26B y =II COS 26C y == 3 COS ¥D y =. 1

' 3 COS ¥

107. REVIEW From a lookout point on a cliff above a lake, the angle of depression to a boat on the water is 1 2 °. The boat is 3 kilometers from the shore just below the cliff. What is the height of the cliff from the surface of the water to the lookout point?

C 3tan 1 2 °

J 3 tan 12°

c.d.e.

108. FREE RESPONSE Use the graph to answer each of the following.a. Write a function of the form f(x ) = a cos (bx + c) + d that

corresponds to the graph.Rewrite/(x) as a sine function.Rewrite/(x) as a cosine function of a single angle.Find all solutions of f(x ) = 0.How do the solutions that you found in part d relate to the graph of/(x)?

354 | Lesson 5-5 j M u ltip le -A n g le a n d P rod uct-to -S um Id en tities

Study Guide and Review

Chapter Summary

KeyConcepts

Trigonometric Identities (Lesson 5 - 1)

• Trigonometric identities are identities that involve trigonometric functions and can be used to find trigonometric values.

• Trigonometric expressions can be simplified by writing the expression in terms of one trigonometric function or in terms of sine and cosine only.

• The most common trigonometric identity is the Pythagorean identity sin2 0 + cos2 0 = 1.

Verifying Trigonometric Identities (Lesson 5-2)

• Start with the more complicated side of the identity and work to transform it into the simpler side.

• Use reciprocal, quotient, Pythagorean, and other basic trigonometric identities.

• Use algebraic operations such as combining fractions, rewriting fractions as sums or differences, multiplying expressions, or factoring expressions.

• Convert a denominator of the form 1 ± u or u ± 1 to a single term using its conjugate and a Pythagorean Identity.

• Work each side separately to reach a common expression.

Solving Trigonometric Equations Lesson 5 3)

• Algebraic techniques that can be used to solve trigonometric equations include isolating the trigonometric expression, taking the square root of each side, and factoring.

• Trigonometric identities can be used to solve trigonometric equations by rewriting the equation using a single trigonometric function orby squaring each side to obtain an identity.

Sum and Difference Identities (Lesson 5-4)

• Sum and difference identities can be used to find exact values of trigonometric functions of uncommon angles.

• Sum and difference identities can also be used to rewrite a trigonometric expression as an algebraic expression.

Multiple-Angle and Product-Sum Identities (Lesson 5-5)

• Trigonometric identities can be used to find the values of expressions that otherwise could not be evaluated.

KeyVocabulary

cofunction p. 314)

difference identity (p. 337)

double-angle identity (p. 346)

half-angle identity (p. 348)

identity (p. 312)

odd-even identity (p. 314)

power-reducing identity p. 347)

product-to-sum identity p. 350)

Pythagorean identity (p. 313)

quotient identity (p. 312)

reciprocal identity (p. 312)

reduction identity (p. 340)

sum identity p. 337)

trigonometric identity (p. 312)

verify an identity (p. 320)

VocabularyCheck

Complete each identity by filling in the blank. Then name the identity.

1. sec0 = ___________

2 . ______

3. ______

COS 9

. + 1 = sec2 1

4. cos (90° - 0 ) = .

5. tan ( - 0 ) = ____

6. sin (a + (3) = sin a . . + cos a .

7.

8 .

9.

1 0 .

. = cos2 a - sin2 a

1 — cos 29

+ COS I

. = l[cos (a - /3) + cos (a + (3)]

355

Study Guide and Review Continued

Lesson-by-Lesson Review

Trigonometric Identities (pp. 3 1 2 -3 1 9 )

Find the value of each expression using the given information.

11. sec 0 and cos 0; tan 9 = 3, cos 9 > 0

12. cot 9 and sin 0; cos 9 = tan 9 < 00

13. csc 9 and tan 0; cos 9 = sin 9 < 0

14. cot 9 and cos 9; tan 9 = y , csc 9 > 0

15. sec 9 and sin 9\ cot 9 = - 2 , csc e < 0

16. cos 9 and sin 0; cot 9 = f , sec 9 < 08

Example 1If sec 0 = - 3 and sin 9 > 0, find sin 9.

Since sin 9 > 0 and sec 9 < 0 ,9 must be in Quadrant II. To find sin 9, first find cos 9 using the Reciprocal Identity for sec 9 and cos 9.

1cos 9 = -sec e

1

Reciprocal Identity

sec 9 = - 3

Now you can use the Pythagorean identity that includes sin 9 and cos 9 to find sin 9.

sin2 9 + cos2 9 = 1


17. sin2 (—x) -f- cos2 (—x)

sec2 x - tan2 x

sin^ 9 + nr-Pythagorean Identity

cos 9 = —\

19.

21.cos ( - X )

1

18. sin2 x -l- cos2 x + cot2 x„„„2 „

20.

1 - sin x 22.tan2 x + 1

cos x 1 + sec x

sin2 9 + 4 - = 1

2 a - :sin M. „ V8 2V 2 sin 9 = - V - or =^-=-

3 3

Multiply.

Subtract.

Simplify.

j

Verifying Trigonometric Identities (pp. 3 2 0 - 3 2 6 )\

Verify each identity. Example 223. sin e ■ + - sin e

1 - cos e 1 + cos e

24 cos 0 + sin 9 _ ^

25.

26.

27.

28.

29.

sec 9 CSC 9

cot 9 1 + csc e1 + CSC $ cot 9

COS 9 _ 1 + sin 9 1 - sin 9 cos 9

cot2 9

= 2 csc 9

= 2 sec 9

Verify that — S !n® _ + I ± £ 2 i » = 2 esc 9.1 + cos e sin e

The left-hand side of this identity is more complicated, so start with that expression.

sin 9 + 1 + cos 9 _ sin2 9 + (1 + cos 0)21 + cos 9 sin 9 sin 0(1 + cos 9)

1 + CSC 9

sec 6> ! csc 0 tan 9 cot 9

sec 9 + csc 9

= CSC 0 — 1

= sec 9 + csc 9

_ sin2 e + 1 + 2 cos 9 + cos2 9 sin 9( 1 + COS 9)

_ sin2 9 + COS2 9 + 1 + 2 COS 9 sin 0(1 + cos 0)

_ 1 + 1 + 2 COS 9

= CSC 01 + tan 9

30. cot 0 csc 0 + sec 0 = csc2 0 sec 0

31. sin 9 tan 9sin 9 + cos 9 1 + tan 9

,4 q _ 0in 4 a _ 1 - tan2 932. cos4 0 - sin4sec‘

sin 0(1 + cos 9)

2 + 2 cos 0 sin 0(1 + cos 0)

2(1 + cos 0) — sin 0(1 + cos 0)

_ 2 sin 0

= 2 csc 0

356 | C h a p te r 5 | Study G u id e a n d R eview


Solving Trigonometric Equations (pp. 327-333)

Find all solutions of each equation on the interval Example 3 4[0 ,2 * ] . Solve the equation sin 0 = 1 — cos 0 for all values of 0 .33. 2 sin x = \Jl 34. 4 cos2 x = 3 sin 0 = 1 - cos 0 Original equation.35. tan2 x- 3 = 0 36. 9 + cot2 x = 12 s in2 0 = (1 - cos 0 )2 Square each side.

37. 2 sin2 x = s in * 38. 3 c o sx+ 3 = s in2 x s in2 0 = 1 - 2 cos 8 + cos2 9

1 - cos2 0 = 1 - 2 cos 9 + cos2 0

Expand.

Pythagorean IdentitySolve each equation for all values of x. 0 = 2 cos2 0 - 2 cos 0 Subtract.39. sin2 x — sin x = 0 0 = 2 cos 0(cos 0 - 1 ) Factor.40. tan2 x = tan x

Solve for x on [0 ,2ix].41. 3 cos x = cos x - 1 cos 0 = 0 or cos 0 = 142. sin2 x = sin x + 2 0 = cos-1 0 0 = cos-1 1

43. sin2 x = 1 - cos x II 0 - Jco

1x9 N

11 0

44. sin x = cos x + 1 A check shows that - y - is an extraneous solution. So the solutions

k-

are 0 = y + 2nix or 0 = 0 + 2 n ir .

.... ....... J

Sum and Difference Identities (pp. 336-343) I

Find the exact value of each trigonometric expression. Example 4

45. cos 15° 46. sin 345° 47. t a n ^23-7T

Find the exact value of tan - j y .

48. s i n ^ | 49. c o s - y y 50. t a n ^ | - t a n f L = t a n ( f + f )23ir _ 5 ir , 2 lt 12 4 3

Simplify each expression.t a n f + t a n ^ -

51 ■ ------- ‘ ------------ 5 -1 - tan y tan y

52. cos 24° cos 36° - sin 24° sin 36°

tan ^ + tan ^ 4 31 - t a n ^ t a n - ^ 4 3

1 - V 3 1 - ( - V 3 )

Sum Identity

Evaluate for tangent.

53. sin 95° cos 50° - cos 95° sin 50°

54. cos ^ cos ^ + s in ^ sin ^

II II

+ I

+ I

I I

Simplify.

Rationalize the denominator.


55. cos (0 + 30°) - sin (0 + 60°) = - s in 84 - 2V 3

1 - 3Multiply.

56. cos (0 + j ) = (cos 0 - sin 0) = 4 “ ^ o r 2 + V 3 Simplify.

57. cos (0 - + cos (0 + j j = cos 0

58. tan (0 + ^ ) = ta n f ~ ■ V 4 / tan 9 + 1

. .... -i

JjponnectS 357

Study Guide and Review Continued


Multiple-Angle and Product-Sum Identities (pp. 346-354)

Find the values of sin 29, cos 20, and tan 20 for the given value

59. cos i ' 3 4 / TT

, (0°, 90° 60. tan 9 = 2, (180°, 270°

61. sin 0 = | , ( f , t t ) 62. sec 9 = 2 ir)

Find the exact value of each expression.

63. sin 75°

65. tan 67.5°

64. cos 11 TT 12

66. cos 4 ?

67. sin 15tt8

68. tan 13-tt12

24fourth quadrant and tan o = —

> is in the fourth quadrant, so cos 9 = and sin 9 = - | f .25 2ocos 29 = 2 cos2 9 - 1sin 29 = 2 sin tfcos

= 2 (—2 1 UI 25/ 2E

336 25 625 - f t ) * - ' -

527625

tan 29 = 2 tan e 1 - tan2 6

i-fi 48 ' 7

/ _ 2 4 \ 2 527I 7 j 49

or 336527

Applications and Problem Solving69. CONSTRUCTION Find the tangent of the angle that the ramp makes

with the building if sin 9 -(Lesson 5-1)

^ . n d c o s . - ' 2^145 145

70. LIGHT The intensity of light that emerges from a system of two

polarizing lenses can be calculated by I = L 'ocscr e

-, where L is

the intensity of light entering the system and 9 is the angle of the axis of the second lens with the first lens. Write the equation for the light intensity using only tan 9. (Lesson 5-1)

71. MAP PROJECTIONS Stenographic projection is used to project the contours of a three-dimensional sphere onto a two-dimensional map. Points on the sphere are related to points on the map usingr - SJP. g . Verify that r = 1 +C0SQ. (Lesson 5-2)

1 - cos a sin a

72. PROJECTILE MOTION A ball thrown with an initial speed i/Q at an angle 9 that travels a horizontal distance d w ill remain in the air tseconds, where t = yncos Suppose a ball is thrown with an

initial speed of 50 feet per second, travels 100 feet, and is in the air for 4 seconds. Find the angle at which the ball was thrown.(Lesson 5-3)

73. BROADCASTING Interference occurs when two waves pass through the same space at the same time. It is destructive if the amplitude of the sum of the waves is less than the amplitudes of the individual waves. Determine whether the interference is destructive when signals modeled by y = 20 sin (31 + 45°) and y = 20 sin (31 + 225°) are combined. (Lesson 5-4

74. TRIANGULATION Triangulation is the process of measuring a distance d using the angles a and /3 and the distance £using £ - d • dtana + tan/3' (Lesson 5-5)

dA a / v \

I

a. Solve the formula for d.

£ sin a sin /3b. Verify that d =

c. Verify that d =

sin a cos (3 + cosasin /3 ' £ sin a sin 0sin (a + 0 ) '

d. Show that if a = /?, then d = 0.5£ tan a.

358 | C h ap te r 5 | Study G u id e an d R eview

• - ■

Practice Test

.

Find the value of each expression using the given information.

1. sin 8 and cos 6>, csc e = - 4 , cos e < 0

2. csc e and sec e, tan 6

Simplify each expression.sin (90° - x)

r, csc 0 < 0

3.

4.

tan (90° - x)

sec2x - 1tan2x + 1

5. sin 9 (1 + cot2 9)

Verify each identity.csc2 0 - 1 , sec2 e - 16.

7.

8.

csc‘

COS <

secz 0= 1

1 + sin 0

1

+ 1 - sin e _ 2 cos f

+

COS 0

11 + COS 0 1 - COS

1 + sin 0

- = 2 csc2 8

9. - s e c 2 8 sin2 9 = cos2 e ~ - -COS‘ 0

10. sin4 x - cos4 x = 2 sin2 x - 1

11. MULTIPLE CHOICE Which expression is noftrue? A tan ( - 9 ) = - ta n 8

1B tan ( - $ ) ■

C tan ( - 8) ■ ,' cos (-0 )

D tan ( - 9 ) + 1 = sec ( - 9 )

cot ( - 0 )

sin ( - 0 )

Find all solutions of each equation on the interval [0, 2tt].

12. \ / 2 sin 9 + 1 = 0

13. sec2 0 = - |

Solve each equation for all values of 0.

14. tan2 9 - tan 8 = 0

1 - sin 0 .15.

16.

COS 0

1

- = COS I

1sec 0 — 1 sec 0 + 1

17. s e c 0 - 2 t a n 0 = O

18. CURRENT The current produced by an alternator is givenb y / = 40 sin 1357vf, where I is the current in amperes and f is the time in seconds. At what time / does the current first reach 20 amperes? Round to the nearest ten-thousandths.

19. tan 165°

Find the exact value of each trigonometric expression.

20. cos 1221. sin 75°

22. cos 465° - cos 15°

23. 6 sin 675° - 6 sin 45°

24. MULTIPLE CHOICE Which identity is true?

F cos {9 + 7r) = - s in tv

G cos ( i t - 9) = cos 9

H sin ^0 - - y - j = cos 9

J sin (tv + 9) = sin 8


25. cos £ cos 4 ? - sin ^ sin -3lT

26.

8 8

tan 1 3 5 ° - t a n 15°1 + tan 135° tan 15°

27. PHYSICS A soccer ball is kicked from ground level with an initial speed of v at an angle of elevation 9.

a. The horizontal distance d the ball w ill travel can be determinedv2 sin 20 where g is the acceleration due to gravity.

2 „2 /■using d = -Verify that this expression is the same as - ^ z(tan 9 - tan sin2 9).

b. The maximum height h the object will reach can be determined

using h = \r sin e 2 9

Find the ratio of the maximum height

attained to the horizontal distance traveled.

Find the values of sin 20, cos 29, and tan 20 for the given value and interval.

28. tan 9 = - 3 , 2tv|

29. cos 9 = -L (0°, 90°) 0

30. cose

5 359

:• Objective

• Approximate rates of change fo r sine and cosine functions using the difference quotient.

In Chapter 4, you learned that many real-world situations exhibit periodic behavior over time and thus, can be modeled by sinusoidal functions.Using transformations of the parent functions sin xand cos x, trigonometric models can be used to represent data, analyze trends, and predict future values.

Hours of Daylight

16

12V)3X 8

4

__

uJ M M J S N

Month

While you are able to model real-world situations using graphs of sine and cosine, differential calculus can be used to determine the rate that the model is changing at any point in time. Your knowledge of the difference quotient, the sum identities for sine and cosine, and the evaluation of limits now makes it possible to discover the rates of change for these functions at any point in time.

Activity 1 Approximate Rate of Change

Approxim ate the rate of change of f(x ) = sin x at several points.

ETITn Substitute/(x) = sin x into the difference quotient.

f(x + h) — f(x) sin (x + h) — sin xm = ; ► m = -----h h

Approximate the rate of change of/(x) at x = y . Let h = 0.1, 0.01, 0.001, and 0.0001.

ETTSTfH Repeat Steps 1 and 2 for x = 0 and for x = tv.

f Analyze the Results

1. Use tangent lines and the graph of f(x ) = sin x to interpret the values found in Steps 2 and 3.

2. What will happen to the rate of change off(x ) as x increases?V_____________

Unlike the natural base exponential function g{x) = e*and the natural logarithmic function h(x) = In x, an expression to represent the rate of change of f(x) = sin x a t any point is not as apparent. However, we can substitute f(x) into the difference quotient and then simplify the expression.

f (x + h) - f(x)m = - h

_ sm(x+ h)-s\nx h

_ (sin x cos /7 + cos x sin h) - sin x h

_ (sin xcosh- sin x) + cos xsin h h

= sinx(“ i ^ l ) + Co s x ( ^ )

Difference quotient

f(x) = sin x

Sine Sum Identity

Group terms with sin x.

Factor sin xand cos x.

360 | C h a p te r 5

We now have two expressions that involve h, sin x — - j and cos x | - y y j . T o obtain an accurate approximation

of the rate of change of f(x) at a point, we want h to be as close to 0 as possible. Recall that in Chapter 1, we were able to substitute h = 0 into an expression to find the exact slope of a function at a point. However, both of the fractional expressions are undefined at h = 0.

sjnx(co s_ ^ l)

= sin x

undefined( T )

Original expressions

h = 0

cosx (t )

= COSX

undefined(V)

We can approximate values for the two expressions by finding the lim it of each as h approaches 0 using techniques discussed in Lesson 1-3.

Activity 2 Calculate Rate of ChangeFind an expression for the rate of change of f(x ) = sin x.

cos h — 1Use a graphing calculator to estimate lim -

Verify the value found in Step 1 by using the TA B LE function of your calculator.

h

Repeat Steps 1 and 2 to estimate lim sin h h^O h

Substitute the values found in Step 2 and Step 3 into the slope equation

H r 1 ) * ' “ H t Im = sm x\

'X

K = .066BH Y=-.033Hi[—tt, tt] scl: - y by [—1.5,1.5l scl: 1

X V i-.0030-.0020BRSna0.0000.00100.00200.00300

.00150 l.OE-3 E.OE-H ERROR -SE-H -IE -3 -.0015

X = - . 0 0 1F3EEI3 Simplify the expression in Step 4.

f Analyze the Results

3. Find the rate of change of f{x ) = sin x at x = 2, and -y-.

4. Make a conjecture as to why the rates of change for all trigonometric functions must be modeled by other trigonometric functions.

Model and Apply5. In this problem, you will find an expression for the rate

of change off(x ) = cos x at any point x.

a. Substitute/(x) = cos x into the difference quotient.

b. Simplify the expression from part a.

C. Use a graphing calculator to find the limit of the two fractional expressions as h approaches 0 .

d. Substitute the values found in part c into the slope equation found in part b.

e. Simplify the slope equation in part d.

f. Find the rate of change of f(x ) = cos x at x = 0, y , 7T, and -y-.

361

Trigonometric Identities and Equations · Pythagorean identity p. 313 Pitagoras identidad ... sm 9...

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Transcript of Trigonometric Identities and Equations · Pythagorean identity p. 313 Pitagoras identidad ... sm 9...