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SECTION 7-1 289

CHAPTER 7 Section 7–1 1. To verify an identity means to prove that both sides of an equation are equal when any values of the

variables for which both sides are defined are substituted into that equation. 3. Values of the variables must be found for which both sides of the equation are defined, but are not

equal. A single set of such values, called a counterexample, suffices to prove that the equation is not an identity.

5. sin θ sec θ = sin θ 1

cos Reciprocal Identity

= sin

cos

Algebra

= tan θ Quotient Identity

7. cot u sec u sin u = cos 1

sin cos

u

u usin u Reciprocal and Quotient Identities

= 1 Algebra

9. sin( )

cos( )

x

x

= sin

cos

x

x

Identities for Negatives

= –tan x Quotient Identity

11. tan cot

csc

= 1

tantan

csc

Reciprocal Identity

= 1

csc Algebra

= 1

sin

1

Reciprocal Identity

= sin α Algebra

13. csc u(cos u + sin u) = 1

sin u(cos u + sin u) Reciprocal Identity

= cos

sin

u

u +

sin

sin

u

u Algebra

= cos

sin

u

u + 1 Algebra

= cot u + 1 Quotient Identity

Key Algebraic Steps: 1

b(a + b) =

a

b +

b

b =

a

b + 1

15. cos sin

sin cos

x x

x x

=

cos

sin cos

x

x x –

sin

sin cos

x

x x Algebra

= 1

sin x –

1

cos x Algebra

= csc x – sec x Reciprocal Identities

290 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS

17. 2sin

cos

t

t + cos t =

2sin

cos

t

t +

cos

1

t Algebra

= 2sin

cos

t

t +

2cos

cos

t

t Algebra

= 2 2sin cos

cos

t t

t

Algebra

= 1

cos t Pythagorean Identity

= sec t Reciprocal Identity

Key Algebraic Steps: 2a

b + b =

2a

b +

1

b=

2a

b +

2b

b=

2 2a b

b

19. 2

cos

1 sin

x

x =

2 2 2

cos

sin cos sin

x

x x x Pythagorean Identity

= 2

cos

cos

x

x Algebra

= 1

cos x Algebra

= sec x Reciprocal Identity 21. (1 – cos u)(1 + cos u) = 1 – cos2 u Algebra

= sin2 u + cos2 u – cos2 u Pythagorean Identity = sin2 u Algebra

23. 1 – 2 sin2 x = cos2 x + sin2 x – 2 sin2 x Pythagorean Identity

= cos2 x – sin2 x Algebra 25. (sec t + 1)(sec t – 1) = sec2 t – 1 Algebra

= tan2 t + 1 – 1 Pythagorean Identity = tan2 t Algebra

27. csc2 x – cot2 x = 1 + cot2 x – cot2 x Pythagorean Identity

= 1 Algebra

29. cos tan

sin

x x

x

=

cos

sin

x

x +

tan

sin

x

x Algebra

= cot x + sincos

sin

xx

x Quotient Identities

= cot x + sin

cos

x

x ÷ sin x Algebra

= cot x + sin

cos

x

x ·

1

sin x Algebra

= cot x + 1

cos x Algebra

= cot x + sec x Reciprocal Identities

Key Algebraic Steps: a + bc

b = a +

b

c ÷ b = a +

b

c ·

1

b= a +

1

c

SECTION 7-1 291

31. Plug in x = –4.

Left side: 2( 4 3) = 2( 1) 1 = 1

Right side: –4 + 3 = –1 The equation is not an identity.

33. Plug in x = –1. Left side: (–1)2 + (–1) = 1 – 1 = 0 Right side: 3(–1)3 – (–1)2 = –3 – 1 = –4 The equation is not an identity.

35. Plug in x = 3

.

Left side: sin 3

=

3

2

Right side: 1 – cos 3

= 1 –

1

2 =

1

2

The equation is not an identity.

37. Plug in x = 4

.

Left side: sec 4

tan

4

= 2 1

Right side: cot 4

= 1


39.

The two graphs do not appear to coincide. The equation 1+ sin x = cos x is not an identity.

41.

The two graphs appear to coincide. The equation cos x tan x = sin x appears to be an identity.

43.

The two graphs do not appear to coincide. The equation sin x – csc x = –cot x csc x does not appear to be an identity.

45. Simplify the left side:

2 9

3

x

x

= ( 3)( 3)

3

x x

x

= x – 3

The equation is an identity.

47. Simplify the left side:

2 4 4x x = ( 2)( 2)x x = 2( 2)x = |x + 2|

The equation is not an identity: the left side must be positive, but the right can be negative. For example, if we plug in x = –4, we get

2( 4) 4( 4) 4 ?

–4 + 2

4 ?

–2

2 ?

–2

49. Plug in x = 4

: sin

4

– cos

4

?

1

1

2 –

1

2 ?

1

Obviously this is false so the equation is not an identity.

51. The equation is an identity. We can verify it as follows: cos3 x = cos x cos2 x Algebra = cos x( 1 – sin2 x) Pythagorean Identity

= cos x – cos x sin2 x Algebra


53. 21 (sin cos )

sin

x x

x

=

2 21 (sin 2sin cos cos )

sin

x x x x

x

Algebra

= 2 21 ( 2sin cos sin cos )

sin

x x x x

x

Algebra

= 1 ( 2sin cos 1)

sin

x x

x

Pythagorean Identity

= 2sin cos

sin

x x

x Algebra

= 2 cos x Algebra Key Algebraic Steps: (a – b)2 = a2 – 2ab + b2 = –2ab + a2 + b2

55. cot 1

csc

= cossin

1sin

1

Reciprocal and Quotient Identities

= cossin

1sin

sin 1 sin

sin

Algebra

= cos θ + sin θ Algebra


1ba

a

=

1

1ba

a

a a

a

= b + a

57. 1 cos

1 cos

y

y

= (1 cos ) (1 cos )

(1 cos ) (1 cos )

y y

y y

Algebra

= 2

2

1 cos

(1 cos )

y

y

Algebra

= 2 2 2

2

sin cos cos

(1 cos )

y y y

y


= 2

2

sin

(1 cos )

y

y Algebra

59. tan2 x – sin2 x = 2

2

sin

cos

x

x – sin2 x Quotient Identity

= sin2 x2

1

cos x – sin2 x · 1 Algebra

= sin2 x2

11

cos x

Algebra

= sin2 x(sec2 x – 1) Reciprocal Identity = sin2 x tan2 x Pythagorean Identity = tan2 x sin2 x Algebra

61. csc

cot tan

= 1

sincos sinsin cos

Reciprocal and Quotient Identities

= 1

sincos sinsin cos

cos sin

cos sin cos sin

Algebra

SECTION 7-1 293

= 2 2

cos

cos sin

Algebra

= cos

1


= cos θ

Key Algebraic Steps: 1a

b aa b

= 1a

b aa b

ab

ab ab

=

2 2

b

b a

63. ln (tan x) = lnsin

cos

x

x

Quotient Identity

= ln (sin x) – ln (cos x) Algebra

65. sec 1

sec 1

A

A

= 1

cos1

cos

1

1A

A

Reciprocal Identity

= 1

cos1

cos

cos 1 cos

cos 1 cosA

A

A A

A A

Algebra

= 1 cos

1 cos

A

A

Algebra


1

1

1a

a

=

1

1

1

1a

a

a a

a a

=

1

1

a

a

67. sin4 w – cos4 w = (sin2 w)2 – (cos2 w)2 Algebra = (sin2 w + cos2 w)(sin2 w – cos2 w) Algebra = 1(sin2 w – cos2 w) Pythagorean Identity = sin2 w – cos2 w Algebra = 1 – cos2 w – cos2 w Pythagorean Identity = 1 – 2 cos2 w Algebra

Key Algebraic Steps: a4 – b4 = (a2)2 – (b2)2 = (a2 + b2)(a2 – b2)

69. sec x – cos

1 sin

x

x =

1

cos x –

cos

1 sin

x

x Reciprocal Identity

= 21 sin cos

(1 sin )cos

x x

x x

Algebra

= 2 2 2sin cos sin cos

(1 sin )cos

x x x x

x x


= 2sin sin

(1 sin ) cos

x x

x x

Algebra

= (1 sin )sin

(1 sin ) cos

x x

x x

Algebra

= sin

cos

x

x Algebra

= tan x Quotient Identity



a –

1

a

b =

21

(1 )

b a

b a

2

(1 )

b b

b a

= (1 )

(1 )

b b

b a

= b

a

71. 2

2

cos 3cos 2

sin

z z

z

=

2

2 2 2

cos 3cos 2

sin cos cos

z z

z z z

Algebra

= 2

2

cos 3cos 2

1 cos

z z

z


= (cos 1)(cos 2)

(1 cos )(1 cos )

z z

z z

Algebra

= (cos 2)

1 cos

z

z

Algebra

= 2 cos

1 cos

z

z

Algebra


2

3 2

1

a a

a

= ( 1)( 2)

(1 )(1 )

a a

a a

= ( 2)

1

a

a

= 2

1

a

a

73. 3 3cos sin

cos sin

= 2 2(cos sin )(cos cos sin sin )

cos sin

Algebra

= cos2 θ + cos θ sin θ + sin2 θ Algebra = 1 + cos θ sin θ Pythagorean Identity = 1 + sin θ cos θ Algebra

Key Algebraic Steps: 3 3a b

a b

= 2 2( )( )a b a ab b

a b

= a2 + ab + b2 Common Error:

3 3

2 2a ba b

a b

75. Graph both sides of the equation in the same viewing

window.

sin( )

cos( ) tan( )

x

x x

= –1 is not an identity, since the graphs

do not match. Try x = –4

.

Left side:

4

4 4

sin

cos tan

= 12

12

1 = 1

Right side: = –1 This verifies that the equation is not an identity.

SECTION 7-1 295

77. Graph both sides of the equation in the same viewing window.

sin

cos tan( )

x

x x = –1 appears to be an identity,

which we now verify:

sin

cos tan( )

x

x x =

sin

cos ( tan )

x

x x Identities for Negatives

= –sin

cos tan

x

x x Algebra

= –sincos

sin

cos xx

x

x Quotient Identity

= –sin

sin

x

x Algebra

= –1 Algebra 79. Graph both sides of the equation in the same

viewing window.

sin x + 2cos

sin

x

x = sec x is not an identity, since

the graphs do not match. Try x = –4

.

15

-15

-5 5

Left side: sin4

+

24

4

cos

sin

= –

1

2 +

212

12

= –

1

2 –

1

2 = – 2

Right side: sec4

= 2

This verifies that the equation is not an identity. 81. Graph both sides of the equation in the same

viewing window.

sin x + 2cos

sin

x

x = csc x appears to be an identity,

which we now verify:

15

-15

-5 5

sin x + 2cos

sin

x

x =

sin

1

x +

2cos

sin

x

x Algebra

= 2sin

sin

x

x +

2cos

sin

x

x Algebra

= 2 2sin cos

sin

x x

x

Algebra

= 1

sin x Pythagorean Identity

= csc x Reciprocal Identity


83. Graph both sides of the equation in the same viewing window.

tan

sin 2 tan

x

x x =

1

cos 2x appears to be an

identity, which we now verify:

tan

sin 2 tan

x

x x =

sincos

sincossin 2

xx

xxx

Quotient Identity

= sincos

sincos

cos

sin cos 2cos

xx

xx

x

x x x

Algebra

= sin

sin cos 2sin

x

x x x Algebra

= sin

sin (cos 2)

x

x x Algebra

= 1

cos 2x Algebra

85. 2

2

2sin 3cos 3

sin

x x

x

=

2

2

2(1 cos ) 3cos 3

sin

x x

x


= 2

2

2 2cos 3cos 3

sin

x x

x

Algebra

= 2

2

2cos 3cos 1

sin

x x

x

Algebra

= 2

2

2cos 3cos 1

1 cos

x x

x


= 2

2

(2cos 3cos 1)

(cos 1)

x x

x

Algebra

= (2cos 1)(cos 1)

(cos 1)(cos 1)

x x

x x

Algebra

= 2cos 1

cos 1

x

x

Algebra

= 2cos 1

1 cos

x

x

Algebra

Key Algebraic Steps:2

2

2 3 1

1

a a

a

= 2

2

(2 3 1)

( 1)

a a

a

= (2 1)( 1)

( 1)( 1)

a a

a a

= 2 1

1

a

a

= 2 1

1

a

a

87. tan sin

tan sin

u u

u u

– sec 1

sec 1

u

u

= sincossincos

sin

sin

uuuu

u

u

–

sec 1

sec 1

u

u

Quotient Identity

= 1

cos1

cos

sin 1 sin

sin 1 sinu

u

u u

u u

–

sec 1

sec 1

u

u

Algebra

=

1cos

1cos

sin 1

sin 1

u

u

u

u

–

sec 1

sec 1

u

u

Algebra

SECTION 7-1 297

= 1

cos1

cos

1

1u

u

–

sec 1

sec 1

u

u

Algebra

= sec 1

sec 1

u

u

– sec 1

sec 1

u

u

Reciprocal Identity

= 0 Algebra

Key Algebraic Steps: baba

b

b

=

1

1a

a

b b

b b

=

1

1

1

1

a

a

b

b

=

1

1

1

1a

a

89. tan cot

tan cot

= tan

tan cot

+ cot

tan cot

Algebra

= 1

cot +

1

tan Algebra

= tan α + cot β Reciprocal Identities 91. Statement Reason

cot2 x + 1 = cos

sin

x

x

2

+ 1 (A) cot x = cos

sin

x

x

= 2

2

cos

sin

x

x + 1 Algebra

= 2 2

2

cos sin

sin

x x

x

Algebra

= 2

1

sin x (B) cos2 x + sin2 x = 1

= 1

sin x

2

Algebra

= csc2 x (C) csc x = 1

sin x

93. Since 21 cos x = 2sin x , the equation will be true when 2sin x = –sin x, that is, when sin x is negative. This occurs in Quadrants III, IV.

95. Since 21 sin x = 2cos x = |cos x| is an identity, this will hold in all quadrants.

97. Since 21 sin x = 2cos x = cos x if cos x is positive, this will hold when

2

sin

1 sin

x

x =

2

sin

cos

x

x =

sin

cos

x

x = tan x, that is, when cos x is positive. This occurs in Quadrants I,

IV.

99. 2 2a u = 2 2( sin )a a x = 2 2 2sina a x = 2 2(1 sin )a x = 2 21 sina x = 2 2cosa x

Common Error: 21 sin 1 sinx x

Since a > 0, 2a = a. cos x will be ≥ 0 if x is a quadrant I or IV angle.

But the restriction –2

< x <

2

requires that x is such an angle.

Therefore 2cos x = cos x 2 2a u = a cos x


101. 2 2a u = 2 2( tan )a a x = 2 2 2tana a x = 2 2(1 tan )a x = 2 21 tana x = 2 2seca x

Since a > 0, 2a = a. sec x will be ≥ 0 if x is a quadrant I or IV angle.

But the restriction 0 < x < 2

requires that x is a quadrant I angle.

Therefore 2sec x = sec x

2 2a u = a sec xExercise 7–2 1. The six trigonometric functions can be paired as sine-cosine, tangent-cotanget, secant-cosecant.

Each function in a pair is the cofunction of the other and they are related by cofunction identities of the type

f 2

= cof(θ).

3. Use x + y = x – (–y).

5. The only difference is that then 90° is used in place of 2

.

7. Plug in x = 1, y = 1. Left side: (1 + 1)2 = 22 = 4 Right side: 12 + 12 = 2 The equation is not an identity.

9. Plug in x = 2, y = 6

.

Left side: 2 sin 6

= 1

Right side: sin 26

= 3

2


11. Plug in x = 6

, y =

6

.

Left side: cos6 6

= cos 3

=

1

2

Right side: cos 6

+ cos

6

=

3

2 +

3

2 = 3


13. Plug in x = 3

, y =

6

.

Left side: tan3 6

= tan 6

=

1

3

Right side: tan 3

– tan

6

= 3 –

1

3 =

2

3


15. Plug in x = 2

, y = 0.

Left side: cos 02

= cos 2

= 0

Right side: cos2

– cos 0 = 0 – 1= –1


17. Expand and simplify the left side:

tan(x – π) = tan tan

1 tan tan

x

x

Difference Identity

= tan 0

1 tan 0

x

x

tan π = sin

cos

= 0

1 = 0

= tan

1

x Algebra

= tan x Algebra The equation is an identity.

SECTION 7-2 299

19. Expand and simplify the left side: sin(x – π) = sin x cos π – cos x sin π Difference Identity = sin x(–1) – cos x(0) cos π = –1; sin π = 0 = –sin x Algebra The equation is not an identity.


csc(2π – x) = 1

sin(2 )x csc x =

1

sin x

= 1

sin 2 cos cos 2 sinx x Difference Identity

= 1

0 cos 1 sinx x sin 2π = 0; cos 2π = 1

= 1

sin x Algebra

= –csc x csc x = 1

sin x



sin2

x

= sin x cos

2

– cos x sin

2

Difference Identity

= sin x · 0 – cos x · 1 cos 2

= 0; sin

2

= 1

= –cos x Algebra The equation is an identity.

25. cot2

x

=

2

2

cos

sin

x

x

Quotient Identity

= 2 2

2 2

cos cos sin sin

sin cos cos sin

x x

x x

Difference Identities

= 0cos 1sin

1cos 0sin

x x

x x

Known Values

= sin

cos

x

x Algebra

= tan x Quotient Identity

27. csc2

x

=

2

1

sin x Reciprocal Identity

= 2 2

1

sin cos cos sinx x Difference Identity


= 1

1cos 0sinx x Known Values

= 1

cos x Algebra

= sec x Reciprocal Identity 29. cos x = cos π cos x + sin π sin x Difference Identity

= (–1) cos x + (0) sin x Known Values = –cos x Algebra 31. sin x y = sin [ x – (–y)] Algebra

= sin x cos(–y) – cos x sin(–y) Difference Identity = sin x cos y – cos x (–sin y) Identities for Negatives = sin x cos y + cos x sin y Algebra 33. sin(30° – x) = sin 30° cos x – cos 30° sin x Difference Identity

= 1

2 cos x –

3

2 sin x Known Values

= 1

2 (cos x – 3 sin x) Algebra

35. sin(180° – x) = sin 180° cos x – cos 180° sin x Difference Identity = 0 cos x – (–1) sin x Known Values = sin x Algebra

37. tan3

x

= 3

3

tan tan

1 tan tan

x

x

Sum Identity

= tan 3

1 3 tan

x

x

Known Values

39. sec 75° = sec(30° + 45°)

= 1

cos(30 45 )

= 1

cos30 cos 45 sin 30 sin 45

= 3 1 1 1

2 22 2

1

= 3 1

2 2 2 2

2 2 1

2 2 2 2

=

2 2

3 1

41. sin 7

12

= sin

3 4

= sin 3

cos

4

+ cos

3

sin

4

= 3

2·

1

2 +

1

2·

1

2

= 3

2 2 +

1

2 2

= 3 1

2 2

43. cos 74° cos 44° + sin 74° sin 44° = cos(74° – 44°) = cos 30° = 3

2

45. tan 27 tan18

1 tan 27 tan18

= tan(27° + 18°) = tan 45° = 1

SECTION 7-2 301

47.

a = 2 25 ( 3) = 4

cos x = 4

5 tan x = –

3

4

sin(x – y) = sin x cos y – cos x sin y

= 3 1

5 3

– 4 8

5 3

= 3 4 8

15

y

(a, )8

b

3

aa

8

a = 2 23 ( 8) = 1

cos y = 1

3 tan y =

8

1 = 8

tan(x + y) = tan tan

1 tan tan

x y

x y

= 34

34

8

1 8

= 3 4 8

4 3 8

= 4 8 3

4 3 8

49.

-3

b

-4 a

(-4, -3)

x

r = 2 2( 4) ( 3) = 5

sin x = –3

5 cos x = –

4

5

sin(x – y) = sin x cos y – cos x sin y

= 3 2 4 1

5 55 5

= –6

5 5 –

4

5 5=

10

5 5

=

2

5

b

-1a

2y

(2, -1)

r = 2 22 ( 1) = 5

sin y = 1

5

cos y =

2

5

tan(x + y) = tan tan

1 tan tan

x y

x y

=

3 14 2

3 14 21

=

6 4

8 3

= 2

11

51. cos 2x = cos(x + x) Algebra = cos x cos x – sin x sin x Sum Identity = cos2 x – sin2 x Algebra

53. cot(x + y) = cos( )

sin( )

x y

x y

Quotient Identity

= cos cos sin sin

sin cos cos sin

x y x y

x y x y

Sum Identities

= cos cos sin sinsin sin sin sin

sin cos cos sinsin sin sin sin

x y x yx y x y

x y x yx y x y

Algebra

= coscos

sin sin

cos cossin sin

1yxx y

y xy x

Algebra

= cot cot 1

cot cot

x y

y x

Quotient Identity


55. tan 2x = tan(x + x) Algebra

= tan tan

1 tan tan

x x

x x

Sum Identity

= 2

2 tan

1 tan

x

x Algebra

57. sin( )

sin( )

v u

v u

= sin cos cos sin

sin cos cos sin

v u v u

v u v u

Sum and Difference Identities

= sin cos cos sinsin sin sin sinsin cos cos sinsin sin sin sin

v u v uv u v uv u v uv u v u

Algebra

= cos cossin sincos cossin sin

u vu vu vu v

Algebra

= cot cot

cot cot

u v

u v

Quotient Identity

59. cot x – tan y = cos

sin

x

x –

sin

cos

y

y Quotient Identities

= cos cos

sin cos

x y

x y –

sin sin

sin cos

x y

x y Algebra

= cos cos sin sin

sin cos

x y x y

x y

Algebra

= cos( )

sin cos

x y

x y

Sum Identity

61. tan(x – y) = tan tan

1 tan tan

x y

x y

Difference Identity

= 1 1

cot cot

1 1cot cot1x y

x y

Reciprocal Identity

=

1 1cot cot

1 1cot cot

cot cot

cot cot 1

x y

x y

x y

x y

Algebra

= cot cot

cot cot 1

y x

x y

Algebra

63. cos( ) cosx h x

h

=

cos cos sin sin cosx h x h x

h

Sum Identity

= cos cos cos sin sinx h x x h

h

Algebra

= cos (cos 1) sin sinx h x h

h

Algebra

= cos xcos 1h

h

– sin x sin h

h Algebra

SECTION 7-2 303

65. x =4

, y =

3

4

cos (x + y) = cos x cos y – sin x sin y

Left side: cos3

4 4

= cos π = –1

Right side: cos4

cos

3

4

– sin

4

sin

3

4

=

2 2 2 2 1 1

2 2 2 2 2 2

= –1

sin (x + y) = sin x cos y + cos x sin y

Left side: sin3

4 4

= sin π = 0

Right side: sin4

cos

3

4

+ cos

4

sin

3

4

=

2 2 2 2 1 1

2 2 2 2 2 2

= 0

67. x =11

6

, y =

5

6

cos (x + y) = cos x cos y – sin x sin y

Left side: cos11 5

6 6

= cos π = –1

Right side: cos11

6

cos

5

6

– sin11

6

sin

5

6

= 3 3 1 1 3 1

2 2 2 2 4 4

= –1

sin (x + y) = sin x cos y + cos x sin y

Left side: sin11 5

6 6

= sin π = 0

Right side: sin11

6

cos

5

6

+ cos11

6

sin

5

6

= 1 3 3 1 3 3

2 2 2 2 4 4

= 0

69. sin(x – y) = sin(5.288 – 1.769) = –0.3685 sin x cos y – cos x sin y = sin 5.288 cos 1.769 – cos 5.288 sin 1.769 = –0.3685 tan(x + y) = tan(5.288 + 1.769) = 0.9771

tan tan

1 tan tan

x y

x y

= tan 5.288 tan1.769

1 tan 5.288 tan1.769

= 0.9771

71. sin(x – y) = sin(42.08° – 68.37°) = –0.4429

sin x cos y – cos x sin y = sin 42.08° cos 68.37° – cos 42.08° sin 68.37° = –0.4429 tan(x + y) = tan(42.08° + 68.37°) = –2.682

tan tan

1 tan tan

x y

x y

= tan 42.08 tan 68.37

1 tan 42.08 tan 68.37

= –2.682

73. Evaluate each side for a particular set of values of x and y for which each side is defined. If the left

side is not equal to the right side, then the equation is not an identity. For example, for x = 0 and y = 0, both sides are defined, but are not equal.


75. Let y1 = sin6

x

Then y2 = sin x cos 6

+ cos x sin

6

= sin x ·3

2 + cos x ·

1

2 =

3

2sin x +

1

2cos x.

The graphs coincide as shown at the right.

77. Let y1 = cos3

4x

Then y2 = cos x cos 3

4

+ sin x sin

3

4

= cos x2

2

+ sin x2

2

= –2

2 cos x +

2

2 sin x


79. Let y1 = tan2

3x

Then y2 = 2323

tan tan

1 tan tan

x

x

=

tan ( 3)

1 tan ( 3)

x

x

= tan 3

1 3 tan

x

x


81. Let u = cos–1 4

5

, v = sin–1 3

5

Then we are asked to evaluate sin (u + v) which is sin u cos v + cos u sin v from the sum identity.

We know sin v = sin 1 3sin

5

= –3

5 and cos u = cos 1 4

cos5

= –4

5 from the function–

inverse function identities. It remains to find cos v and sin u. Note: 0 ≤ u ≤ π and –2

≤ v ≤

2

b = 2 25 ( 4) = 3 a = 2 25 ( 3) = 4

sin u = 3

5 cos v =

4

5

Then sin 1 14 3cos sin

5 5

= sin(u + v) = sin u cos v + cos u sin v = 3 4 4 3

5 5 5 5

= 12

25 +

12

25=

24

25

SECTION 7-2 305

83. We could proceed as in problem 81. Alternatively, we can shorten the process by recognizing

arccos1

2=

3

and arcsin(–1) = –

2

.

Then sin[arccos 1

2 + arcsin(–1)] = sin

3 2

= sin 3

cos

2

+ cos 3

sin

2

= 3

2(0) +

1

2

(–1) = –1

2.

85. Let u = sin–1 x, v = cos–1 y. Then x = sin u, –2

≤ u ≤

2

, y = cos v, 0 ≤ y ≤ π.

Then cos u = 21 sin u (in Quadrants I, IV) = 21 x

sin v = 21 cos v (in Quadrants I, II) = 21 y

Hence sin (sin–1 x + cos–1 y) = sin (u + v) = sin u cos v + cos u sin v = xy + 21 x 21 y

87. cos(x + y + z) = cos[(x + y) + z]

= cos(x + y)cos z – sin(x + y)sin z = (cos x cos y – sin x sin y)cos z – (sin x cos y + cos x sin y)sin z = cos x cos y cos z – sin x sin y cos z – sin x cos y sin z – cos x sin y sin z

89. tan(x – y) = sin( )

cos( )

x y

x y

91.

sin cos cos sinsin cos cos sin cos cos

cos cos sin sincos cos sin sincos cos

x y x yx y x y x y

x y x yx y x yx y

93.

sin cos cos sin sin sin

cos cos cos cos cos coscos cos sin sin sin sin

1cos cos cos cos cos cos

x y x y x y

x y x y x yx y x y x y

x y x y x y

95. tan(θ2 – θ 1) = 2 1

2 1

tan tan

1 tan tan

Difference Identity

= 2 1

2 11

m m

m m

Given

= 2 1

1 21

m m

m m

Algebra

97.

Note: In the text figure we have drawn EF perpendicular to CF, the track of the incident ray. ∆CDE and ∆CEF are right triangles. Angle DCF = β + γ = α. Hence, γ = α – β. Denote EC by x. Then,

in ∆CDE M

x = cos β = sin(90° – β)

in ∆CEF N

x = sin γ = sin(α – β)

Therefore x = sin(90 )

M

=

sin( )

N


M sin(α – β) = N sin(90° – β) Algebra M(sin α cos β – cos α sin β) = N(sin 90° cos β – cos 90° sin β) Difference Identities

M(sin α cos β – cos α sin β) = N(1 cos β – 0 sin β) Known Values M(sin α cos β – cos α sin β) = N cos β Algebra M sin α cos β – M cos α sin β = N cos β Algebra M sin α cos β – N cos β = M cos α sin β Algebra cos β (M sin α – N) = M cos α sin β Algebra

sin

cos

M N

M

= sin

cos

Algebra

tan β = sin

cos

M N

M

Quotient Identity

tan β = sin

cos

M

M

– cos

N

M Algebra

tan β = sin

cos

– 1

cos

N

M Algebra

tan β = tan α – N

M sec α Quotient and Reciprocal Identities

99. (A) From the text figure:

In right triangle ABE, we have (1) cot α = AB

AE =

AB

h

In right triangle BCD, we have (2) cot α = BC

CD =

BC

H

In right triangle EE'D, we have (3) tan β = '

'

E D

EE =

H h

AC

=

H h

AB BC

From (3), H – h = (AB + BC)tan β From (1) and (2), AB = h cot α and BC = H cot α Hence, substituting, we have (4) H – h = (h cot α + H cot α)tan β, or = (h + H)cot α tan β

Solving for H in terms of h yields:

H – h = h cot α tan β + H cot α tan β H – H cot α tan β = h cot α tan β + h H(1 – cot α tan β) = h(cot α tan β + 1)

H = h 1 cot tan

1 cot tan

(B) H = h sincos

sin cos

sincossin cos

1

1

Quotient Identities

H = h

sincossin cos

sincossin cos

1 sin cos

1 sin cos

Algebra

H = h sin cos cos sin

sin cos cos sin

Algebra

H = h sin( )

sin( )

Sum and Difference Identities

SECTION 7-2 307

(C) Substitute the given values to obtain

H = 4.90 sin(46.23 46.15 )

sin(46.23 46.15 )

= 4.90 sin(92.38 )

sin(0.08 )

= 3510 ft (to three significant digits)

Section 7-3

1. Substitute x for y in the equation for sin(x + y). 3. Substitute x for y in the equation for cos(x + y). 5. Solve the formulas cos 2x = 1 – 2sin2 x for sin x and cos 2x = 2 cos2 x – 1 for cos x, then replace x

with 1

2x.

7. cos 2(30°) = cos 60° = 1

2 cos2 30° – sin2 30° =

3

2

2

– 1

2

2

= 3

4 –

1

4 =

2

4 =

1

2

9. tan 23

= tan 2

3

= – 3

3 3

2

cot tan =

13

2

3 =

13

2 3

3 3 =

2 3

1 3 =

2 3

2 = – 3

11. sin 2

= 1

1 cos

2

=

1 ( 1)

2

=

2

2 = 1 = 1

13. sin 22.5° = sin1

452

= +1 cos 45

2

=

12

1

2

=

2 1

2 2

=

2( 2 1)

2 2

= 2 2

4

=

2 2

2

(We use + since 22.5° is a I quadrant angle)

15. cos 67.5° = cos1

1352

= +1 cos135

2

=

221

2

=

221

2

=

2 2

4

=

2 2

2

(We use + since 67.5° is a I quadrant angle)

17. tan8

= tan

1

2 4

=

111 cos 2 124

1 1sin4 2

or 2 1

19. cos5

12

= cos

1 5

2 6

=

35 11 cos 2 2 3 2 362 2 4 2

(We use + since 5

12

is a I quadrant angle)

21. (sin x + cos x)2 = sin2 x + 2 sin x cos x + cos2 x Algebra = 1 + 2 sin x cos x Pythagorean Identity = 1 + sin 2x Double–angle Identity

23. 1

2(1 – cos 2x) =

1

2 [1 – (1 – 2 sin2 x)] Double–angle Identity

= 1

2 [1 – 1 + 2 sin2 x)] Algebra

= 1

2 (2 sin2 x) Algebra

= sin2 x Algebra


25. tan x sin 2x = sin

cos

x

x2 sin x cos x Quotient and Double–angle Identities

= 2 sin2 x Algebra = 2 sin2 x – 1 + 1 Algebra = 1 – (1 – 2 sin2 x) Algebra = 1 – cos 2x Double–angle Identity

27. sin2 2

x =

1 cos

2

x

2

Half–angle Identity

= 1 cos

2

x

2

Algebra

= 1 cos

2

x Algebra

29. cot 2x = 1

tan 2x Reciprocal Identity

= 2

2tan1 tan

1xx

Double–angle Identity

= 21 tan

2 tan

x

x

Algebra

31. cot 2

= 2

2

cos

sin

Quotient Identity

= 1 cos

2

1 cos2

Half–angle Identities

= ±1 cos

1 cos

Algebra

cot2

=

1 cos

1 cos

Algebra

= 1 cos 1 cos

1 cos 1 cos

Algebra

= 2

2

1 cos

(1 cos )

Algebra

= 2

2

sin

(1 cos )


= sin

1 cos

Algebra

Since 1 – cos θ ≥ 0 and sin θ has the same sign as cot 2

, we may drop the absolute value signs to

obtain

cot 2

=

sin

1 cos

SECTION 7-3 309

[To show that sin θ has the same sign as cot2

, we note the following cases:

If 0 < θ < π, sin θ > 0 then 0 < 2

<

2

, cot

2

> 0.

If π = θ, cot 2

= cot

2

= 0,

sin

1 cos

= sin

1 cos

= 0

1 ( 1) = 0

If π < θ < 2π, sin θ < 0, then 2

<

2

< π, cot

2

< 0.

The truth of the statement for other values of θ follows since 2sin( 2 ) sin and cot cot .

2 2

kk

Key Algebraic Steps:

If y = ±1

1

a

a

, then |y | = 1

1

a

a

= (1 )(1 )

(1 )(1 )

a a

a a

= 2

2

1

(1 )

a

a

An alternative proof can be given that avoids dealing with the sign ambiguity:

cot 2

= 2

2

cos

sin

Quotient Identity

= 2sin cos

2 2

2sin sin2 2

Algebra

= 2

2sin cos2 2

2sin2

Algebra

= 2

sin 22

2sin2

Double-angle Identity

= 2

sin

1 1 2sin2

Algebra

= sin

1 cos 22

Double-angle Identity

= sin

1 cos

Algebra

33. cos 2u = cos2 u – sin2 u Double–angle Identity

= 2 2cos sin

1

u u Algebra

= 2 2

2 2

cos sin

cos sin

u u

u u


=

2 2

2 2

2 2

2 2

cos sincos cos

cos sincos cos

u uu u

u uu u

Algebra


=

2

2

2

2

sincos

sincos

1

1

uu

uu

Algebra

= 2

2

1 tan

1 tan

u

u

Quotient Identity

35. 2 csc 2x = 2 · 1

sin 2x Reciprocal Identity

= 2

sin 2x Algebra

= 2

2sin cosx x Double–angle Identity

= 1

sin cosx x Algebra

= 2 2cos sin

sin cos

x x

x x


=

2 2

2 2

2

cos sincos cos

sin coscos

x xx x

x xx

Algebra

=

2

2sincos

sincos

1 xx

xx

Algebra

= 21 tan

tan

x

x

Quotient Identity

37. 2

22

2

1 tan

1 tan

=

21 cos1 cos

21 cos1 cos

1

1

Half–angle Identity

= 1 cos1 cos1 cos1 cos

1

1

Algebra

=

1 cos1 cos

1 cos1 cos

(1 cos ) 1

(1 cos ) 1

Algebra

= 1 cos (1 cos )

1 cos 1 cos

Algebra

= 1 cos 1 cos

1 cos 1 cos

Algebra

= 2cos

2

Algebra

= cos α

Key Algebraic Steps:

211

211

1

1

xx

xx

=

1111

1

1

xxxx

=

11

11

(1 ) 1

(1 ) 1

xx

xx

x

x

=

1 (1 )

1 1

x x

x x

= 1 1

1 1

x x

x x

= 2

2

x = x

SECTION 7-3 311

39. Plug in x = 6

.

Left side: tan 26

= tan 3

= 3

Right side: 2 tan 6

=

2

3


41. Plug in x = 3

.

Left side: sin 3

2

= sin

6

=

1

2

Right side: 1

2 sin

3

=

3

4


43. Plug in x = 4

3

.

Left side: cos 4 3

2

= cos

2

3

= –

1

2

Right side: 1 cos 4 3

2

=

121

2

=

1

2


45. Plug in x = 4

3

.

Left side: tan 4 3

2

= tan

2

3

= – 3

Right side: 43

43

1 cos

1 cos

=

12

12

1

1

= 3


47. Plug in x = 2

.

Left side: sin 22

= sin π = 0

Right side: 2 sin2

= 2·1 = 2


49. The equation is not an identity. Plug in x = 4

:

sin 44

?

4 sin 4

cos

4

sin π ?

4 · 1

2 ·

1

2

0 ?

4 · 1

2

This is false. 51. Expand the left side using a double–angle formula:

cot 2x = 1

tan(2 )x cot x =

1

tan x

= 2

2tan1 tan

1xx

Double–angle formula

= 21 tan

2 tan

x

x

Algebra

Now simplify the right side:

2tan (cot 1)

2

x x =

21

tantan 1

2x

x cot2 x =

2

1

tan x

= 1

tan tan

2x x

Algebra

= tan

tan

x

x · 1

tan tan

2x x

Algebra

= 21 tan

2 tan

x

x

Algebra

Both sides can be simplified to the same expression, so the equation is an identity.


53. The equation is not an identity. Plug in x = π.

cos(2π) ?

1 – 2 cos2 π

1 ?

1 – 2(–1)2

1 ?

1 – 2 This is false.

55. a = – 2 25 3 = –4

cos x = –4

5 tan x = –

3

4

sin 2x = 2 sin x cos x = 23

5

4

5

= –24

25

cos 2x = 1 – 2 sin2 x = 1 – 23

5

2

= 1 – 18

25 =

7

25

tan 2x = sin 2

cos 2

x

x =

24

25

÷ 7

25

= –24

25 ·

25

7 = –

24

7

57. r = 2 212 ( 5) = 13

sin x = –5

13 cos x =

12

13

sin 2x = 2 sin x cos x = 25

13

12

13

= –120

169

cos 2x = 2 cos2 x – 1 = 212

13

2 – 1 =

288

169 – 1 =

119

169

tan 2x = sin 2

cos 2

x

x =

120

169

÷ 119

169

= –120

169 ·

169

119 = –

120

119

59. Since π < x < 3

2

,

2

<

2

x <

3

4

,

sin 2

x will be positive, cos

2

x, tan

2

x will be negative.

a = – 2 23 ( 1) = – 8

cos x = 8

3

3(a, -1)

-1

b

ax

sin 1

2x =

1 cos

2

x =

831

2

=

831

2

=

3 8

6

=

3 2 2

6

cos 1

2x = –

1 cos

2

x = –

831

2

= –

831

2

= –

3 8

6

= –

3 2 2

6

tan 1

2x = –

1 cos

1 cos

x

x

= –

83

83

1

1

= –

3 8

3 8

= –(3 8)(3 8)

(3 8)(3 8)

= – 2(3 8) = –(3 + 8 ) = –3 – 8 = –3 – 2 2

SECTION 7-3 313

61. Since –π < x < –2

, –

2

<

2

x < –

4

, cos

2

x will be positive, sin

2

x, tan

2

x will be negative.

r = 2 2( 3) ( 4) = 5 sin x = –4

5 cos x = –

3

5

sin1

2x = –

1 cos

2

x = –

351

2

= –

351

2

= –

85

2

= –4

5 = –

2 5

5

-4

b

a

(-3, -4)

r

-3x

cos 1

2x =

1 cos

2

x =

351

2

=

25

2 =

1

5 =

5

5

tan 1

2x =

1212

sin

cos

x

x =

2 5

5

÷

5

5 = –

2 5

5 ·

5

5 = –2

63. (A) 2θ is a second quadrant angle, since θ is a first quadrant angle

and tan 2θ is negative for 2θ in the second quadrant and not for 2θ in the first.

(B) Construct a reference triangle for 2θ in the second quadrant with (a, b) = (–3, 4). Use the Pythagorean theorem to find r = 5.

Thus, sin 2θ = 4

5 and cos 2θ = –

3

5.

(C) The double angle identities cos 2θ = 1 – 2 sin2 θ and cos 2θ = 2 cos2 θ – 1.

4

b

a

(-3, 4)

r

-3

2

(D) Use the identities in part (C) in the form

sin θ = 1 cos 2

2

and cos θ =

1 cos 2

2

The positive radicals are used because θ is in quadrant one.

(E) sin θ = 3

51

2

=

5 3

10

=

8

10 =

4

5 =

2

5 =

2 5

5

cos θ = 3

51

2

=

5 3

10

=

2

10 =

1

5 =

1

5 =

5

5

65. (A) tan[2(252.06°)] = –0.72335 (B) cos252.06

2

= –0.58821

2

2 tan

1 tan

x

x =

2

2 tan(252.06 )

1 tan (252.06 )

= –0.72335 –1 cos 252.06

2

= –0.58821

67. (A) tan[2(0.93457)] = –3.2518 (B) cos0.93457

2 = 0.89279

2

2 tan(0.93457)

1 tan (0.93457) = –3.2518

1 cos 0.93457

2

= 0.89279


69.

The graphs appear to coincide on the interval [–π, π].

71.

The graphs appear to coincide on the interval [–2π, 0].

73. cos 3x = cos(2x + x) Algebra = cos 2x cos x – sin 2x sin x Sum Identity = (2 cos2 x – 1)cos x – 2 sin x cos x sin x Double–angle Identities = 2 cos3 x – cos x – 2 sin2 x cos x Algebra = 2 cos3 x – cos x – 2(1 – cos2 x)cos x Pythagorean Identity = 2 cos3 x – cos x – 2 cos x + 2 cos3 x Algebra = 4 cos3 x – 3 cos x Algebra

75. cos 4x = cos 2(2x) Algebra

= 2 cos2 2x – 1 Double–angle Identity = 2(2 cos2 x – 1)2 – 1 Double–angle Identity = 2(4 cos4 x – 4 cos2 x + 1) – 1 Algebra = 8 cos4 x – 8 cos2 x + 2 – 1 Algebra = 8 cos4 x – 8 cos2 x + 1 Algebra

77. Let u = cos–1 3

5. Then cos u =

3

5, 0 < u < π.

cos 1 32cos

5

= cos 2u = 2 cos2 u – 1 = 23

5

2

– 1 = –7

25

79. Let u = cos–1 4

5

. Then cos u = –4

5, 0 < u < π.

b = 2 25 ( 4) = 3 tan u = –3

4

tan 1 42cos

5

= tan 2u = 2

2 tan

1 tan

u

u =

34

234

2

1

= 3

29

161

=

327

16

= 3

2

÷ 7

16= –

3

2 ·

16

7 = –

24

7

-4

b(-4, b)

b5

u

81. Let u = cos–1 3

5

. Then cos u = –3

5, 0 < u < π. Since 0 <

1

2u <

2

, cos

1

2u is positive.

So cos 11 3cos

2 5

= cos 1

2u =

1 cos

2

u =

351

2

=

25

2 =

1

5 =

5

5

SECTION 7-3 315

83. cos 2x = 1 – 2sin2 x Subtract 1 from both sides: cos 2x – 1 = – 2sin2 x Multiply both sides by –1: 1 – cos 2x = 2sin2 x Divide both sides by 2:

21 cos 2sin

2

xx

Finally, reverse the sides of the equation:

2 1 cos 2sin

2

xx

85. tan(2x) = tan (x + x) Algebra

= tan tan

1 tan tan

x x

x x

Sum Identity for tangent

= 2

2 tan

1 tan

x

x Algebra

87. tan(2x) = 2

2 tan

1 tan

x

x Problem 85

= 2

2 tan cot

(1 tan )cot

x x

x x Algebra

= 2

2 tan cot

cot tan cot

x x

x x x Algebra

= 2

12 tan

tan1

cot tantan

xx

x xx

Reciprocal Identity

= 2

cot tanx x Algebra

89. (A) Since tan2

x=

1 cos

1 cos

x

x

, tan

2

x and

1 cos

1 cos

x

x

are always equal other than possibly their sign.

(B) 2 21 cos sinx x ; Pythagorean identity

(C) 2sin sinx x because 2a a for any real number a; 2(1 cos ) 1 cosx x because

1 + cos x is never negative. (D) Since tan(x/2) and sin x always have the same sign, and since 1 + cos x is never negative,

tan(x/2) and sin x/(1+cosx) always have the same sign for any x.

91. From the figure we see: ACD is a right triangle, hence cos θ = 7

8, θ = cos–1

7

8

ABC is a right triangle, with angle BAC = 2θ. Hence cos 2θ = 7

x. Using the hint,

cos 2θ = 2 cos2 θ – 1= 27

8

2

– 1= 2 · 49

64 – 1=

34

64=

17

32

Thus 17

32 =

7

x

17x = 32(7)

x = 32(7)

17 =

224

17= 13.176 meters (To three decimal places)

θ = cos–1 7

8 = 28.955° (calculator in degree mode)

8 meters

A

B CD

x

7 meters


93. (A) Since 2 sin θ cos θ = sin 2θ by the double-angle identity, we can write

d = 20

2

2 sin cos

32 ft/sec

v =

20

2

(2sin cos )

32 ft/sec

v =

20

2

sin 2

32 ft/sec

v

(B) For fixed v0, the quantity 20 sin 2

32

v will be maximum when sin 2θ achieves

its maximum value, namely 1. Then sin 2θ = 1 2θ = 90° θ = 45°

95. (A) Here is a drawing of a typical situation. (The case n = 8 is shown,

but the reasoning is independent of n): We note: Area of polygon = n(Area of triangle OAB)

= 2n(Area of triangle OAC)

= 2n1

2OC AC

= n·OC·AC

O RA

B

C

In triangle OAC, angle AOC = 1

2θ =

1

2 ·

2

n

=

n

OC

R= cos AOC = cos

n

AC

R = sin AOC = sin

n

Thus, OC = R cos n

, AC = R sin

n

Hence, Area of polygon = n R cos n

R sin

n

= nR2 sin

n

cos

n

=

1

2nR2 · 2 sin

n

cos

n

= 1

2nR2 sin

2

n

by the double-angle identity for sine

(B) TABLE 1 n 10 100 1,000 10,000 An 2.93893 3.13953 3.14157 3.14159

(C) An appears to approach π, the area of the circle with radius 1. (D) An will not exactly equal the area of the circumscribing circle for any n no matter how large n is

chosen; however, An can be made as close to the area of the circumscribing circle as we like by making n sufficiently large.

Section 7–4 1. A product-sum identity expresses a product of two trigonometric functions as a sum of two other

trigonometric functions. 3. The identity for sine of a sum. 5. The double-angle identities for sine and cosine.

7. sin x cos y = 1

2 [sin(x + y) + sin(x – y)]

sin 3m cos m = 1

2 [sin(3m + m) + sin(3m – m)]

= 1

2 (sin 4m + sin 2m)

= 1

2sin 4m +

1

2sin 2m

9. sin x sin y = 1

2 [cos(x – y) – cos(x + y)]

sin u sin 3u = 1

2 [cos(u – 3u) – cos(u + 3u)]

= 1

2 (cos(–2u) – cos 4u)

= 1

2(cos 2u – cos 4u) =

1

2cos 2u –

1

2cos 4u

SECTION 7-4 317

11. sin x + sin y = 2 sin 2

x y cos

2

x y

sin 3t + sin t = 2 sin 3

2

t t cos

3

2

t t

= 2 sin 2t cos t

13. cos x – cos y = –2 sin 2

x y sin

2

x y

cos 5w – cos 9w = –2 sin 5 9

2

w w sin

5 9

2

w w

= –2 sin 7w sin(–2w) = 2 sin 7w sin 2w

15. sin x cos y = 1

2 [sin(x + y) + sin(x – y)]

sin 195° cos 75° = 1

2 [sin(195° + 75°) + sin(195° – 75°)]=

1

2 [sin 270° + sin 120°] =

1

2

31

2

= 1

2

3 2

2

= 3 2

4

17. cos x cos y = 1

2[cos(x + y) + cos(x – y)]

cos 157.5° cos 67.5° = 1

2 [cos(157.5° + 67.5°) + cos(157.5° – 67.5°)] =

1

2 [cos 225° + cos 90°]

= 1

2

10

2

= 1 2 2

42 2 2

19. cos x sin y = 1

2[sin(x + y) – sin(x – y)]

cos 37.5° sin 7.5° = 1

2[sin(37.5° + 7.5°) – sin(37.5° – 7.5°)] =

1

2 [sin 45° – sin 30°] =

1

2

2 1

2 2

2 1

4

21. sin x sin y = 1

2[cos(x – y) – cos(x + y)]

sin5

8

sin

8

=

1

2

5 5cos cos

8 8 8 8

= 1

2

3cos cos

2 4

=

1

2

20

2

2

4

23. cos x cos y = 1

2[cos(x – y) + cos(x + y)]

cos11

12

cos

12

=

1

2

11 11cos cos

12 12 12 12

= 1

2

5cos cos

6

=

1

2

31

2

3 2 3 21

2 2 4

25. sin x cos y = 1

2[sin(x + y) + sin(x – y)]

sin13

24

cos

5

24

=

1

2

13 5 13 5sin sin

24 24 24 24

= 1

2

3sin sin

4 3

=

1

2

2 3

2 2

2 3

4


27. sin x – sin y = 2 cos 2

x y sin

2

x y

sin 195° – sin 105° = 2 cos 195 105

2

sin

195 105

2

= 2 cos 150° sin 45° =

3 22

2 2 =

6

2

29. cos x – cos y = –2 sin 2

x y sin

2

x y

cos 75° – cos 15° = –2 sin 75 15

2

sin

75 15

2

= –2 sin 45° sin 30° =

2 12

2 2 =

2

2

31. cos x + cos y = 2 cos 2

x y cos

2

x y

cos17

12

+ cos

12

= 2 cos

17

12 122

cos

17

12 122

= 2 cos3

4

cos

2

3

=

2 12

2 2 =

2

2

33. sin x + sin y = 2 sin 2

x y cos

2

x y

sin12

+ sin

5

12

= 2 sin

5

12 122

cos

5

12 122

= 2 sin4

cos

6

= 2 3

22 2

= 6

2

35. cos(x + y) = cos x cos y – sin x sin y

cos(x – y) = cos x cos y + sin x sin y cos(x + y) + cos(x – y) = 2 cos x cos y (adding the above)

cos x cos y = 1

2[cos(x + y) + cos(x – y)]

37. Let x = u + v and y = u – v and solve for u and v in terms of x and y:

x = u + v y = u – v

x + y = 2u x – y = 2v

u = 2

x y v =

2

x y

Substitute these results into sin u sin v = 1

2 [cos(u – v) – cos(u + v)] to obtain:

sin 2

x y sin

2

x y =

1

2 [cos y – cos x] or

–sin 2

x y sin

2

x y =

1

2 [cos x – cos y] or

cos x – cos y = –2 sin 2

x y sin

2

x y

39. sin 2 sin 4

cos 2 cos 4

t t

t t

= 2 4 2 4

2 22 4 2 4

2 2

2sin cos

2sin sin

t t t t

t t t t

Sum–product Identities

= 2sin 3 cos( )

2sin 3 sin( )

t t

t t

Algebra

SECTION 7-4 319

= 2sin 3 cos

2sin 3 sin

t t

t t Identities for Negatives

= cos

sin

t

t Algebra

= cot t Quotient Identity

41. sin sin

cos cos

x y

x y

= 2 2

2 2

2cos sin

2sin sin

x y x y

x y x y


= – 2

2

cos

sin

x y

x y

Algebra

= –cot 2

x y Quotient Identity

43. cos cos

sin sin

x y

x y

= 2 2

2 2

2cos cos

2cos sin

x y x y

x y x y


= 2

2

cos

sin

x y

x y

Algebra

= cot 2


45. cos cos

cos cos

x y

x y

= 2 2

2 2

2cos cos

2sin sin

x y x y

x y x y


= – 2 2

2 2

cos cos

sin sin

x y x y

x y x y

Algebra

= –cot 2

x y cot

2


47. Let x = y = 0

Left side: sin 0 cos 0 = 0 Right side: sin 0 + cos 0 = 1 The equation is not an identity.

49. Let x = 0 and y = 2

.

Left side: sin 0 sin 2

= 0

Right side: sin 02

= 1


51. Let x = 3

and y =

3

.

Left side: cos 3

+ cos

3

= 1

Right side: cos 3

cos

3

=

1

4


53. Let x = 2

and y = 0.

Left side: sin 2

– sin 0 = 1

Right side: cos 2 0

2

sin 2 0

2

= cos

4

sin

4

= 1

2 ·

1

2 =

1

2




x y sin

2

x y Sum–product Identity

sin 3x – sin x = 2 cos 3

2

x x sin

3

2

x x Sum–product Identity

= 2 cos 4

2

x sin

2

2

x Algebra

= 2 cos 2x sin x Algebra The equation is an identity. 57. Try using a sum-product identity:

cos x – cos y = –2 sin 2

x y sin

2

x y

cos 3x – cos x = –2 sin 3

2

x x sin

3

2

x x

= –2 sin 2x sin x This is not an identity--the right side lacks the negative sign.


x y cos

2

x y Sum–product Identity

cos x + cos 5x = 2 cos 5

2

x x cos

5

2

x x Sum–product Identity

= 2 cos 6

2

x cos

4

2

x

Algebra

= 2 cos 3x cos(–2x) Algebra = 2 cos 3x cos 2x cos(–2x) = cos(2x)

The equation is an identity. 61. (A) cos 172.63° sin 20.177° = –0.34207

1

2[sin(172.63° + 20.177°) – sin(172.63° – 20.177°)] = –0.34207

(B) cos 172.63° + cos 20.177° = –0.05311

2 cos 172.63 20.177

2

cos

172.63 20.177

2

= –0.05311 (calculator in degree mode)

63. (A) cos 1.1255 sin 3.6014 = –0.19115

1

2 [sin(1.1255 + 3.6014) – sin(1.1255 – 3.6014)] = –0.19115

(B) cos 1.1255 + cos 3.6014 = –0.46541

2 cos 1.1255 3.6014

2

cos

1.1255 3.6014

2

= –0.46541 (calculator in radian mode)

65. sin 2x + sin x = 2 sin 2

2

x x cos

2

2

x x = 2 sin

3

2

x cos

2

x

Graph y1 = sin 2x + sin x and y2 = 2 sin 3

2

x cos

2

x

SECTION 7-4 321

67. cos 1.7x – cos 0.3x = –2 sin 1.7 0.3

2

x x sin

1.7 0.3

2

x x

= –2 sin x sin 0.7x

Graph y1 = cos 1.7x – cos 0.3x and y2 = –2 sin x sin 0.7x

69. sin 3x cos x = 1

2[sin(3x + x) + sin(3x – x)]

= 1

2 (sin 4x + sin 2x)

Graph y1 = sin 3x cos x and y2 = 1

2 (sin 4x + sin 2x)

71. sin 2.3x sin 0.7x = 1

2 [cos(2.3x – 0.7x) – cos(2.3x + 0.7x)]

= 1

2 (cos 1.6x – cos 3x)

Graph y1 = sin 2.3x sin 0.7x and y2 = 1

2 (cos 1.6x – cos 3x)

73. cos x cos y cos z = cos x 1

2 [cos (y + z) + cos(y – z)] Product-sum Identity

= 1

2 cos x cos(y + z) +

1

2 cos x cos(y – z) Algebra

= 1

2

1[cos( ) cos( { })]

2x y z x y z

+ 1

2

Product-sum1[cos( ) cos( { })]

Identity2x y z x y z

= 1

4 cos(x + y + z) +

1

4 cos(x – y – z) +

1

4 cos(x + y – z) +

1

4 cos(x – y + z) Algebra

= 1

4 [cos(x + y – z) + cos(x – y – z) + cos(z + x – y) + cos(x + y + z)] Algebra

= 1

4 [cos(x + y – z) + cos{–(y + z – x)} + cos (z + x – y) + cos(x + y + z)] Algebra

= 1

4 [cos(x + y – z) + cos(y + z – x) + cos(z + x – y) + cos(x + y + z)] Identity for Negatives

75. (A)

(B) 2 cos(28πx) cos(2πx) = 2 · 1

2[cos(28πx + 2πx) + cos(28πx – 2πx)]

= cos 30πx + cos 26πx Graphing y1 = cos 30πx + cos 26πx will yield the same result as before.

77. (A)

(B) 2 sin(20πx) cos(2πx) = 2 · 1

2[sin(20πx + 2πx) + sin(20πx – 2πx)]

= sin 22πx + sin 18πx Graphing y1 = sin 22πx + sin 18πx will yield the same result as before.


79. (A) cos x – cos y = –2 sin 2

x y sin

2

x y

In this case

cos 128πt – cos 144πt = –2 sin 128 144

2

t t sin

128 144

2

t t

cos 128πt – cos 144πt = –2 sin 136πt sin(–8πt) cos 128πt – cos 144πt = 2 sin 136πt sin 8πt

Multiplying both sides by 0.5, we have 0.5 cos 128πt – 0.5 cos 144πt = sin 136πt sin 8πt (B) y = 0.5 cos 128πt y = –0.5 cos 144πt

0 0.25

1

-1

0 0.25

1

-1 y = 0.5 cos 128πt – 0.5 cos 144πt y = sin 8πt sin 136πt

0 0.25

1

-1

0 0.25

1

-1 The latter two graphs are the same, illustrating the identity proved in part (A). Section 7–5 1. An identity is an equation that is true for all replacements of the variables for which both sides are

defined. A conditional equation is true for some replacements, but false for others. To solve a conditional equation is to find all values of the variable(s) for which it is true.

3. Add all integer multiples of 2π to the solutions already found.

5. sin x = 3

2

Sketch a graph of y = sin x and y = 3

2 on the

Interval [0, 2π). x = 3

,

2

3

Or use a unit circle diagram.

1 2 3 4 5 6

��3

��2

SECTION 7-5 323

7. We have found all solutions of sin x =3

2 over one period in problem 5. Since the sine function is

periodic with period 2π, all solutions are given by

2

3 any integer2

23

x kk

x k

9. sin x + 1 = 0

sin x = –1 Sketch a graph of y = sin x and y = –1 on the interval [0, 2π).

x = 3

2

Or use a unit circle diagram (not shown).

1 2 3 4 5 6

-1

-0.5

0.5

1

11. We have found all solutions of sin x + 1 = 0 over one period in problem 9. Since the sine function is periodic with period 2π, all solutions are given by

x = 3

2

+ 2kπ, k any integer

13. tan θ – 1 = 0

tan θ = 1 Sketch a graph of y = tan θ and y = 1 on the interval [0, 360˚) θ = 45˚, 225˚ 90 180 270 360

-4

-2

2

4

15. We have found all solutions of tan θ – 1 = 0 over two periods in problem 13. Since the tangent

function is periodic with period 180˚, all solutions are given by θ = 45˚ + k180˚ k any integer

17. 2 sin x – 1 = 0

sin x = 1

2

Sketch a graph of y = sin x and y = 1

2 on the interval [0, 2π)

x = 6

,

5

6

1 2 3 4 5 6

-1

-0.5

0.5

1

19. We have found all solutions of 2 sin x – 1 = 0 over one period in problem 17. Since the sine

function is periodic with period 2π, all solutions are given by

656

2

2

x k

x k

k any integer

21. 2 sin θ + 3 = 0

sin θ = –3

2

Sketch a graph of y = sin θ and y = –3

2 on the interval [0, 360˚).

θ = 240˚, 300˚

90 180 270 360

��3

��2


23. We have found all solutions of 2 sin θ + 3 = 0 over one period in problem 21. Since the sine function is periodic with period 360˚, all solutions are given by

240 360

300 360

k

k

k any integer

25. tan x – 3 = 0

tan x = 3

Sketch a graph of y = tan x and y = 3 on the interval [0, 2π).

x = 3

,

4

3

1 2 3 4 5 6

��3

27. 2 cos x + 4 = 0

2 cos x = –4 cos x = –2

No solution. cos x has values only between –1 and 1.

29. 7 cos x – 3 = 0 0 ≤ x < 2π

cos x = 3

7

Sketch a graph of y = cos x and y = 3

7, x on the interval [0, 2π).

x

y

1

-1š2

3š2

2šš

37

x =

1

1

3cos 1.1279 First quadrant solution

73

2 cos 5.1553 Fourth quadrant solution7

31. 2 tan θ – 7 = 0 0° ≤ θ < 180°

tan θ = 7

2

Sketch a graph of y = tan θ and y = 7

2on

the interval [0°, 180°)

θ = tan–1 7

2 = 74.0546°

y

x0 180°

72

33. 10 cos 2x + 6 = 0

10 cos 2x = –6

cos 2x = 6

10

Since 6

10

≈ –1.9, there is no solution. cos t has values only between –1 and 1.

35. 5 sin x – 2 = 0

5 sin x = 2

sin x = 2

5

1

1

2sin 1.1071

52

sin 2.03445

x

are the solutions over one period 0 ≤ x < 2π

If x can range over all real numbers, 1.1071 2

2.0344 2

kx

k

k any integer

1 2 3 4 5 6

SECTION 7-5 325

37. Here is a graph of y1 = 1 – x and y2 = 2 sin x, on the interval 0 ≤ x ≤ 2π.

To four decimal places, the only solution is 0.3376. Check: 1 – x = 1 – 0.3376 = 0.6624 2 sin x = 2 sin(0.3376) = 0.6624

39. Here is a graph of y1 = tan 2

x and y2 = 8 – x, on

the interval 0 ≤ x ≤ π.

To four decimal places, the only solution is 2.7642.

Check: tan 2

x = tan

2.7642

2 = 5.2358

8 – x = 8 – 2.7642 = 5.2358

41. 2 sin2 θ + sin 2θ = 0 all θ 2 sin2 θ + 2 sin θ cos θ = 0 2 sin θ (sin θ + cos θ) = 0 Either 2 sin θ = 0 sin θ = 0 Solutions over 0° ≤ θ < 360° are θ = 0°, 180°

Thus, solutions, if θ is allowed to range over all possible values, are θ = 0° + k 180° or θ = k 180° k any integer. Or sin θ + cos θ = 0 sin θ = –cos θ

sin

cos

= –1

tan θ = –1 Solutions over 0° < θ < 180° are θ = 135° Thus, solutions, if θ is allowed to range over all possible values, are θ = 135° + k 180° k any integer Solutions: k 180°, 135° + k 180° k any integer

43. tan x = –2 sin x 0 ≤ x < 2π

sin

cos

x

x = –2 sin x

sin x = –2 sin x cos x cos x ≠ 0 2 sin x cos x + sin x = 0 sin x(2 cos x + 1) = 0 sin x = 0 2 cos x + 1 = 0 x = 0, π 2 cos x = –1

cos x = –1

2

x = 2

3

,

4

3

Solutions: 0,

2

3

, π,

4

3


45. sec 2

x + 2 = 0 0 ≤ x ≤ 2π is equivalent to

sec 2

x + 2 = 0 0 ≤

2

x ≤ π

sec 2

x = –2

cos 2

x = –

1

2

2

x =

2

3

x = 4

3

47. 2 cos2 θ + 3 sin θ = 0 0° ≤ θ < 360° 2(1 – sin2 θ) + 3 sin θ = 0 2 – 2 sin2 θ + 3 sin θ = 0 –2 sin2 θ + 3 sin θ + 2 = 0 2 sin2 θ – 3 sin θ – 2 = 0 (2 sin θ + 1)(sin θ – 2) = 0 2 sin θ + 1 = 0 sin θ – 2 = 0 2 sin θ = –1 sin θ = 2

sin θ = –1

2 No solution

θ = 210°, 330° Solutions: θ = 210°, 330°

49. cos 2θ + cos θ = 0 0° ≤ θ < 360° 2 cos2 θ – 1 + cos θ = 0 2 cos2 θ + cos θ – 1 = 0 (2 cos θ – 1)(cos θ + 1) = 0 2 cos θ – 1 = 0 cos θ + 1 = 0 2 cos θ = 1 cos θ = –1 θ = 180°

cos θ = 1

2

θ = 60°, 300° Solutions: θ = 60°, 180°, 300°

SECTION 7-5 327

51. 2 sin2 2

x – 3 sin

2

x + 1 = 0 0 ≤ x ≤ 2π is equivalent to

2 sin2 2

x – 3 sin

2

x + 1 = 0 0 ≤

2

x ≤ π

(2 sin 2

x – 1)(sin

2

x – 1) = 0

2 sin 2

x – 1 = 0

2 sin sin 2 x x

Common Error: sin

2

x – 1 = 0

2 sin 2

x = 1 sin

2

x = 1

sin 2

x =

1

2

2

x =

2

2

x =

6

,

5

6

x = π

x = 3

,

5

3

Solutions: x =

3

,

5

3

, π

53. Since cos2 x + sin2 x = 1 is an identity, the solutions of this equation are all x, 0 ≤ x < 2π. 55. Since |2 sin θ| ≤ 2 and cos θ ≥ –1 for all θ, the left side of this equation cannot be greater than 2 and

the right side cannot be less than 4. The equation has no solutions. 57. 6 sin2 θ + 5 sin θ = 6 0° ≤ θ ≤ 90°

6 sin2 θ + 5 sin θ – 6 = 0 (3 sin θ – 2)(2 sin θ + 3) = 0

3 sin θ– 2 = 0 0° ≤ θ ≤ 90° 2 sin θ + 3 = 0 0° ≤ θ ≤ 90° 3 sin θ = 2 2 sin θ = –3

sin θ = 2

3 sin θ = –

3

2

θ = sin–1 2

3 (calculator in

degree mode) No solution

θ = 41.81° Solution: θ = 41.81°

59. 3 cos2 x – 8 cos x = 3 0 ≤ x ≤ π 3 cos2 x – 8 cos x – 3 = 0

(3 cos x + 1)(cos x – 3) = 0 3 cos x + 1 = 0 0 ≤ x ≤ π cos x – 3 = 0 0 ≤ x ≤ π 3 cos x = –1 cos x = 3

cos x = –1

3 No solution

x = cos–1 1

3

(calculator in radian mode)

x = 1.911 Solution: x = 1.911


61. 2 sin x = cos 2x 0 ≤ x < 2π 2 sin x = 1 – 2 sin2 x 2 sin2 x + 2 sin x – 1 = 0 Quadratic in sin x

sin x = 2 4

2

b b ac

a

a = 2, b = 2, c = –1

sin x = 22 2 4(2)( 1)

2(2)

=

2 12

4

sin x = 0.3660 sin x = –1.366 No solution

x = 1

1

sin 0.3660 0.3747

2.767sin 0.3660

Solutions: x = 0.3747, 2.767

63. 2 sin2 x = 1 – 2 sin x 2 sin2 x + 2 sin x – 1 = 0 x any real number See Problem 61.

x = 1

1

sin 0.3660 0.3747

2.767sin 0.3660

are the solutions over one period 0 ≤ x < 2π

If x can range over all real numbers, x = 0.3747 2

2.767 2

k

k

k any integer

65. Examining the graphs of y1 = 2 sin x and y2 = cos 2x, 0 ≤ x < 2π, we obtain

To four decimal places, the solutions are 0.3747 and 2.7669.

Check: 0.3747: 2 sin x = 0.7321 2.7669: 2 sin x = 2 sin 2.7669 = 0.7321 cos 2x = 0.7321 cos 2x = cos 2.7669 = 0.7321

In later problems, the checking is left to the student.

67. Examining the graphs of y1 = 2 sin2 x and y2 = 1 – 2 sin x, 0 ≤ x ≤ 2π, we obtain the graphs at the right.

To four decimal places, the solutions are 0.3747 and 2.7669. Thus, the solutions over all real x are given by

x = 0.3747 2

2.7669 2

k

k

k any integer

SECTION 7-5 329

69. Examining the graphs of y1 = cos 2x and y2 = x2 – 2, we obtain the graphs at the right. The graphs intersect at x = –1.1530 and x = 1.1530. The cosine graph is above the parabola, and cos 2x > x2 – 2,

on the interval (–1.1530, 1.1530).

71. Examining the graphs of y1 = cos(2x + 1) and y2 = 0.5x – 2, we obtain

There is clearly one point of intersection at x = 3.5424. Examining the graphs again on a smaller interval, we obtain

Combining information from all three graphs, the graph of the cosine function is below or intersects

the straight line, cos(2x + 1) ≤ 0.5x – 2, on the two intervals [3.5424, 5.3778] and [5.9227, ∞).

73. 2 cos x – 3 < 0 2 cos x < 3

cos x < 3

2

Since |cos x | ≤ 1, the statement cos x < 3

2 is true for all

replacements of the variable x and the solution is all real x.

75. tan–1(–5.377) has exactly one value, –1.387; the equation tan x = –5.377 has infinitely many solutions, which are found by adding kπ, k any integer, to each solution in one period of tan x.

77. cos x – sin x = 1 0 ≤ x < 2π

± 21 sin x – sin x = 1

± 21 sin x = 1 + sin x 1 – sin2 x = (1 + sin x)2 Squaring both sides 1 – sin2 x = 1 + 2 sin x + sin2 x Common Error: (1 + sin x)2 ≠ 1 + sin2 x 0 = 2 sin x + 2 sin2 x 0 = 2 sin x(1 + sin x) 2 sin x = 0 1 + sin x = 0 sin x = 0 sin x = –1

x = 0, π x = 3

2

In squaring both sides we may have introduced extraneous solutions; hence, it is necessary to check solutions of these equations in the original equation.


x = 0 x = π x = 3

2

cos x – sin x = 1 cos x – sin x = 1 cos x – sin x = 1

cos 0 – sin 0 ?= 1 cos π – sin π

?= 1 cos

3

2

– sin

3

2

?= 1

1 – 0 √= 1 –1 – 0 ≠ 1 0 – (–1)

√= 1

A solution Not a solution A solution

Solutions: x = 0, 3

2

79. tan x – sec x = 1 0 ≤ x < 2π

± 2sec 1x – sec x = 1

± 2sec 1x = 1 + sec x sec2 x – 1 = (1 + sec x)2 Squaring both sides sec2 x – 1 = 1 + 2 sec x + sec2 x 0 = 2 + 2 sec x –2 sec x = 2 sec x = –1 x = π

In squaring both sides we may have introduced extraneous solutions; hence, it is necessary to check this apparent solution in the original equation.

tan x – sec x = 1

tan π – sec π ?= 1

0 – (–1) √= 1

A solution Solution: x = π

81. (A) Here are graphs of f(x) = sin 1

x on [0.1, 0.5] and [0.2, 5].

–1.5

0.1

1.5

0.5

–1.5

0.2

1.5

5

The largest zero for f is 0.3183. As x increases without bound, 1

x tends to 0 through positive

numbers, and sin 1

x tends to 0 through positive numbers. y = 0 is a horizontal asymptote for the

graph of f.

SECTION 7-5 331

(B) Here is a computer–generated graph of

f(x) = sin 1

x on [0.015, 0.1].

The student may provide other exploratory graphs. Infinitely many zeros exist between 0 and b, for any b, however small. The exploration graphs suggest this conclusion, which is reinforced by the following reasoning: Note that for each interval (0, b], however small, as x tends to zero through

positive numbers, 1

x increases without bound, and

as 1

x increases without bound, sin

1

x will cross the

x axis an unlimited number of times. The function f does not have a smallest zero, because, between 0 and b, no matter how small b is,

there is always an unlimited number of zeros.

83. The starting point is represented by y = 0, so we solve –8 cos 2t = 0 cos 2t = 0

2t = 3 5 7

, , ,2 2 2 2

(first four solutions)

t =3 5 7

, , ,4 4 4 4

t ≈ 0.785 sec, 2.36 sec, 3.93 sec, 5.50 sec

85. I = 30 sin 120πt I = –10 –10 = 30 sin 120πt

sin 120πt = –10

30

120πt = π + sin–1 10

30 (third quadrant) will yield

the least positive solution of the equation.

t = 11 10sin

120 30

= 0.009235 sec (calculator in radian mode)

87. We want the least positive θ so that I cos2 θ is 40% of I, that is I cos2 θ = 0.40I 0° ≤ θ ≤ 180° cos2 θ = 0.40 cos θ = ± 0.40

θ = cos–1 0.40 will yield the least positive solution of the equation θ = 50.77° (calculator in degree mode)

89. We are to solve 3.09 × 107 = 73.44 10

1 0.206cos

For convenience, we can divide both sides of this equation by 107

3.09 = 3.44

1 0.206cos

3.09(1 – 0.206 cos θ) = 3.44 3.09 – (3.09)(0.206)cos θ = 3.44 –(3.09)(0.206)cos θ = 3.44 – 3.09

cos θ = 3.44 3.09

(3.09)(0.206)

cos θ = –0.5498 θ = 180° – cos–1(0.5498) (second quadrant) will yield the least positive solution of this equation.

or θ = cos–1(–0.5498) = 123°


91. Use A = 1

2R2(θ – sin θ) with R = 8 and A = 48.

48 = 1

2 · 82(θ – sin θ)

48 = 32(θ – sin θ) 1.5 = θ – sin θ Examining the graph of y1 = 1.5 and y2 = θ – sin θ, 0 ≤ θ ≤ π, we obtain to three decimal places, θ = 2.267 radians.

93. Using the figure in the text, we can write:

sin θ = a

R (1)

a2 + (R – b)2 = R2 (2) applying the Pythagorean theorem to the right triangle with sides R, a, and R – b. L = R · 2θ (3)

Thus, given a and b, we can solve equation (2) for R: a2 + R2 – 2Rb + b2 = R2 –2Rb = –a2 – b2

R = 2 2

2

a b

b

We can solve equation (1) for θ to obtain θ = sin–1 a

R (since θ is an acute angle)

Hence, from equation (3),

L = R · 2θ = 2 2

2

a b

b

· 2 sin–1

2 2

2a b

b

a

L = 2 2a b

b

sin–1

2 2

2ab

a b (4)

(A) Substituting the given values a = 5.5 and b = 2.5,

L = 2 2(5.5) (2.5)

2.5

sin–1

2 2

2(5.5)(2.5)

(5.5) (2.5)= 12.4575 mm.

(B) Substituting L = 12.4575 and a = 5.4 mm yields

12.4575 = 2 2(5.4) b

b

sin–1

2 2

2(5.4)

(5.4)

b

b

12.4575 = 229.16 b

b

sin–1

2

10.8

29.16

b

b (5)

Examining the graph of y1 = 12.4575

and y2 = 229.16 x

x

sin–1

2

10.8

29.16

x

x,

2 ≤ x ≤ 3, we obtain the graph at the right. The solution is b = 2.6496 mm.

10

2

13

3

SECTION 7-5 333

95. r = 2 sin θ 0° ≤ θ ≤ 360° r = sin 2θ We solve this system of equations by equating the right sides: 2 sin θ = sin 2θ 2 sin θ = 2 sin θ cos θ 0 = 2 sin θ cos θ – 2 sin θ 0 = 2 sin θ(cos θ – 1) 2 sin θ = 0 cos θ – 1 = 0 sin θ = 0 cos θ = 1 θ = 0°, 180°, 360° θ = 0°, 360° If we substitute these values of θ in either of the original equations, we obtain r = 2 sin 0° r = 2 sin 180° r = 2 sin 360° r = 0 r = 0 r = 0 Thus, the solutions of the system of equations are

(r, θ) = (0, 0°), (r, θ) = (0, 180°), and (r, θ) = (0, 360°) 97. We are to solve 2.818 sin(0.5108x – 1.605) + 12.14 = 12

Examining the graph of y1 = 2.818sin(0.5108x – 1.605) + 12.14 and y2 = 12, we obtain

Using x = 3 for March 15, we can interpret x = 3.0448295 as 0.0448295(30) = 1.34 day later, March 16 or 17. Using x = 9 for September 15, we can interpret x = 9.3897685 as 0.3897685(30) = 11.69 days later,

September 26 or 27. CHAPTER 7 REVIEW

1. tan x + cot x = sin

cos

x

x +

cos

sin

x


= 2 2sin cos

cos sin

x x

x x

Algebra

= 1

cos sinx x Pythagorean Identity

= 1 1

cos sinx x Algebra

= sec x csc x Reciprocal Identity (7–1) 2. sec4 x – 2 sec2 x tan2 x + tan4 x = (sec2 x – tan2 x)2 Algebra = (1 + tan2 x – tan2 x)2 Pythagorean Identity = 12 Algebra = 1 (7–1)


3. 1

1 sin x +

1

1 sin x =

1 sin 1 sin

(1 sin )(1 sin )

x x

x x

Algebra

= 2

2

1 sin x Algebra

= 2 2 2

2

sin cos sinx x x Pythagorean Identity

= 2

2

cos x Algebra

= 2 sec2 x Reciprocal Identity (7–1)

4. cos3

2x

= cos x cos 3

2

+ sin x sin

3

2

Difference Identity

= cos x(0) + sin x(–1) Known Values = –sin x Algebra (7–2)

5. Plug in x = π, y = 1

2

Left side: sin1

2

= sin2

= 1

Right side: sin π sin1

2= 0 sin

1

2 = 0

The equation is not an identity. (7–1)

6. Plug in x = 0 Left side: cos 4 0 = cos 0 = 1

Right side: 4 cos 0 = 4·1 = 4 The equation is not an identity. (7–1)

7. Plug in x = 3π

Left side: sin3

2

= –1

Right side:1 cos3 1 ( 1)

1 12 2

The equation is not an identity. (7–1, 7–3)

8. sin x cos y = 1

2[sin(x + y) + sin(x – y)]

sin 5α cos 3 α = 1

2 [sin(5α + 3α) + sin(5α – 3α)]

= 1

2 [sin 8α + sin 2α]

= 1

2 sin 8α +

1

2 sin 2α (7–4)

9. cos x – cos y = –2 sin 2

x y sin

2

x y

cos 7x – cos 5x = –2 sin 7 5

2

x x sin

7 5

2

x x = –2 sin 6x sin x (7–4)

10. sin9

2x

= sin x cos 9

2

+ cos x sin

9

2

Sum Identity

= sin x cos 2

+ cos x sin

2

Periodicity of sine and cosine

= sin x(0) + cos x(1) Known Values = cos x (7–2)

CHAPTER 7 REVIEW 335

11. 2 cos θ + 1 = 0 all θ Solve over one period [0°, 360°):

cos θ = –1

2

Sketch a graph of y = cos θ and y = –1

2, θ in [0°, 360°).

θ = 135°, 225°

Or use a unit circle graph.

Since the cosine function is periodic with period 360°, all solutions are given by θ = 135° + k360°, θ = 225° + k360°, k any integer. (7–5) 12. sin x tan x – sin x = 0 all x

Solve over one period, [0, 2π) for sin x, [0, π) for tan x: sin x(tan x – 1) = 0 sin x = 0 tan x – 1 = 0 x = 0, π tan x = 1

x = 4

Or use unit circle graphs (not shown).

Thus, solutions if x is allowed to range over all values, are

x = 0 + 2kπ, π + 2kπ, k any integer, and x = 4

+ kπ, k any integer.

These solutions can also be written as x = kπ or x = 4

+ kπ, k any

integer.

(7–5)

13. sin x = 0.7088 all real x

Solve over one period [0, 2π): Sketch a graph of y = sin x and y = 0.7088, x in [0, 2π).

x = 1

1

sin 0.7088 0.7878 First quadrant solution

sin 0.7088 2.3538 Second quadrant solution

(calculator in radian mode) Since the sine function is periodic with period 2π, all solutions are given by

x = 0.7878 2

2.3538 2

k

k

k any integer (7–5)


14. cos θ = 0.2557 all θ Solve over one period, [0°, 360°):

Sketch a graph of y = cos θ and y = 0.2557, θ in [0°, 360°).

x = 1

1

cos 0.2557 75.1849 First quadrant solution

360 cos 0.2557 284.8151 Second quadrant solution

(calculator in degree mode)

Since the cosine function is periodic with period 360°, all solutions are given by

x = 75.1849° + k360°284.8151° + k360°


15. cot x = –0.1692 –2

< x <

2

Sketch a graph of y = cot x and y = –0.1692, x in ,2 2

x = –cot–1 (0.1692) = –1.4032 Fourth quadrant solution, (calculator in radian mode)

(7–5)

16. 3 tan(11 – 3x) = 23.46 –2

< x <

2

tan(11 – 3x) = 23.46

3

11 – 3x = tan–1 23.46

3

–3x = tan–1 23.46

3

– 11

x = 23.46

3

1tan 11

3

x = 3.1855 (7–5)

17. (A) Graph both sides of the equation in the same viewing window. (sin x + cos x)2 = 1 – 2 sin x cos x is not an identity, since the graphs

do not match. Try x = 4

.

Left side: sin cos4 4

2

= 1 1

2 2

2

= 2

2

2

= 2

Right side: 1 – 2 sin 4

cos

4

= 1 – 2

1 1

2 2

= 0

This verifies that the equation is not an identity.

(B) Graph both sides of the equation in the same viewing window. cos2 x – sin2 x = 1 – 2 sin2 x appears to be an identity, which we now verify: cos2 x – sin2 x = 1 – sin2 x – sin2 x Pythagorean Identity = 1 – 2 sin2 x Algebra (7–1)

18. 2

2

1 2cos 3cos

sin

x x

x

=

2

2

1 2cos 3cos

1 cos

x x

x


= (1 3cos )(1 cos )

(1 cos )(1 cos )

x x

x x

Algebra

= 1 3cos

1 cos

x

x

Algebra (7–1)


19. (1 – cos x)(csc x + cot x) = (1 – cos x)1 cos

sin sin

x

x x

Reciprocal andQuotient Identities

= (1 – cos x)(1 cos )

sin

x

x

Algebra

= (1 cos )(1 cos )

sin

x x

x

Algebra

= 21 cos

sin

x

x

Algebra

= 2sin

sin

x

x Pythagorean Identity

= sin x Algebra

Key Algebraic Steps: (1 – a)1 a

b b

= (1 – a)1 a

b

= (1 )(1 )a a

b

=

21 a

b

(7–1)

20. 1 sin

cos

x

x

=

(1 sin )(1 sin )

cos (1 sin )

x x

x x

Algebra

= 21 sin

cos (1 sin )

x

x x

Algebra

= 2cos

cos (1 sin )

x

x x Pythagorean Identity

= cos

1 sin

x

x Algebra (7–1)

21. 2

2

1 tan

1 tan

x

x

= 2

2

1 tan

sec

x

x


=

2

2

2

sincos1

cos

1 xx

x


=

2

2

2

2 sincos

2 1cos

cos 1

cos

xx

x

x

x

Algebra

= 2 2cos sin

1

x x Algebra

= cos2 x – sin2 x Algebra = cos 2x Double–angle Identity (7–3)

22. cot x – tan x = cos

sin

x

x –

sin

cos

x


= 2 2cos sin

sin cos

x x

x x

Algebra

= 2 22(cos sin )

2sin cos

x x

x x

Algebra

= 2 22[cos (1 cos )]

2sin cos

x x

x x



= 22[2cos 1]

2sin cos

x

x x

Algebra

= 24cos 2

2sin cos

x

x x

Algebra

= 24cos 2

sin 2

x

x

Double–angle Identity (7–3)

23. 1 cot

csc

x

x

2

= 2

2

(1 cot )

csc

x

x

Algebra

= 2

2

1 2cot cot

csc

x x

x

Algebra

= 2

2

1 cot 2cot

csc

x x

x

Algebra

= 2

2

csc 2cot

csc

x x

x


= 2

2

csc

csc

x

x –

2

2cot

csc

x

x Algebra

= 1 – 2

2cot

csc

x

x Algebra

= 1 – 2

cossin1

sin

2 xx

x


= 1 – 2 sin x cos x Algebra = 1 – sin 2x (7–3)

24. tan m + tan n = sin

cos

m

m +

sin

cos

n

n Quotient Identity

= sin cos cos sin

cos cos

m n m n

m n

Algebra

= sin( )

cos cos

m n

m n

Sum Identity (7–2)

25. tan(x + y) = tan tan

1 tan tan

x y

x y

Sum Identity

= 1 1

cot cot

1 1cot cot1x y

x y

Reciprocal Identity

= cot cot

cot cot 1

y x

x y

Algebra

= cot cot

cot cot 1

x y

x y

Algebra (7–2)


26. tan 75° = tan(45° + 30°) = tan 45 tan 30

1 tan 45 tan 30

=

11

31

1 13

=

3 1

3

3 1

3

=

3 1

3 1

(7–2)

27. cos12

= cos

3 4

= cos3

cos

4

+ sin

3

sin

4

=

1 2 3 2

2 2 2 2 =

2 6

4 4 =

2 6

4

(7–2)

28. sin 105° = sin210

2

=1 cos 210

2

=

31

2

2

=

2 322

=2 3

4

=

2 3

2

(7–3)

(We choose + because 105° is in Quadrant II)

29. cos7

8

= cos

7

42

=

71 cos

42

=

21

22

=

2 2

22

=2 2

4

=

2 2

2

(7–3)

(We choose – because 7

8

is in Quadrant II)

30. cos x sin y = 1

2[sin(x + y) – sin(x – y)]

cos 195° sin 75° = 1

2[sin(195° + 75°) – sin(195° – 75°)] =

1

2[sin 270° – sin 120°]

= 1

2

31

2

= 1

2

2 3

2

= 2 3

4

(7–4)


x y cos

2

x y

cos 195° + cos 105° = 2 cos 195 105

2

cos

195 105

2

= 2 cos 150° cos 45°

= 23 2

2 2

= –6

2 (7–4, 6–2)

32. sin x sin y = 1

2[cos(x – y) – cos(x + y)]

sin11

24

sin

5

24

=

1

2

11 5 11 5cos cos

24 24 24 24

= 1

2

2cos cos

4 3

= 1

2

2 1

2 2

= 1

2

2 1

2

= 2 1

4

(7–4)



x y sin

2

x y

sin 5

12

– sin

12

= 2 cos

5

12 122

sin

5

12 122

= 2 cos4

sin

6

=

2 12

2 2 =

2

2 (7–4)

34. We suspect that this is not an identity since cot2 x = csc2 x – 1 is a Pythagorean Identity. (Note the

difference in sign.) To verify, plug in x = 2

.

cot2

2

?

csc2

2

+ 1

22

22

cos

sin

?

2

2

1

sin + 1

0

1 ?

1 + 1

Not true, so the equation is not an identity. (7–1) 35. Graphing y

1 = cos 3x, y

2 = cos x(cos 2x – 2 sin2 x), we see that the graphs match so we suspect that

this is an identity.

cos 3x = cos(2x + x) Algebra = cos 2x cos x – sin 2x sin x Sum Identity = cos 2x cos x – 2 sin x cos x sin x Double-angle Identity for sine = cos 2x cos x – 2 sin2 x cos x Algebra = cos x(cos 2x – 2 sin2 x) Factoring The identity is verified. (7–2)

36. sin3

2x

= sin x cos 3

2

+ cos x sin

3

2

Sum Identity

= sin x(0) + cos x(–1) Evaluation = –cos x Algebra The equation is not an identity due to an incorrect sign. (7–2)

37. cos3

2x

= cos x cos 3

2

+ sin x sin

3

2

Difference Identity

= cos x(0) + sin x(–1) Evaluation = –sin x Algebra The equation is not an identity due to an incorrect sign. (7–2)


38. 4 sin2 x – 3 = 0 0 ≤ x < 2π 4 sin2 x = 3

sin2 x = 3

4

sin x = ±3

2

sin x = 3

2 sin x = –

3

2

x = 3

,

2

3

x =

4

3

,

5

3

Solutions: x = 3

,

2

3

,

4

3

,

5

3

(7–5)

39. 2 sin2 θ + cos θ = 1 0° ≤ θ ≤ 180° 2(1 – cos2 θ) + cos θ = 1

2 – 2 cos2 θ + cos θ = 1 1 – 2 cos2 θ + cos θ = 0 –2 cos2 θ + cos θ + 1 = 0 2 cos2 θ – cos θ – 1 = 0 (2 cos θ + 1)(cos θ – 1) = 0 2 cos θ + 1 = 0 cos θ – 1 = 0

cos θ = –1

2 cos θ = 1

θ = 120° θ = 0° Solutions: θ = 0°, 120°

(7–5)

40. 2 sin2 x – sin x = 0 all real x Solve over one period [0, 2π): sin x(2 sin x – 1) = 0 sin x = 0 2 sin x – 1 = 0

x = 0, π sin x = 1

2

x = 6

,

5

6

Since the sine function is periodic with period 2π, all solutions are given by

x = 0 + 2kπ, x = π + 2kπ, x = 6

+ 2kπ, x =

5

6

+ 2kπ, k any integer.

The first two can also be written together as x = kπ, k any integer. (7–5)

41. sin 2x = 3 sin x all real x Solve over one period [0, 2π): sin 2x – 3 sin x = 0

2 sin x cos x – 3 sin x = 0

sin x(2 cos x – 3 ) = 0

sin x = 0 2 cos x – 3 = 0

x = 0, π cos x = 3

2

x = 6

,

11

6


Since the sine and cosine functions are periodic with period 2π, all solutions are given by

x = 0 + 2kπ, x = π + 2kπ, x = 6

+ 2kπ, x =

11

6

+ 2kπ, k any integer.

The first two can also be written together as x = kπ, k any integer. (7–5) 42. 2 sin2 θ + 5 cos θ + 1 = 0 all θ Solve over one period [0°, 360°)

2(1 – cos2 θ) + 5 cos θ + 1 = 0 2 – 2 cos2 θ + 5 cos θ + 1 = 0 –2 cos2 θ + 5 cos θ + 3 = 0 2 cos2 θ – 5 cos θ – 3 = 0 (2 cos θ + 1)(cos θ – 3) = 0

2 cos θ + 1 = 0 cos θ – 3 = 0

cos θ = –1

2 cos θ = 3

θ = 120°, 240° No solution Since the cosine function is periodic with period 360°, all solutions are given by θ = 120° + k360°, θ = 240° + k360°, k any integer. (7–5)

43. tan θ = 0.2557 all θ Solve over one period [0°, 180°) Sketch a graph of y = tan θ and y = 0.2557, on the interval [0°, 180°) θ = tan–1 0.2557 = 14.34° Since the tangent function is periodic with period 180°, all solutions are given by θ = 14.34° + k180°

(7–5)

44. sin2 x + 2 = 4 sin x all real x Solve over on period, [0, 2π): sin2 x – 4 sin x + 2 = 0 Quadratic in sin x

sin x = 2 4

2

b b ac

a

a = 1, b = –4, c = 2

sin x = 2( 4) ( 4) 4(1)(2)

2(1)

=

4 8

2

sin x = 4 8

2

sin x =

4 8

2

sin x = 3.414 sin x = 0.5858 No solution

x =1

1

sin 0.5858 0.6259 First quadrant solution

2.516 Second quadrant solutionsin 0.5858

Since the sine function is periodic with period 2π, all solutions are given by

x = 0.6259 2

2.516 2

k

k



45. tan2 x = 2 tan x + 1 0 ≤ x < π tan2 x – 2 tan x – 1 = 0 Quadratic in tan x

tan x = 2 4

2

b b ac

a

a = 1, b = –2, c = –1

tan x = 2( 2) ( 2) 4(1)( 1)

2(1)

=

(2) 8

2

tan x = 2 8

2

tan x =

2 8

2

tan x = 2.414 tan x = –0.4142 x = tan–1 2.414 x = π + tan–1(–0.4142) x = 1.178 x = 2.749 Solutions: 1.178, 2.749

(7–5)

46. Examining the graph of y1 = 3 sin 2x and y2 = 2x – 2.5, we obtain the graph at the right.

To four decimal places, the solution in [0, 2π] is given by x = 1.4903, and there is no other solution.

(7–5) 47. From the graph in the previous problem, the graph of y = 3 sin 2x is above the graph of y = 2x – 2.5,

that is, 3 sin 2x > 2x – 2.5, for x in the interval (–∞, 1.4903). (7–5) 48. Examining the graph of y1 = 2 sin2 x – cos 2x and y2 = 1 – x2, –π ≤ x ≤ π, we obtain

To four decimal places, the solutions in [–π, π] are given by –0.6716 and 0.6716, and there are no other solutions. (7–5)

49. From the graph in the previous problem, the graph of y = 2 sin2 x – cos 2x is below or intersects the graph of y = 1 – x2, that is, 2 sin2 x – cos 2x ≤ 1 – x2, for x in the interval [–0.6716, 0.6716].

(7–5)

50. (A) Testing x = 0 and y = 4

yields

tan 04

?= tan 0 + tan

4

tan 4

?= 0 + tan

4

1 √= 0 + 1

Yes, x = 0 and y = 4

is a solution.


(B) Consider x = 3

and y =

3

. Then the left side = tan

3 3

= tan 2

3

= – 3 .

The right side = tan 3

+ tan

3

= 3 + 3 = 2 3 . Since the left side is not equal to the

right side for at least one set of values for which both are defined, the equation is a conditional equation. (7–1)

51. sin–1 0.3351 has exactly one value, while the equation sin x = 0.3351 has infinitely many solutions. (6–6, 7–5) 52. (A) Graph both sides of the equation in the same viewing window.

tan

sin 2 tan

x

x x =

1

cos 2x is not an identity, since the graphs do not match. Try x =

4

.

Left side: 4

4 4

tan

sin 2 tan

=

12

1

2 1 =

2

1 2 2

Right side: 4

1

cos 2 =

12

1

2 =

2

1 2 2

This verifies that the equation is not an identity.

(B) Graph both sides of the equation in the same viewing window.

tan

sin 2 tan

x

x x =

1

cos 2x appears to be an identity, which we now

verify. 1

cos 2x =

tan

tan (cos 2)

x

x x Algebra

= tan

tan cos 2 tan

x

x x x Algebra

= sincos

tan

cos 2 tanxx

x

x x Quotient Identity

= tan

sin 2 tan

x

x x Algebra

(7–1)

53. (A) tan 2

x = 2 sin x 0 ≤ x < 2π

1 cos

sin

x

x

= 2 sin x

1 – cos x = 2 sin2 x sin x ≠ 0 1 – cos x = 2(1 – cos2 x ) 1 – cos x = 2 – 2 cos2 x 2 cos2 x – cos x – 1 = 0

(2 cos x + 1)(cos x – 1) = 0 2 cos x + 1 = 0 cos x – 1 = 0

cos x = –1

2 cos x = 1

x = 2

3

,

4

3

x = 0


Note that even though we multiplied both sides of the original equation by sin x, which is 0 when x = 0, x = 0 is still a solution of the original equation, as is shown by checking:

Left side: tan 0

2 = tan 0 = 0

Right side: 2 sin 0 = 2·0 = 0

(B) Examining the graphs of y1 = tan 2

x and y2 = 2 sin x, 0 ≤ x ≤ 2π (drawn in dot mode), we obtain

The graph confirms the solution 0 (checked in the previous part of the problem). The other intersections are found at x = 2.0944 and x = 4.1888. (7–5)

54. Examining the graph of y1 = 3 cos(x – 1) and y2 = 2 – x2, –π ≤ x ≤ π, we obtain

To three decimal places, the solutions on [–π, π] are 0.149 and –2.233, and there are no other solutions. (7–5)

55. Since 2

≤ x ≤ π,

4

≤

2

x ≤

2

, hence sin

2

x is positive.

sin x = 3

5 cos x = –

4

5

(A) sin 2

x =

1 cos

2

x =

451

2

=

95

2 =

9

10=

3

10 or

3 10

10

(B) cos 2x = cos2 x – sin2 x = 4

5

2

– 3

5

2

= 16

25 –

9

25 =

7

25 (7–3)

56. Let u = tan–1 3

4

.

Then –3

4 = tan u –

2

< u <

2

.

sin u = –3

5 cos u =

4

5

sin 1 32 tan

4

= sin 2u = 2 sin u cos u = 23 4

5 5

= –24

25

5

(4, -3)

u

4

-3

a

b

(7–3)


57. Let u = sin–1 3

5

Then 3

5 = sin u, –

2

≤ u ≤

2

.

Ordinarily we would also set v = cos–1 4

5, but a glance at the reference

triangle indicates that cos–1 4

5 = sin–1

3

5 = u

sin 1 13 4sin cos

5 5

= sin(u + u) = sin 2u = 2 sin u cos u = 2

3 4

5 5

= 24

25

a

b

5

u

4

3

(7–3)

58. (A) cos2 2x = cos 2x + sin2 2x 0 ≤ x < π is equivalent to cos2 2x = cos 2x + sin2 2x 0 ≤ 2x < 2π

cos2 2x = cos 2x + 1 – cos2 2x 2 cos2 2x – cos 2x – 1 = 0 (2 cos 2x + 1)(cos 2x – 1) = 0 2 cos 2x + 1 = 0 or cos 2x – 1 = 0

cos 2x = –1

2 cos 2x = 1

2x = 2

3

,

4

3

2x = 0

x = 3

,

2

3

x = 0 Solutions: x = 0,

3

,

2

3

(B) Examining the graph of y1 = cos2 2x and y2 = cos 2x + sin2 2x, 0 ≤ x < π, we obtain

The solutions on [0, π] are 0, and, to four decimal places, 1.0472 and 2.0944. (7–5)

59. We note that tan θ = 3

x and tan 2θ =

3 6

x

=

9

x (see figure).

Then, tan 2θ = 2

2 tan

1 tan

9

x =

3

23

2

1

x

x =

2

6

91x

x

= 2

2 6

2 2 91x

x

x

x x

=

2

6

9

x

x

x(x2 – 9) 9

x = x(x2 – 9)

2

6

9

x

x x ≠ 0, 3, –3

9(x2 – 9) = 6x · x 9x2 – 81 = 6x2 3x2 – 81 = 0 x2 = 27

x = 27 (we discard the negative solution)

To 3 decimal places, x = 5.196 cm. Since tan θ = 3

x =

3

27 =

1

3, θ = 30.000°. (7–3)


60. I = 50 sin 120π(t – 0.001) I = 40 40 = 50 sin 120π(t – 0.001)

sin 120π(t – 0.001) = 40

50

120π(t – 0.001) = sin–1 40

50 will yield the least positive solution of the equation

t – 0.001 = 1

120 sin–1

40

50

t = 0.001 + 1

120 sin–1

40

50= 0.00346 sec (calculator in radian mode) (7–5)

61. (A) 0.6 cos 184πt – 0.6 cos 208πt = 0.6(cos 184πt – cos 208πt) Algebra

= 0.6184 208 184 208

( 2)sin sin2 2

t t t t Sum-Product Identity

= –1.2 sin 196πt sin(–12πt) Algebra = –1.2 sin 196πt(–sin 12πt) Identities for Negatives = 1.2 sin 12πt sin 196πt Algebra (B) The required graphs are as shown.

y = 0.6 cos 184πt y = –0.6 cos 208πt

y = 0.6 cos 184πt – 0.6 cos 208πt y = 1.2 sin 12πt sin 196πt

(7–4)


62.

From the figure, Rθ = 18 and sin θ = 16

R. The definition of radian measure is θ =

S

R where S is the

arc length and R the radius. From these two equations, solving each for R in terms of θ and setting the results equal to each other, we obtain

R = 18

R =

16

sin

18

=

16

sin

18 sin θ = 16θ

sin θ = 8

9θ as required.

Examining the graph of y1 = sin θ and y2 = 8

9θ, 0 ≤ θ ≤

2

, we obtain

To four decimal places, θ = 0.8307.

Then R = 18

=

18

0.8307 = 21.668 ft.

Since 16

R h = cot θ, h = R – 16 cot θ = 7.057 ft.

(7–5)63. We are to solve 58.1 + 24.5 sin(0.524x + 4.1) = 70 Examining the graph of y1 = 58.1 + 24.5 sin(0.524x + 4.1) and y2 = 70 we obtain

Using x = 5 for May 15 we can interpret x = 5.1342852 as 0.1342852(30) = 4.03 days later, May 19. Using x = 9 for September 15 we can interpret x = 9.193889 as 0.193889(30) = 5.8 days later, September 21 or 22.

The average high temperature is above 70° between May 19 and September 22. (7–5) 64. The minimum value of y = 58.1 + 24.5 sin(0.524x + 4.1) is ymin = 58.1 – 24.5 or ymin = 33.6. Therefore the average high temperature is never as low as 30°. (7–5)

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