Young CAT2e SSM Ch7 Part2

Chapter 7

466

Section 7.6 Solutions -------------------------------------------------------------------------------- 1.

[ ] [ ]1 1sin 2 cos sin(2 ) sin(2 ) sin 3 sin2 2

x x x x x x x x= + + − = +

3.

[ ]

[ ] [ ]

15sin 4 sin 6 5 cos(4 6 ) cos(4 6 )2

5 5cos( 2 ) cos10 cos 2 cos102 2

x x x x x x

x x x x

= ⋅ − − +

= − − = −

5.

[ ]

[ ] [ ]

14cos( ) cos 2 4 cos( 2 ) cos( 2 )2

2 cos cos( 3 ) 2 cos cos3

x x x x x x

x x x x

− = ⋅ − + + − −

= + − = +

7.

[ ] [ ]

3 5 1 3 5 3 5sin sin cos cos2 2 2 2 2 2 2

1 1cos( ) cos 4 cos cos 42 2

x x x x x x

x x x x

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

= − − = −

9. 2 4 1 2 4 2 4cos cos cos cos3 3 2 3 3 3 3

1 2 1 2cos 2 cos cos 2 cos2 3 2 3

x x x x x x

xx x x

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= + − = +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

11. [ ] [ ]3 32 23cos(0.4 )cos(1.5 ) cos(1.9 ) cos( 1.1 ) cos(1.9 ) cos(1.1 )x x x x x x− = − + − = − +

13.

( ) ( ) ( ) ( ) ( ) ( )4sin 3 cos 3 3 2 sin 2 3 sin 4 3 2 sin 2 3 sin 4 3x x x x x x⎡ ⎤ ⎡ ⎤− = + − = −⎣ ⎦ ⎣ ⎦

15. 5 3 5 3cos5 cos3 2cos cos 2cos 4 cos

2 2x x x xx x x x+ −⎛ ⎞ ⎛ ⎞+ = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

17. 3 3sin 3 sin 2sin cos 2sin cos 2

2 2x x x xx x x x− +⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Section 7.6

467

19. 5 5

5 3 32 2 2 2sin sin 2sin cos 2sin( ) cos 2sin cos2 2 2 2 2 2

x x x xx x x xx x

⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

21. 2 7 2 7

2 7 3 3 3 3cos cos 2cos cos3 3 2 2

3 5 3 52cos cos 2cos cos2 6 2 6

x x x xx x

x x x x

⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

23.

( ) ( )

( ) ( ) ( ) ( )

0.4 0.6 0.4 0.6sin 0.4 sin 0.6 2sin cos2 2

2sin 0.5 cos 0.1 2sin 0.5 cos 0.1

x x x xx x

x x x x

+ −⎛ ⎞ ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − =

25.

( ) ( )

( ) ( ) ( ) ( )

5 3 5 5 3 5sin 5 sin 3 5 2sin cos2 2

2sin 5 cos 2 5 2sin 5 cos 2 5

x x x xx x

x x x x

⎛ ⎞ ⎛ ⎞− +− = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= − = −

27.

4 6 4 6cos cos 2cos cos4 6 2 2

5 52cos cos 2cos cos24 24 24 24

x x x xx x

x x x x

π π π ππ π

π π π π

⎛ ⎞ ⎛ ⎞− + − −⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞− + = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

29.

2cos3 cossin 3 sin

x xx x

−−

=+

3sin2

x x+⎛ ⎞⎜ ⎟⎝ ⎠

3sin2

2

x x−⎛ ⎞⎜ ⎟⎝ ⎠

3sin2

x x+⎛ ⎞⎜ ⎟⎝ ⎠

3tan tan23cos

2

x x xx x

−⎛ ⎞= − = −⎜ ⎟− ⎝ ⎠⎛ ⎞

⎜ ⎟⎝ ⎠

Chapter 7

468

31.

2cos cos3sin 3 sin

x xx x

−−

=−

3 3sin sin2 2

2

x x x x+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ sin 2 sin( ) sin 2 sin3 3 sin cos 2sin cos

2 2

x x x xx x x x x x

− −= =

− +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

sin xtan 2

cos 2x

x=

33.

2cos5 cos 2sin 5 sin 2

x xx x+

=−

5 2 5 2cos cos2 2

x x x x− +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 5 2 5 2sin cos2 2

x x x x− +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

5 2 3cot cot2 2

x x x−⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

35.

2sin sincos cos

A BA B+

=+

sin cos2 2

A B A B+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 cos cos2 2

A B A B+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

tan2

A B+⎛ ⎞= ⎜ ⎟⎝ ⎠

37.

2cos cossin sin

A BA B

−−

=+

sin2

A B+⎛ ⎞⎜ ⎟⎝ ⎠

sin2

2

A B−⎛ ⎞⎜ ⎟⎝ ⎠

sin2

A B+⎛ ⎞⎜ ⎟⎝ ⎠

tan2

cos2

A BA B

−⎛ ⎞= − ⎜ ⎟− ⎝ ⎠⎛ ⎞

⎜ ⎟⎝ ⎠

39.

2sin sinsin sin

A BA B+

=−

sin cos2 2

2

A B A B+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

sin cos2 2

sin cos cos sin2 2 2 2

tan cot2 2

A B A B

A B A B A B A B

A B A B

+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= ⋅

− + + −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

+ −⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

41.

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

A B A BA B A B+ + −

=+ + −

cos cos2 2

A B A B A B A B+ + − + − +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 sin cos2 2

A B A B A B A B+ + − + − +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cos cotsin

A AA

= =

Section 7.6

469

43. Description of G note: ( )cos 2 (392)tπ Description of B note: ( )cos 2 (494)tπ

Combining the two notes: ( ) ( )cos 784 cos 988t tπ π+ . Using the sum-to-product identity then yields:

( ) ( )

( ) ( )( ) ( )

784 988 784 988cos 784 cos 988 2cos cos2 2

2cos 886 cos 1022cos 886 cos 102 (since cosine is even)

t t t tt t

t tt t

π π π ππ π

π ππ π

+ −⎛ ⎞ ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −=

The beat frequency is 494 – 392 = 102 Hz.

The average frequency is 494 392 443 Hz2+

= .

45. The resulting signal is

6 6

6 6 6 6

6 6 6

2 2sin sin1.55 10 0.63 10

1 1 1 11.55 10 0.63 10 1.55 10 0.63 102sin 2 cos 2

2 2

10 10 10 11.55 0.63 1.55

2sin 2 cos 22

tc tc

tc tc

tc tc

π π

π π

π π

− −

− − − −

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟× × × ×⎝ ⎠ ⎝ ⎠⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞

+ −⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟=

⎜ ⎟⎜ ⎟⎝ ⎠

6

6 6

00.63

2

2 1 1 2 1 12sin 10 cos 102 1.55 0.63 2 1.55 0.63tc tcπ π

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

47.

( ) ( )

( ) ( ) ( ) ( )

1540 2418 1540 2418sin 2 (770) sin 2 (1209) 2sin cos2 2

2sin 1979 cos 439 2sin 1979 cos 439

t t t tt t

t t t t

π π π ππ π

π π π π

+ −⎛ ⎞ ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − =

Chapter 7

470

49. Note that 52.5 7.5 180A+ + = , so that 120A = . So, the area is

( ) ( ) ( )( )

( ) ( )( )

( ) ( )( )

( ) ( )

( )

2

2

2

2 2 2

2 2

1100 cos 52.5 7.5 cos 52.5 7.510 ft. sin 52.5 sin 7.5 2 ft.2sin 120 2sin 120

50 cos 45 cos 60ft.

2sin 120

2 125 25 2 1 25 2 1 32 2ft. ft. ft.

33 32

25 6 3ft. 5.98 ft.

3

⎡ ⎤⋅ − − +⎣ ⎦=

⎡ ⎤−⎣ ⎦=

⎡ ⎤−⎢ ⎥ − −⎣ ⎦= = =

−= ≈

51. In the final step of the computation, note that cos cos cosA B AB≠ and sin sin sinA B AB≠ . Should have used the product-to-sum identities. 53. False. From the product-to-sum identities, we have

[ ]1cos cos cos( ) cos( )2

A B A B A B= + + − ,

and the right-side is not, in general, expressible as the cosine of a product. 55. True. From the product-to-sum identities, we have

[ ]1cos cos cos( ) cos( )2

A B A B A B= + + − .

57. Observe that [ ]

( )( )

( )

( ) ( )( )

( ) ( )( )

[ ]

sin sin sin sin sin sin

1 cos cos( ) sin2

1 cos sin cos( )sin21 1 sin sin2 2

1 sin sin2

1 sin( ) sin( ) sin( ) sin( )4

A B C A B C

A B A B C

A B C A B C

C A B C A B

C A B C A B

A B C C A B A B C A B C

=

⎡ ⎤= − − +⎢ ⎥⎣ ⎦

= − − +⎡ ⎤⎣ ⎦

⎧ ⎡ ⎤= + − + − −⎨ ⎣ ⎦⎩⎫⎡ ⎤− + + + − + ⎬⎣ ⎦⎭

= − + + − + − + + − + −

At this point, depending on which terms you decide to apply the odd identity for sine, the answer can take on a different form.

Section 7.6

471

59. [ ] [ ]1 1

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

A B A B A B A B A B A BA B

− = + + − − − + +

= +

61. Observe that ( ) ( ) ( )

( ) ( )6 6 6

3 7 52 6 6

1 3sin( )sin 1 3 cos cos

1 cos cos

y x x x x x x

x x

π π π

π π

π π π⎡ ⎤= − − = − + − −⎣ ⎦⎡ ⎤= − −⎣ ⎦

The graph is as follows:

63. Observe that

( ) ( )( ) ( )( ) ( ) ( ) ( )

2 53 6

2 5 2 512 3 6 3 6

3 31 12 2 6 2 2 6

cos cos

cos cos

cos cos cos cos

y x x

x x x x

x x x x

π π

π π π π

π π π π

= −

⎡ ⎤= − + + −⎣ ⎦⎡ ⎤ ⎡ ⎤= − + − = − +⎣ ⎦ ⎣ ⎦

The graph is as follows:

Chapter 7

472

65. Consider the graph of 4sin cos cos 2y x x x= , as seen below:

From the graph, it seems as though this function is equivalent to sin 4x . We prove this identity below:

( )sin 2

4sin cos cos 2 2 2sin cos cos 2 2sin 2 cos 2 sin(2 2 ) sin 4x

x x x x x x x x x x=

= = = ⋅ =

67. To the right are the graphs of the following functions:

[ ]

1

2

13 2

sin 4 sin 2 (solid)sin 6 (dashed)

cos 2 cos 6

y x xy xy x x

==

= −

Note that the graphs of 1y and 3y are the same.

Section 7.7

473

Section 7.7 Solutions --------------------------------------------------------------------------------

1. The equation 2arccos2

θ⎛ ⎞

=⎜ ⎟⎜ ⎟⎝ ⎠

is equivalent to 2cos2

θ = . Since the range of

arccosine is [ ]0,π , we conclude that 4πθ = .

3. The equation 3arcsin2

θ⎛ ⎞− =⎜ ⎟⎜ ⎟⎝ ⎠

is equivalent to 3sin2

θ = − . Since the range of arcsine

is ,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, we conclude 3πθ = − .

5. The equation ( )1cot 1 θ− − = is equivalent to 2 cos2cot 1

sin22

θθθ

= − = − = . Since the

range of inverse cotangent is ( )0,π , we conclude that 34πθ = .

7. The equation 2 3arcsec3

θ⎛ ⎞

=⎜ ⎟⎜ ⎟⎝ ⎠

is equivalent to 2 3sec3

θ = , which is further the same

as 3 3cos22 3

θ = = . Since the range of inverse secant is 0, ,2 2π π π⎡ ⎞ ⎛ ⎤∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦

, we conclude

that6πθ = .

9. The equation ( )1csc 2 θ− = is equivalent to csc 2θ = , which is further the same as 1sin2

θ = . Since the range of inverse cosecant is ,0 0,2 2π π⎡ ⎞ ⎛ ⎤− ∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦

, we conclude

that6πθ = .

11. The equation ( )arc tan 3 θ− = is equivalent to 3 sin2tan 3 1 cos2

θθθ

= − = − = . Since

the range of inverse tangent is ,2 2π π⎛ ⎞−⎜ ⎟

⎝ ⎠, we conclude that

3πθ = − .

13. The equation ( )arcsin 0 θ= is equivalent to sin 0θ = . Since the range of arcsine is

,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, we conclude 0θ = .

15. The equation ( )1sec 1 θ− − = is equivalent to sec 1θ = − , which is further the same as

cos 1θ = − . Since the range of inverse secant is 0, ,2 2π π π⎡ ⎞ ⎛ ⎤∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦

, we conclude that θ π= .

Chapter 7

474

17. The equation 1 1cos2

θ− ⎛ ⎞ =⎜ ⎟⎝ ⎠

is equivalent to 1cos2

θ = . Since the range of arccosine is

[ ]0,π , we conclude that 3πθ = , which corresponds to 60θ = .

19. The equation 1 2sin2

θ− ⎛ ⎞=⎜ ⎟⎜ ⎟

⎝ ⎠ is equivalent to 2sin

2θ = . Since the range of arcsine

is ,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, we conclude that 4πθ = , which corresponds to 45θ = .

21. The equation 1 3cot3

θ− ⎛ ⎞− =⎜ ⎟⎜ ⎟⎝ ⎠

is equivalent to 13 cos2cot

3 sin32

θθθ

= − = − = . Since

the range of inverse cotangent is ( )0,π , we conclude that 23πθ = , which corresponds to

120θ = .

23. The equation 3arc tan3

θ⎛ ⎞

=⎜ ⎟⎜ ⎟⎝ ⎠

is equivalent to 13 sin2tan

3 cos32

θθθ

= = = . Since the

range of inverse tangent is ,2 2π π⎛ ⎞−⎜ ⎟

⎝ ⎠, we conclude that

6πθ = , which corresponds to

30θ = . 25. The equation ( )arccsc 2 θ− = is equivalent to csc 2θ = − , which is further the same as

1sin2

θ = − . Since the range of inverse cosecant is ,0 0,2 2π π⎡ ⎞ ⎛ ⎤− ∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦

, we conclude

that6πθ = − , which corresponds to 30θ = − .

27. The equation ( )arcsec 2 θ− = is equivalent to sec 2θ = − , which is further the

same as 1cos2

θ = − . Since the range of inverse secant is 0, ,2 2π π π⎡ ⎞ ⎛ ⎤∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦

, we conclude

that 34πθ = , which corresponds to 135θ = .

29. The equation ( )1sin 1 θ− − = is equivalent to sin 1θ = − . Since the range of arcsine is

,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, we conclude that 2πθ = − , which corresponds to 90θ = − .

Section 7.7

475

31. The equation ( )arccot 0 θ= is equivalent to coscot 0sin

θθθ

= = , which implies

cos 0θ = . Since the range of inverse cotangent is ( )0,π , we conclude that 2πθ = , which

corresponds to 90θ = .

33. ( )1cos 0.5432 57.10− ≈ 35. ( )1tan 1.895 62.18− ≈ 37.

( )1 1 1sec 1.4973 cos 48.101.4973

− − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

39.

( )1 1 1csc 3.7893 sin 15.303.7893

− − ⎛ ⎞− = ≈ −⎜ ⎟−⎝ ⎠

41.

( )1 1 1cot 4.2319 tan4.2319

180 13.30 166.70

π− − ⎛ ⎞− = + ⎜ ⎟−⎝ ⎠

≈ − =

43. ( )1sin 0.5878 0.63− − ≈ −

45. ( )1cos 0.1423 1.43− ≈ 47. ( )1tan 1.3242 0.92− ≈ 49.

( )1 1 1cot 0.5774 tan 2.090.5774

π− − ⎛ ⎞− = + ≈⎜ ⎟−⎝ ⎠

51.

( )1 1 1csc 3.2361 sin 0.313.2361

− − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

53. 1 5 5sin sin12 12π π− ⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

since 52 12 2π π π

− ≤ ≤ .

55. ( )( )1sin sin 1.03− is undefined since 1.03 is not in the domain of inverse sine.

57. Note that we need to use the angle θ in ,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

such that 7sin sin6πθ ⎛ ⎞= −⎜ ⎟

⎝ ⎠ . To

this end, observe that 1 17sin sin sin sin6 6 6π π π− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

.

59. Note that we need to use the angle θ in [ ]0,π such that 4cos cos3πθ ⎛ ⎞= ⎜ ⎟

⎝ ⎠ . To this

end, observe that 1 14 2 2cos cos cos cos3 3 3π π π− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

.

61. Since ( )1cot cot x x− = for all x−∞ < < ∞ , we see that ( )1cot cot 3 3− = .

63. Note that we need to use the angle θ in 0, ,2 2π π π⎡ ⎞ ⎛ ⎤∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦

such that sec sec .3πθ ⎛ ⎞= −⎜ ⎟

⎝ ⎠

To this end, observe that 1 1sec sec sec sec3 3 3π π π− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

.

65. 1 1csc csc2

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

is undefined since 12

is not in the domain of inverse cosecant.

Chapter 7

476

67. Since ( )1cot cot x x− = for all x−∞ < < ∞ , we see that ( )1cot cot 0 0− = .

69. Since ( )1tan tan x x− = for all 2 2

xπ π− < < , we see that 1tan tan

4 4π π− ⎛ ⎞⎛ ⎞− = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

.

71. Not possible 73. ( )( ) ( )1 18 21

3 33cot cot cotπ π− −= − =

75. ( ) ( )1 1 2154 2 4csc csc cscπ π− −= − = −

77. Let 1 3sin4

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 3sin4

θ = , as

shown in the diagram:

Using the Pythagorean Theorem, we see that 2 2 23 4z + = , so that 7z = .

Hence, 1 3 7cos sin cos4 4

θ−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

79. Let 1 12tan5

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 12tan5

θ = , as


Using the Pythagorean Theorem, we see that

2 2 212 5 z+ = , so that 13z = .

Hence, 1 12 12sin tan sin5 13

θ−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

θ θ

Section 7.7

477

81. Let 1 3sin5

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 3sin5

θ = , as


Using the Pythagorean Theorem, we see that 2 2 23 5z + = , so that 4z = .

Hence, 1 3 3tan sin tan5 4

θ−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

83. Let 1 2sin5

θ − ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠. Then, 2sin

5θ = , as



( )22 22 5z + = , so that 23z = . So,

1 3 1 5 5 23sec sin sec4 cos 2323

θθ

−⎛ ⎞⎛ ⎞ = = = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

85. Let 1 1cos4

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 1cos4

θ = , as


Using the Pythagorean Theorem, we see that 2 2 21 4z + = , so that 15z = . Hence,

1 1 1 4 4 15csc cos csc4 sin 1515

θθ

−⎛ ⎞⎛ ⎞ = = = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

.

87. Let 1 60sin61

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 60sin61

θ = , as



2 2 260 61z + = , so that 11z = . Hence,

1 60 11cot sin cot61 60

θ−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

θθ

2

θθ

Chapter 7

478

89. Use ( )sin 2i I f tπ= with f = 5 and I = 115. Find the smallest positive value of t for which i = 85. To this end, observe

( )

( )

1

115sin 2 5 8585sin 10

11585sin

115 0.02647610

t

t

t

π

π

π

−

⋅ =

=

⎛ ⎞⎜ ⎟⎝ ⎠= ≈

So, 0.026476 sec. 26 mst ≈ = . 91. Given that ( ) 12 2.4sin(0.017 1.377)H t t= + − , we must find the value of t for which

( ) 14.4H t = . To this end, we have

1

12 2.4sin(0.017 1.377) 14.4sin(0.017 1.377) 1

0.017 1.377 sin (1) 21.3772 173.4

0.017

tt

t

t

π

π

−

+ − =− =

− = =

+= ≈

Now, note that 173.4 151 22.4− = . As such, this corresponds to June 22-23. 93. We need to find the smallest value of t for which

( )12.5cos 0.157 2.5 0t + = , and the graph of the left-side is decreasing prior to this value. We solve this graphically. The solid graph corresponds to the left-side of the equation. We have:

Notice that the solution is approximately t = 11.3, or about 11 years.

Section 7.7

479

95. Consider the following diagram:

Let θ α β= + . Then,

( )

2 2

2

tan tantan tan1 tan tan

1 7 88

1 7 7 71

xx x xx x

x x x

α βθ α βα β+

= + =−

+= = =

− −⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

97. Use the formula

12 tan1

2

kf dM

π

−⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥= −

⎢ ⎥⎢ ⎥⎣ ⎦

with 2m 0.002 km and 4 kmf d= = = .

We have the following specific calculation:

k = 2 km: 1

1

22 tan0.002 0.001 141 2 tan 0.0007048 km 0.70m

2 2M π

π π

−

−

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥ ⎡ ⎤⎛ ⎞⎝ ⎠⎢ ⎥= − = − ≈ ≈⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥ ⎣ ⎦

⎢ ⎥⎣ ⎦

k = 10 km: 1

1

102 tan0.002 0.001 541 2 tan 0.00024 km 0.24m

2 2M π

π π

−

−

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥ ⎡ ⎤⎛ ⎞⎝ ⎠⎢ ⎥= − = − ≈ ≈⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥ ⎣ ⎦

⎢ ⎥⎣ ⎦


Since 300tan200 x

α =−

, we have

1 300tan200 x

α − ⎛ ⎞= ⎜ ⎟−⎝ ⎠.

Also, since 150tanx

β = , we have

1 150tanx

β − ⎛ ⎞= ⎜ ⎟⎝ ⎠

.

Therefore, since α β θ π+ + = , we see that

1 1300 150tan tan200 x x

θ π − −⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠.

101. The identity ( )1sin sin x x− = is valid only for x in the interval ,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, not [ ]0,π .

α

β

θα β

Chapter 7

480

103. In general, 11

1cottan

xx

−−≠ .

105. False. Upon inspection of the graphs, the portion to the right of the y-axis, when reflected over the y-axis does not match up identically with the left portion, as seen below:

More precisely, note that for instance 1 1 1sec (1) cos 01

− − ⎛ ⎞= =⎜ ⎟⎝ ⎠

, while

1 1 1sec ( 1) cos1

π− − ⎛ ⎞− = − =⎜ ⎟⎝ ⎠

. As such, ( ) ( )1 1sec secx x− −≠ − , for all x in the domain of

inverse secant. 107. False. This holds only on a subset of the domain to which cosecant is restricted in order to define its inverse.

109. 1 1sec2

− ⎛ ⎞⎜ ⎟⎝ ⎠

does not exist since 12

is not in the domain of the inverse secant function

(which coincides with the range of the secant function).

111. In order to compute 1 12 1sin cos sin2 2

− −⎡ ⎤⎛ ⎞ ⎛ ⎞+ −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦

, we first simplify both 1 2cos2

− ⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

and 1 1sin2

− ⎛ ⎞−⎜ ⎟⎝ ⎠

:

1 2 2If cos , then cos2 2

θ θ− ⎛ ⎞= =⎜ ⎟⎜ ⎟

⎝ ⎠. So,

4πθ = .

1 1 1If sin , then sin2 2

β β− ⎛ ⎞− = = −⎜ ⎟⎝ ⎠

. So, 6πβ = − .

Hence,

1 12 1sin cos sin sin sin cos sin cos2 2 4 6 4 6 6 4

2 3 1 2 6 22 2 2 2 4

π π π π π π− −⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − = − = −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞⎛ ⎞ ⎛ ⎞ −⎛ ⎞= − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠

113. In order to compute ( )1sin 2sin (1)− , first observe that

1If =sin (1), then sin 1.θ θ− = So, 2πθ = .

Hence, ( )1sin 2sin (1) sin 2 sin 02π π− ⎛ ⎞= ⋅ = =⎜ ⎟

⎝ ⎠.

Section 7.7

481

115. Consider the function ( ) 3 cos4

f x x π⎛ ⎞= + −⎜ ⎟⎝ ⎠

.

a. Note that this function has a phase shift of 4π units to the right. So, we take the

interval used to define 1cos x− , namely [ ]0,π and add 4π to both endpoints to get

the interval 5,4 4π π⎡ ⎤⎢ ⎥⎣ ⎦

. Note that f is, in fact, one-to-one on this interval.

b. Now, we determine a formula for 1f − , along with its domain:

( )

( )

1

1

3 cos4

3 cos4

cos 34

cos 34

y x

y x

y x

y x

π

π

π

π

−

−

⎛ ⎞= + −⎜ ⎟⎝ ⎠

⎛ ⎞− = −⎜ ⎟⎝ ⎠

− = −

+ − =

So, the equation of the inverse of f is given by: ( )1 1( ) cos 34

f x xπ− −= + −

The domain of 1f − is equal to the range of f. Since the amplitude of f is 1 and there is a vertical shift up of 3 units, we see that the range of f is [ ]2, 4 . Hence, the domain of 1f −

is [ ]2, 4 .

117. Consider the function ( )14 6( ) 2 cot 2f x x π= + − .

(a) We know that cot 2y x= is 1-1 on ( )20, π . As such, since ( )6cot 2y x π= − is simply a horizontal shift of coty x= to the right 12

π units, we conclude that it is 1-1 on the interval

( ) ( )512 2 12 12 120 , ,π π π π π− − = − . Since multiplying by a constant and shifting it vertically do not

affect whether or not it is 1-1, we conclude that f is 1-1 on this interval as well. (b) Restricting our attention to x values in ( )5

12 12,π π− , we determine the inverse as follows:

( )( )

( )( )

( )( )( )

14 6

14 6

14 6

6

16

16

1112 2

2 cot 2

2 cot 2

2 cot 2

4( 2) cot 2

cot 4( 2) 2

cot 4 8 2

cot 4 8

y x

x y

x y

x y

x y

x y

x y

π

π

π

π

π

π

π

−

−

−

= + −

= + −

− = −

− = −

− = −

+ − =

+ − =

Hence, the inverse is ( )1 1112 2( ) cot 4 8f x xπ− −= + − with domain .

Chapter 7

482

119. The graphs of the following two functions on the interval [ ]3,3− is below:

( )11 sin sinY x−= , 2Y x=

The graphs are different outside the interval [ ]1,1− because the identity ( )1sin sin x x− = only holds for 1 1x− ≤ ≤ . The results are different outside the interval [ ]1,1− because the identity ( )1cos cos x x− = only holds for 1 1x− ≤ ≤ .

121. The graphs of the following two functions on the interval ,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

is below:

( )11 csc cscY x−= , 2Y x=

Observe that the graphs do indeed coincide on this interval. This occurs since

( )1csc csc x x− = holds when ,0 0,2 2

x π π⎡ ⎞ ⎛ ⎤∈ − ∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦.

Section 7.8

483

123. From the given information, we have the following diagram:

a. ( )( )40 9 720

41 41 1681sin 2 2sin cos 2x x x= = − − =

b. ( )1 409tan 1.34948x −= ≈ . So,

( )sin 2 sin 2.69896 0.42832x = = c. Yes, the results in parts a. and b. are the same.

Section 7.8 Solutions --------------------------------------------------------------------------------

1. The values of θ in [ ]0, 2π that satisfy 2cos2

θ = − are θ = 3 5,4 4π π .

3. First, observe that csc 2θ = − is equivalent to 1sin2

θ = − . The values of θ in [ ]0,4π

that satisfy 1sin2

θ = − are

7 7 11 11 7 11 19 23, 2 , , 2 , , ,6 6 6 6 6 6 6 6π π π π π π π πθ π π= + + = .

5. The only way tan 0θ = is for sin 0θ = . The values of θ in that satisfysin 0θ = (and hence, the original equation) are θ = , where is an integern nπ .

7. The values of θ in [ ]0, 2π that satisfy 1sin 22

θ = − must satisfy

7 7 11 11 7 11 19 232 , 2 , , 2 , , ,6 6 6 6 6 6 6 6π π π π π π π πθ π π= + + = .

So, dividing all values by 2 yields the following values ofθ which satisfy the original equation:

7 11 19 23, , ,12 12 12 12π π π πθ =

Chapter 7

484

9. The values of θ in that satisfy 1sin2 2θ⎛ ⎞ = −⎜ ⎟⎝ ⎠

must satisfy

7 112 , 22 6 6

n nθ π ππ π= + + , where n is an integer.

So, multiplying all values by 2 yields the following values ofθ which satisfy the original equation:

7 11 7 112 2 , 2 2 4 , 46 6 3 3

7 114 , 4 , where is an integer .3 3

n n n n

n n n

π π π πθ π π π π

π ππ π

⎛ ⎞ ⎛ ⎞= + + = + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + +

11. The values of θ in [ ]2 , 2π π− that satisfy tan 2 3θ = must satisfy 4 4 2 2 5 52 , 2 , , 2 , , 2 , , 2

3 3 3 3 3 3 3 34 7 10 2 5 8 11, , , , , , ,

3 3 3 3 3 3 3 3

π π π π π π π πθ π π π π

π π π π π π π π

= + + − − − − − −

= − − − −.


4 7 10 2 5 8 11, , , , , , ,6 6 6 6 6 6 6 6

2 7 5 5 4 11, , , , , , ,6 3 6 3 3 6 3 6

π π π π π π π πθ

π π π π π π π π

= − − − −

= − − − −

13. First, observe that sec 2θ = − is equivalent to 1cos2

θ = − . The values of θ in

[ ]2 ,0π− that satisfy 1cos2

θ = − are 2 4,3 3π πθ = − − .

15. The values of θ in that satisfy123

2

3cot 43

θ = − = − must satisfy

2 54 2 , 23 3

n nπ πθ π π= + + , where n is an integer.

This reduces to 243

nπθ π= + , where n is an integer.


( )2 3, where is an integer

6 4 12nn n

ππ πθ+

= + = .

Section 7.8

485

17. First, observe that sec3 1θ = − is equivalent to cos3 1θ = − . The values of θ in for which this is true must satisfy

3 2 (2 1)n nθ π π π= + = + , where n is an integer. So, dividing all values by 5 yields the following values ofθ which satisfy the original equation:

(2 1) , where is an integer3

n nπθ += .

19. First, observe that csc3 1θ = is equivalent to sin 3 1θ = . The values of θ in for which this is true must satisfy

(4 1)3 22 2

nnπ πθ π += + = , where n is an integer.

So, dividing all values by 5 yields the following values ofθ in [ ]2 ,0π− which satisfy the original equation:

7 11, ,2 6 6π π πθ = − − −

21. First, observe that 2sin 2 3θ = is equivalent to 3sin 22

θ = . The values of θ in

[ ]0, 2π that satisfy 3sin 22

θ = must satisfy

2 2 2 7 82 , 2 , , 2 , , ,3 3 3 3 3 3 3 3π π π π π π π πθ π π= + + = .


2 7 8 7 4, , , , , ,6 6 6 6 6 3 6 3π π π π π π π πθ = = .

23. First, observe that 3 tan 2 3 0θ − = is equivalent to 3tan 23

θ = . The values of θ in

[ ]0, 2π that satisfy 3tan 23

θ = must satisfy

7 7 7 13 192 , 2 , , 2 , , ,6 6 6 6 6 6 6 6π π π π π π π πθ π π= + + = .


7 13 19, , ,12 12 12 12π π π πθ = .

Chapter 7

486

25. First, observe that ( )2cos 2 1 0θ + = is equivalent to ( ) 1cos 22

θ = − . The values of θ

in [ ]0, 2π that satisfy ( ) 1cos 22

θ = − must satisfy

2 2 4 4 2 4 8 102 , 2 , , 2 , , ,3 3 3 3 3 3 3 3π π π π π π π πθ π π= + + = .


2 4 8 10 2 4 5, , , , , ,6 6 6 6 3 3 3 3π π π π π π π πθ = =

27. First, observe that 3 cot 3 02θ⎛ ⎞ − =⎜ ⎟⎝ ⎠

is equivalent to

cos13 22cot2 3 3 sin2 2

θθ

θ

⎛ ⎞⎜ ⎟⎛ ⎞ ⎝ ⎠= = =⎜ ⎟ ⎛ ⎞⎝ ⎠⎜ ⎟⎝ ⎠

.

The value of θ in[ ]0,2π that satisfy this equation must satisfy 2 3θ π= , so that 2

3πθ = .

29. Factoring the left-side of 2tan 1 0θ − = yields the equivalent equation ( )( )tan 1 tan 1 0θ θ− + = which is satisfied when either tan 1 0 or tan 1 0θ θ− = + = . The

values of θ in [ ]0, 2π that satisfy tan 1θ = are 5,4 4π πθ = , and those which satisfy

tan 1θ = − are 3 7,4 4π πθ = . Thus, the solutions to the original equation are

3 5 7, , ,4 4 4 4π π π πθ = .

31. Factoring the left-side of 22cos cos 0θ θ− = yields the equivalent equation ( )cos 2cos 1 0θ θ − = which is satisfied when either cos 0 or 2cos 1 0θ θ= − = . The

values of θ in [ ]0, 2π that satisfy cos 0θ = are 3,2 2π πθ = , and those which satisfy

1cos2

θ = are 5,3 3π πθ = . Thus, the solutions to the original equation are

3 5, , ,2 2 3 3π π π πθ = .

Section 7.8

487

33. Factoring the left-side of 2csc 3csc 2 0θ θ+ + = yields the equivalent equation ( )( )csc 2 csc 1 0θ θ+ + = which is satisfied when either csc 2 0 or csc 1 0θ θ+ = + = . The

values of θ in [ ]0, 2π that satisfy csc 2θ = − (or equivalently 1sin2

θ = − ) are

7 11,6 6π πθ = , and those which satisfy csc 1θ = − (or equivalently sin 1θ = − ) are 3

2πθ =

. Thus, the solutions to the original equation are 7 11 3, ,6 6 2π π πθ = .

35. Factoring the left-side of 2sin 2sin 3 0θ θ+ − = yields the equivalent equation ( )( )sin 3 sin 1 0θ θ+ − = which is satisfied when either sin 3 0 or sin 1 0θ θ+ = − = . Note that the equationsin 3 0θ + = has no solution since –3 is not in the range of sine. The

value of θ in [ ]0, 2π that satisfies sin 1θ = is2πθ = .

37. Factoring the left-side of 2sec 1 0θ − = yields the equivalent equation ( )( )sec 1 sec 1 0θ θ− + = , which is satisfied whenever sec 1θ = ± , or equivalently

cos 1θ = ± . The values of θ in ( )0, 2π for which this is true are 0,θ π= .

39. Factoring the left-side of 2 43sec (2 ) 0θ − = yields the equivalent equation

2 2sec 2 sec 2 03 3

θ θ⎛ ⎞⎛ ⎞− + =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

, which is satisfied whenever 2sec 23

θ = ± , or

equivalently 3cos 22

θ = ± . The values of θ for which this is true must satisfy

5 7 11 13 17 19 232 , , , , , , ,6 6 6 6 6 6 6 6π π π π π π π πθ =

So, dividing by 2 then yields the values of θ in [ ]0, 2π for which this is true:

5 7 11 13 17 19 23, , , , , , ,12 12 12 12 12 12 12 12π π π π π π π πθ =

Chapter 7

488

41. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy sin 2 0.7843θ = − , we proceed as follows: Step 1: Find the values of 2θ whose sine is 0.7843− . Indeed, observe that one solution is ( )1sin 0.7843 51.655− − ≈ − , which is in QIV. Since the angles we seek have positive measure, we use the representative 360 51.655 308.345− ≈ . A second solution occurs in QIII, and has value 180 51.655 231.655+ = . Step 2: Use periodicity to find all values of θ that satisfy the original equation. Using periodicity with the solutions obtained in Step 1, we see that

2 308.345 , 308.345 360 , 231.655 , 231.655 360308.345 , 668.345 , 231.655 , 591.655

θ = + +

=

and so, the solutions to the original equation are approximately: 115.83 , 295.83 , 154.17 , 334.17θ ≈

43. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy tan 0.23432θ⎛ ⎞ = −⎜ ⎟⎝ ⎠

,

we proceed as follows:

Step 1: Find the values of 2θ whose tangent is 0.2343− .

Indeed, observe that one solution is ( )1tan 0.2343 13.187− − ≈ − , which is in QIV. Since the angles we seek have positive measure, we use the representative 360 13.187 346.813− ≈ . A second solution occurs in QII, and has value 346.813 180 166.813− = . Step 2: Use Step 1 to find all values of θ that satisfy the original equation, and exclude any value of θ that satisfies the equation, but lies outside the interval 0 ,360⎡ ⎤⎣ ⎦ . The solutions obtained in Step 1 are

166.813 , 346.8132θ= .

When multiplied by 2, the solution corresponding to346.813 will no longer be in the interval. So, the solution to the original equation in 0 ,360⎡ ⎤⎣ ⎦ is approximately

333.63θ ≈ .

Section 7.8

489

45. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 5cot 9 0θ − = , we proceed as follows:

Step 1: First, observe that the equation is equivalent to 9cot5

θ = .

Step 2: Find the values of θ whose cotangent is 95

.

Indeed, observe that one solution is 1 1 19 1 5cot tan tan 29.054695 95

− − −⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟= = ≈⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

,

which is in QI. A second solution occurs in QIII, and has value 29.0546 180 209.0546+ = . Since the input of cotangent is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately 29.05 , 209.05θ ≈ .

47. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 4sin 2 0θ + = , we proceed as follows:

Step 1: First, observe that the equation is equivalent to 2sin4

θ = − .

Step 2: Find the values of θ whose sine is 24

− .

Indeed, observe that one solution is 1 2sin 20.70484

− ⎛ ⎞− ≈ −⎜ ⎟⎜ ⎟⎝ ⎠

, which is in QIV.

Since the angles we seek have positive measure, we use the representative 360 20.7048 339.30− ≈ . A second solution occurs in QIII, and has value 20.7048 180 200.70+ = . Since the input of sine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately 200.70 , 339.30θ ≈ .

Chapter 7

490

49. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 24cos 5cos 6 0θ θ+ − = , we proceed as follows:

Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields:

( )( )

No solution

4cos 3 cos 2 04cos 3 0 or cos 2 0

3cos or cos 24

θ θθ θ

θ θ

− + =

− = + =

= = −

Step 2: Find the values of θ whose cosine is 34

.

Indeed, observe that one solution is 1 3cos 41.40964

− ⎛ ⎞ ≈⎜ ⎟⎝ ⎠

, which is in QI. A

second solution occurs in QIV, and has value 360 41.4096 318.59− = . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately 41.41 , 318.59θ ≈ .

Section 7.8

491

51. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 26 tan tan 12 0θ θ− − = , we proceed as follows:

Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields:

( )( )3tan 4 2 tan 3 03tan 4 0 or 2 tan 3 0

4 3tan or tan3 2

θ θθ θ

θ θ

+ − =

+ = − =

= − =

Step 2: Solve 4tan3

θ = − .

To do so, we must find the values of θ whose tangent is 43

− .

Indeed, observe that one solution is 1 4tan 53.133

− ⎛ ⎞− ≈ −⎜ ⎟⎝ ⎠

, which is in QIV. Since

the angles we seek have positive measure, we use the representative 360 53.13 306.87− ≈ . A second solution occurs in QII, and has value 180 53.13 126.87− = . Since the input of tangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 126.87 , 306.87θ ≈ .

Step 3: Solve 3tan2

θ =

To do so, we must find the values of θ whose tangent is 32

.

Indeed, observe that one solution is 1 3tan 56.312

− ⎛ ⎞ ≈⎜ ⎟⎝ ⎠

, which is in QI. A second

solution occurs in QIII, and has value 180 56.31 236.31+ = . Since the input of tangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 56.31 , 236.31θ ≈ . Step 4: Conclude that the solutions to the original equation are

56.31 , 126.87 , 236.31 , 306.87θ ≈ .

Chapter 7

492

53. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 215sin 2 sin 2 2 0θ θ+ − = , we proceed as follows:

Step 1: Simplify the equation algebraically. Factoring the left-side of the equation yields: ( )( )5sin 2 2 3sin 2 1 0

5sin 2 2 0 or 3sin 2 1 02 1sin 2 or sin 25 3

θ θθ θ

θ θ

+ − =

+ = − =

= − =

Step 2: Solve 2sin 25

θ = − .

Step a: Find the values of 2θ whose sine is 25

− .


− ⎛ ⎞− ≈ −⎜ ⎟⎝ ⎠

, which is in QIV. Since the

angles we seek have positive measure, we use the representative 360 23.578 336.42− ≈ . A second solution occurs in QIII, namely 180 23.578 203.578+ = .

Step b: Use periodicity to find all values of θ that satisfy 2sin 25

θ = − .

Using periodicity with the solutions obtained in Step a, we see that 2 203.578 , 203.578 360 , 336.42 , 336.42 360

203.578 , 563.578 , 336.42 , 696.42θ = + +

=

and so, the solutions are approximately: 101.79 , 281.79 , 168.21 , 348.21θ ≈

Step 3: Solve 1sin 23

θ = .

Step a: Find the values of 2θ whose sine is 13

.


− ⎛ ⎞ ≈⎜ ⎟⎝ ⎠

, which is in QI.

A second solution occurs in QII, and has value 180 19.4712 160.528− = .

Step b: Use periodicity to find all values of θ that satisfy 1sin 23

θ = .

Using periodicity with the solutions obtained in Step a, we see that 2 19.4712 , 19.4712 360 , 160.528 , 160.528 360

19.4712 , 160.528 , 379.4712 , 520.528θ = + +

=

and so, the solutions are approximately: 9.74 , 189.74 , 80.26 , 260.26θ ≈ Step 4: Conclude that the solutions to the original equation are

101.79 , 281.79 , 168.21 , 348.21 , 9.74 , 189.74 , 80.26 , 260.26θ ≈ .

Section 7.8

493

55. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 2cos 6cos 1 0θ θ− + = , we proceed as follows: Step 1: Simplify the equation algebraically. Since the left-side does not factor nicely, we apply the quadratic formula (treating

cosθ as the variable): ( ) ( ) ( )( )( )

26 6 4 1 1 6 4 2cos 3 2 22 1 2

θ− − ± − − ±

= = = ±

So, θ is a solution to the original equation ifNo solution since 3 2 2 1

cos 3 2 2 or cos 3 2 2θ θ+ >

= + = − .

Step 2: Find the values of θ whose cosine is 3 2 2− . Indeed, observe that one solution is ( )1cos 3 2 2 80.1207− − ≈ , which is in QI. A

second solution occurs in QIV, and has value 360 80.1207 279.88− = . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 80.12 , 279.88θ ≈ .

Step 3: Conclude that the solutions to the original equation are 80.12 , 279.88θ ≈ .

Chapter 7

494

57. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 22 tan tan 7 0θ θ− − = , we proceed as follows: Step 1: Simplify the equation algebraically. Since the left-side does not factor nicely, we apply the quadratic formula (treating cosθ as the variable):

( ) ( ) ( )( )( )

21 1 4 2 7 1 57tan2 2 4

θ− − ± − − − ±

= =

So, θ is a solution to the original equation if either 1 57 1 57tan or tan

4 4θ θ− += = .

Step 2: Solve 1 57tan4

θ −= .

To do so, we must find the values of θ whose tangent is 1 574

− .

Indeed, observe that one solution is 1 1 57tan 58.5874

− ⎛ ⎞−≈ −⎜ ⎟⎜ ⎟

⎝ ⎠, which is in QIV.

Since the angles we seek have positive measure, we use the representative 360 58.587 301.41− ≈ . A second solution occurs in QII, and has value 180 58.587 121.41− = . Since the input of tangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 121.41 , 301.41θ ≈ .

Step 3: Solve 1 57tan4

θ +=

To do so, we must find the values of θ whose tangent is 1 574

+ .

Indeed, observe that one solution is 1 1 57tan 64.934

− ⎛ ⎞+≈⎜ ⎟⎜ ⎟

⎝ ⎠, which is in QI. A

second solution occurs in QIII, and has value 180 64.93 244.93+ = . Since the input of tangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 64.93 , 244.93θ ≈ . Step 4: Conclude that the solutions to the original equation are

64.93 , 121.41 , 244.93 , 301.41θ ≈ .

Section 7.8

495

59. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 2csc (3 ) 2 0θ − = , we proceed as follows: Step 1: Simplify the equation algebraically. Factoring the left-side of 2csc (3 ) 2 0θ − = yields the equivalent equation

( )( )csc3 2 csc3 2 0θ θ− + = , which is satisfied whenever csc3 2θ = ± , or

equivalently 2sin 32

θ = ± . The values of θ for which this is true must satisfy

3 5 73 2 , 2 , 2 , 24 4 4 4

n n n nπ π π πθ π π π π= + + + + , where n is an integer,

which reduces to

34 2

nπ πθ = + , where n is an integer.

So, dividing by 3 then yields the values of θ in [ ]0, 2π for which this is true. Step 2: Convert the values obtained in Step 1 to degrees.

15 , 45 , 75 , 105 , 135 , 165 , 195 , 225 ,255 , 285 ,315 , 345θ =

61. By inspection, the values of x in [ ]0, 2π that satisfy the equation sin cosx x= are

5,4 4

x π π= .

63. Observe that

2

2

2

sec cos 21 cos 2

cos1 cos 2cos

cos 2cos 1 0(cos 1) 0

cos 1 0cos 1

x x

xx

x xx x

xx

x

+ = −

+ = −

+ = −

+ + =

+ =+ =

= −

The value of x in [ ]0, 2π that satisfies the equation cos 1x = − is x π= . Substituting this value into the original equation shows that it is, in fact, a solution to the original equation.

Chapter 7

496

65. Observe that

( )

( )( )

( )

2

2

2

3sec tan3

1 sin 3cos cos 3

1 sin 3cos 3

1 sin 1cos 3

1 sin 1 sin 11 sin 3

1 sin

x x

xx x

xx

xx

x xx

x

− =

− =

−=

−=

− −=

−− ( )

( )1 sin

1 sin

x

x

−

− ( )131 sin

3(1 sin ) 1 sin4sin 2

1sin2

x

x xx

x

=+

− = +=

=

The values of x in [ ]0,2π that satisfy the

equation 1sin2

x = are 5,6 6

x π π= .

Substituting these values into the original

equation shows that while 6π is a solution,

56π is extraneous. Indeed, note that

15 5 1 2sec tan6 6 3 3

2 23 3

3 3

π π⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ − −

= − ≠

So, the only solution is 6π .

67. Observe that

( )

( )( )

( )

2

2

2

csc cot 31 cos 3

sin sin1 cos 3

sin1 cos

3sin

1 cos 1 cos3

1 cos1 cos

x xx

x xx

xxx

x xx

x

+ =

+ =

+=

+=

+ +=

−+ ( )

( )1 cos

1 cos

x

x

+

+ ( )( )

31 cos

3 1 cos 1 cos3 3cos 1 cos

4cos 21cos2

x

x xx xx

x

=−

− = +

− = +=

=

The values of x in [ ]0,2π that satisfy the

equation 1cos2

x = are 5,3 3

x π π= .

Substituting this value into the original

equation shows that while3π is a solution,

53π is extraneous. Indeed, note that

15 5 1 2csc cot3 3 3 3

2 23 3 33

π π⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ − −

= − = − ≠

So, the only solution is 3π .

Section 7.8

497

69. Observe that

2

2

2

2sin csc 012sin 0

sin2sin 1 0

sin2sin 1 0

1sin2

x x

xx

xxx

x

− =

− =

−=

− =

=

1 1sin or sin2 2

x x= − =

The solutions to these equations in [ ]0, 2π

are x = 3 5 7, , ,4 4 4 4π π π π . Substituting

these into the original equation shows that they are all, in fact, solutions to the original equation.

71. Observe that

( )

sin 2 4cos2sin cos 4cos

2cos sin 2 0

x xx x x

x x

==

− =

No solution

cos 0 or sin 2 0cos 0 or sin 2

x xx x= − == =


are x = 3,2 2π π . Substituting these into the

original equation shows that they are all, in fact, solutions to the original equation.

73. Observe that

( )

2 sin tansin2 sincos

2 sin cos sin

sin 2 cos 1 0

x xxxx

x x x

x x

=

=

=

− =

sin 0 or 2 cos 1 01sin 0 or cos2

x x

x x

= − =

= =

The solutions to these equations in [ ]0, 2π are x = 70, , 2 , ,4 4π ππ π . Substituting these

into the original equation shows that they are all, in fact, solutions to the original equation.

Chapter 7

498

75. Observe that

( ) ( )( )

tan 2 cotsin 2 coscos 2 sin

cos cos 2 sin 2 sin 0

cos 2 0cos3 0

x xx xx x

x x x x

x xx

=

=

− =

+ =

=

Note that the solutions of cos3 0x = are 3 3 33 , 2 , 4 , , 2 , 4

2 2 2 2 2 2x π π π π π ππ π π π= + + + +

so that 5 3 7 11, , , , ,

6 6 2 2 6 6x π π π π π π= .

Substituting these into the original equation shows that they are all, in fact, solutions to the original equation. 77. Observe that

( )

3 sec 4sin

3 4sincos

3 4sin cos

3 2 2sin cos

3 2sin 2

3 sin 22

x x

xx

x x

x x

x

x

=

=

=

=

=

=

Next, the solutions of 3sin 22

x = are

2 22 , 2 , , 23 3 3 3

x π π π ππ π= + +

so that 7 4, , ,

6 3 6 3x π π π π= .

Substituting all of these into the original equation shows that they are all, in fact, solutions to the original equation.

79. Observe that

( )

( )

2

2 2 2

2 2 2

2

2

1sin cos 241sin cos sin41sin 1 sin sin413sin 14

1sin4

x x

x x x

x x x

x

x

− = −

− − = −

− − − = −

− = −

=

1 1sin or sin2 2

x x= − =


are x = 5 7 11, , ,6 6 6 6π π π π . Substituting

these into the original equation shows that they are all solutions to original equation.

Section 7.8

499

81. Observe that

( )

( )( )

2

2

2

cos 2sin 2 0

1 sin 2sin 2 0

sin 2sin 3 0sin 1 sin 3 0

x x

x x

x xx x

+ + =

− + + =

− − =

+ − =

No solution

sin 1 0 or sin 3 0sin 1 or sin 3

x xx x+ = − == − =

The solution to these equations in [ ]0,2π

is x = 32π . Substituting this into the

original equation shows that it is, in fact, a solution to the original equation.

83. Observe that

( )

( )( )

2

2

2

2

2sin 3cos 0

2 1 cos 3cos 0

2 2cos 3cos 02cos 3cos 2 0

2cos 1 cos 2 0

x x

x x

x xx x

x x

+ =

− + =

− + =

− − =

+ − =

No solution

2cos 1 0 or cos 2 01cos or cos 22

x x

x x

+ = − =

= − =


are x = 2 4,3 3π π . Substituting these into the

original equation shows that they are, in fact, solutions to the original equation.

85. Observe that

( )( )( )

( )( )

2 2

2 2

2

cos 2 cos 0

cos sin cos 0

cos 1 cos cos 0

2cos cos 1 02cos 1 cos 1 0

x x

x x x

x x x

x xx x

+ =

− + =

− − + =

+ − =

− + =

2cos 1 0 or cos 1 0

1cos or cos 12

x x

x x

− = + =

= = −


are x = 5, ,3 3π π π . Substituting these into

the original equation shows that they are, in fact, solutions to the original equation.

87. Observe that 14

12

sec 2 sin 21 sin 2

4cos 21 4sin 2 cos 21 2sin 4

sin 4

x x

xx

x xx

x

=

=

===

The solutions to this equation must satisfy 5 13 17 25 29 37 414 , , , , , , ,

6 6 6 6 6 6 6 6x π π π π π π π π=

and so, the solutions are 5 13 17 25 29 37 41, , , , , , ,

24 24 24 24 24 24 24 24x π π π π π π π π=

Chapter 7

500

89. In order to find all of the values of x in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 1cos(2 ) sin 02

x x+ = , we

proceed as follows: Step 1: Simplify the equation algebraically.

( )

2 2

2 2

2

2

1 1cos(2 ) sin 0 cos sin sin 02 2

11 sin sin sin 02

4sin sin 2 0

( 1) ( 1) 4(4)( 2) 1 33sin2(4) 8

x x x x x

x x x

x x

x

+ = ⇒ − + =

⇒ − − + =

⇒ − − =

− − ± − − − ±⇒ = =

Step 2: Solve 1 33sin8

x += .

To do so, we must find the values of x whose sine is 1 338

+ .

Indeed, observe that one solution is 1 1 33sin 57.478

− ⎛ ⎞+≈⎜ ⎟⎜ ⎟

⎝ ⎠, which is in QI. A

second solution occurs in QII, and has value 180 57.47 122.53− = . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately 57.47 , 122.53x ≈ .

Step 3: Solve 1 33sin8

x −= .

To do so, we must find the values of x whose sine is 1 338

− .

Indeed, observe that one solution is 1 1 33sin 36.388

− ⎛ ⎞−≈ −⎜ ⎟⎜ ⎟

⎝ ⎠, which is in QIV.

Since the angles we seek have positive measure, we use the representative 360 36.38 323.62− = . A second solution occurs in QIII, and has value 180 36.38 216.38+ = . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately 216.38 , 323.62x ≈ . Step 4: Conclude that the solutions to the original equation are

57.47 , 122.53 , 216.38 , 323.62x ≈ . Substituting these into the original equation shows that they are, in fact, solutions to the original equation.

Section 7.8

501

91. In order to find all of the values of x in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 26cos sin 5x x+ = , we proceed as follows: Step 1: Simplify the equation algebraically.

( )

( )( )

2

2

2

6cos sin 5

6 1 sin sin 5

6 6sin sin 53sin 1 2sin 1 0

x x

x x

x xx x

+ =

− + =

− + =

+ − =

1 13sin 1 0 or 2sin 1 0 so that sin or sin3 2

x x x x+ = − = = − =

Step 2: Solve 1sin3

x = − .

To do so, we must find the values of x whose sine is 13

− .


− ⎛ ⎞− ≈ −⎜ ⎟⎝ ⎠

, which is in QIV.

Since the angles we seek have positive measure, we use the representative 360 19.47 340.53− = . A second solution occurs in QIII, and has value 180 19.47 199.47+ = . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are 199.47 , 340.53x ≈ .

Step 3: Solve 1sin2

x = .

Indeed, observe that one solution is 1 1sin 302

− ⎛ ⎞ =⎜ ⎟⎝ ⎠


solution occurs in QII, and has value 180 30 150− = . Since the input of sine is simply x, and not some multiple thereof, we see that the solutions are 30 , 150x = . Step 4: Conclude that the solutions to the original equation are

30 , 150 , 199.47 , 340.53x ≈ . Substituting these into the original equation shows that they are all solutions.

Chapter 7

502

93. In order to find all of the values of x in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 2cot 3csc 3 0x x− − = , we proceed as follows: Step 1: Simplify the equation algebraically.

( )

( )( )

2

2

2

cot 3csc 3 0

csc 1 3csc 3 0

csc 3csc 4 0csc 1 csc 4 0

x x

x x

x xx x

− − =

− − − =

− − =

+ − =

csc 1 0 or csc 4 0csc 1 or csc 4

1sin 1 or sin4

x xx x

x x

+ = − == − =

= − =

Step 2: Solve sin 1x = − . To do so, we must find the values of x whose sine is 1− . Indeed, the only solution is ( )1sin 1 270− − = , which is in QIII.

Step 3: Solve 1sin4

x = .


− ⎛ ⎞ ≈⎜ ⎟⎝ ⎠


solution occurs in QII, and has value 180 14.48 165.52− = . Since the input of sine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are 14.48 , 165.52x ≈ . Step 4: Conclude that the solutions to the original equation are

14.48 , 165.52 ,270x ≈ . Substituting these into the original equation shows that they are, in fact, solutions to the original equation.

Section 7.8

503

95. In order to find all of the values of x in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 22sin 2cos 1 0x x+ − = , we proceed as follows: Step 1: Simplify the equation algebraically.

( )2

2

2

2

2

2sin 2cos 1 0

2 1 cos 2cos 1 0

2 2cos 2cos 1 02cos 2cos 1 0

( 2) ( 2) 4(2)( 1) 2 2 3 1 3cos2(2) 4 2

x x

x x

x xx x

x

+ − =

− + − =

− + − =

− − =

− − ± − − − ± ±= = =

1 3No solution since 12

1 3 1 3cos or cos2 2

x x

+>

− += = .

Step 2: Solve 1 3cos2

x −= .

To do so, we must find the values of x whose cosine is 1 32− .

Indeed, the only solution is 1 1 3cos 111.472

− ⎛ ⎞−≈⎜ ⎟⎜ ⎟

⎝ ⎠, which is in QII. A second

solution occurs in QIII and has value 360 111.47 248.53− = . Since the input of cosine is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately 111.47 , 248.53x ≈ . Step 3: Conclude that the solutions to the original equation are

111.47 , 248.53x ≈ . Substituting these into the original equation shows that they are, in fact, solutions to the original equation.

Chapter 7

504

97. Observe that ( ) ( )

( ) ( )( ) ( ) ( )( )( ) ( )

( ) ( )( )

( )

2 2116 4 4

1 14 4 4 4 4 4

14 4 4

14 4 4

1 14 2 4

12 2

csc cos 0

csc cos csc cos 0

csc cos

sin cos

sin 2

sin

x x

x x x x

x x

x x

x

x

− =

− + =

= ±

=±

=± ⋅

± =

The values of x that satisfy these equation must satisfy 5,2 6 6x π π= . So, the solutions are

5,3 3

x π π= .

99. Solving for x yields:

2,400 400sin 2,0006

400 400sin6

1 sin6

x

x

x

π

π

π

⎛ ⎞= +⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

This equation is satisfied when 6 2

xπ π= , so that 6 3

2x π

π= ⋅ = . So, the sales reach 2,400

in March.

Section 7.8

505


Let h = height of the trapezoid, and x = length of one base and two edges of the trapezoid, as labeled above. Note that α β θ= = since they are alternate interior angles. As such,

sin hx

θ = , so that sinh x θ= .

Furthermore, using the Pythagorean Theorem enables us to find z: 2 2 2z h x+ = , so that ( )

2

2 2 2 2 2 2

cos

( sin ) 1 sin cosz x h x x x xθ

θ θ θ

=

= − = − = − = .

Since 0 θ π≤ ≤ , we conclude that cosz x θ= . Hence, 1b x= and ( )2 2 2 cos 1 2cosb x z x x xθ θ= + = + = + .

Thus, the area A of the cross-section of the rain gutter is

( ) ( ) ( )

( )[ ][ ]

( )

1 2

2 (1 cos )

2

2

2

1 1 sin 1 2cos2 2

sin (1 cos )

sin sin cos

1sin 2sin cos2sin 2sin

2

x

A h b b x x x

x x

x

x

x

θ

θ θ

θ θ

θ θ θ

θ θ θ

θθ

= +

= + = + +⎡ ⎤⎣ ⎦

= +

= +

⎡ ⎤= +⎢ ⎥⎣ ⎦⎡ ⎤= +⎢ ⎥⎣ ⎦

θθ

α β

Chapter 7

506

103. Solving the equation 200 100sin 3002

xπ⎛ ⎞+ =⎜ ⎟⎝ ⎠

, for 2,000x > yields:

200 100sin 3002

100sin 1002

sin 12

x

x

x

π

π

π

⎛ ⎞+ =⎜ ⎟⎝ ⎠⎛ ⎞ =⎜ ⎟⎝ ⎠⎛ ⎞ =⎜ ⎟⎝ ⎠

Observe that this equation is satisfied when 22 2

x nπ π π= + , where n is an integer, so that

21 2 1 4x n nππ⎛ ⎞= + = +⎜ ⎟⎝ ⎠

, where n is an integer. So, the first value of n for which

1 4 2,000n+ > is 500n = . The resulting year is 1 4(500) 2001x = + = . 105. Use ( ) ( )sin sini i r rn nθ θ= with the given information to obtain

( ) ( )1.00sin 75 2.417sin rθ= so that

( ) ( )

( )1

1.00sin 75sin

2.4171.00sin 75

24 sin2.417

r

r

θ

θ−

=

⎛ ⎞⎜ ⎟≈ =⎜ ⎟⎝ ⎠

107. Observe that using the identity

sin 2 2sin cosA A A= with 3

A xπ= yields

2sin cos 3 43 3

sin 2 13

x x

x

π π

π

⎛ ⎞ ⎛ ⎞ + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⋅ =⎜ ⎟⎝ ⎠

This equation is satisfied when 23 2

xπ π= , so

that 34

x = . So, it takes 3 sec.4

for the

volume of air to equal 4 liters.

Section 7.8

507

109. First, we solve 2sin 2sin 2 0x x− + = on [ ]0, 2π :

[ ]( )

2sin 2sin 2 02sin 2 2sin cos 0

2sin 1 2cos 0

x xx x x

x x

− + =

− + =

− + =

so that 2sin 0 or 1 2cos 0

12sin 0 or cos2

x x

x x

= − + =

= =

These equations are satisfied when 50, , 2 , ,3 3

x π ππ π= . We now need to determine the

corresponding y-coordinates. x ( ) 2cos cos 2y x x x= − Point 0 ( ) ( ) ( )0 2cos 0 cos 2 0 2 1 1y = − ⋅ = − = ( )0,1

3π ( ) ( ) ( ) 1 1 32cos cos 2 23 3 3 2 2 2

y π π π ⎛ ⎞ ⎛ ⎞= − ⋅ = − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3,3 2π⎛ ⎞⎜ ⎟⎝ ⎠

53

π ( ) ( ) ( ) 1 1 35 5 52cos cos 2 23 3 3 2 2 2y π π π ⎛ ⎞ ⎛ ⎞= − ⋅ = − − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ 5 3,

3 2π⎛ ⎞

⎜ ⎟⎝ ⎠

π ( ) ( ) ( )2cos cos 2 2( 1) 1 3y π π π= − ⋅ = − − = − ( ), 3π − 2π ( ) ( ) ( )2 2cos 2 cos 2 2 2(1) 1 1y π π π= − ⋅ = − = ( )2 ,1π

So, the turning points are ( )0,1 , 3,3 2π⎛ ⎞⎜ ⎟⎝ ⎠

, 5 3,3 2π⎛ ⎞

⎜ ⎟⎝ ⎠

, ( ), 3π − , and ( )2 ,1π .

111. The value 32

πθ = does not satisfy the original equation. Indeed, observe that

32 sin 2 1 12π⎛ ⎞+ = − =⎜ ⎟

⎝ ⎠, while 3sin 1

2π⎛ ⎞ = −⎜ ⎟

⎝ ⎠. So, this value of θ is an extraneous

solution. 113. Cannot divide by cos x since it could be zero. Rather, should factor as follows:

( )

6sin cos 2cos6sin cos 2cos 0

2cos 3sin 1 0

x x xx x x

x x

=− =

− =

Now, proceed…

115. False. For instance, 3sin2

θ = has two solutions on [ ]0,2π , namely 2,3 3π πθ = .

117. True. This follows by definition of an identity.

Chapter 7

508

119. Solving the equation 4 216sin 8sin 1 0θ θ− + = on [ ]0, 2π yields

( )( )( )

4 2

22

16sin 8sin 1 0

4sin 1 0

2sin 1 2sin 1 0

θ θ

θ

θ θ

− + =

− =

− + =

So, θ satisfies the original equation if either 2sin 1 0 or 2sin 1 0θ θ− = + = .

Observe that 2sin 1 0θ − = is equivalent to 1sin2

θ = , which is satisfied when 5,6 6π πθ = .

Also, 2sin 1 0θ + = is equivalent to 1sin2

θ = − , which is satisfied when 7 11,6 6π πθ = .

So, we conclude that the solutions to the original equation are 5 7 11, , ,6 6 6 6π π π πθ = .

121. First, observe that using the addition formulae for sin( )A B± yields

sin sin sin cos cos sin4 4 4 4

x x x xπ π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + − = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

sin cos cos sin4 4

x xπ π⎡ ⎤ ⎛ ⎞ ⎛ ⎞+ −⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

( )2 2sin22 sin

x

x

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

=

=

We substitute this for the left-side of the given equation to obtain: 2sin sin

4 4 2

22 sin2

1sin2

x x

x

x

π π⎛ ⎞ ⎛ ⎞+ + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

=

The smallest positive value of x for which this is true is or 306

x π= .

123. Observe that using a half-angle formula, we see that ( )( ) ( )3 2

63

1 costan

1 cos

xx

x

−=

+, thereby

resulting in the equivalent equation ( )26tan 1x = − , which has no solution.

Section 7.8

509

125. Factoring the left-side of ( )44csc 4 0π θ π− − = yields the equivalent equation

( )( ) ( )( )2 24 4csc 2 csc 2 0π πθ π θ π− − − + = , which is satisfied whenever either

( )24csc 2 0π θ π− − = or ( )2

4csc 2 0π θ π− + = . The second equation has no solution since the left-side is always greater than or equal to 2. The first equation is equivalent to

( )2 14 2sin π θ π− = , which holds whenever ( ) 2

4 2sin π θ π− = ± . The values of θ for which this is true must satisfy

4 4 2nπ π πθ π− = + , where n is an integer.

Solving this equation for θ yields the values of θ in for which this is true: 5 2 , where is an integer .n nθ = +

127. Consider the graphs below of 1 2sin , cos 2y yθ θ= = .

Observe that the solutions of the equation sin cos 2θ θ= on [ ]0,π are approximately

( ) ( )1 0.524,0.5 , 2 2.618,0.5P P .

The exact solutions are 5,6 6π π .

129. Consider the graphs below of 1 2sin , secy yθ θ= = .

Since the curves never intersect, there are no solutions of the equation sin secθ θ= on [ ]0,π .

Chapter 7

510

131. Consider the graphs below of 1 2sin ,y y eθθ= = .

First, note that while there are no positive solutions of the equation sin eθθ = , there are infinitely many negative solutions (at least one between each consecutive pair of x-intercepts). They are all irrational, and there is no apparent closed-form formula that can be used to generate them.

133. To determine the smallest positive solution (approximately) of the equation sec3 csc 2 5x x+ = graphically, we search for the intersection points of the graphs of the following two functions:

1 2sec3 csc 2 , 5y x x y= + =

The x-coordinate of the intersection point is in radians. Observe that the smallest positive solution, in degrees, is approximately 7.39 .

Young CAT2e SSM Ch7 Part2

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