Section 7.6 Solutionsfaculty.oprfhs.org/cavalos/Book Chapters and Book Solutions/CATA...cos 3 7 cos 2 7 2sin sin()() 37 2 7 37 2 7 22 757 7 57 2sin sin 2sin sin 22 2 2 xx x x xx x

Chapter 7

892

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= + + − = +

2.

[ ]

[ ] [ ]

1cos10 sin 5 sin(5 10 ) sin(5 10 )21 1sin15 sin( 5 ) sin15 sin 52 2

x x 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

= ⋅ − − +

= − − = −

4.

[ ]

[ ] [ ]

13sin 2 sin 4 3 cos(2 4 ) cos(2 4 )2

3 3cos( 2 ) cos 6 cos 2 cos 62 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

− = ⋅ − + + − −

= + − = +

6.

[ ]

[ ] [ ]

18cos3 cos5 8 cos(3 5 ) cos(3 5 )2

4 cos8 cos( 2 ) 4 cos8 cos 2

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

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

= − − = −

Section 7.6

893

8.

[ ] [ ]

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

1 1cos( 2 ) cos3 cos 2 cos32 2

x x x x x x

x x x x

π π π π π π

π π π π

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

= − − = −

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

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

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

10. 1sin cos sin sin

4 2 2 4 2 4 2

1 3 1 3sin sin sin sin2 4 4 2 4 4

x x x x x x

x x x x

π π π π π π

π π π π

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

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

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− = − + − = − +

12. 2sin(2.1 )sin(3.4 ) cos( 1.3 ) cos(5.5 ) cos(1.3 ) cos(5.5 )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⎡ ⎤ ⎡ ⎤− = + − = −⎣ ⎦ ⎣ ⎦

14.

( )2 5 2 5 4 2 6 2 5 4 25cos sin sin sin sin sin 2 23 3 2 3 3 2 3

x x x x x x⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

− − = − + − = − −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

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

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

⎝ ⎠ ⎝ ⎠

16. 2 4 2 4cos 2 cos 4 2sin sin 2sin(3 )sin( ) 2sin 3 sin

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

⎝ ⎠ ⎝ ⎠

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

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

⎝ ⎠ ⎝ ⎠

Chapter 7

894

18. 10 5 10 5 15 5sin10 sin 5 2sin cos 2sin cos

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

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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

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

⎝ ⎠ ⎝ ⎠

20.

( )5 5

5 3 32 2 2 2cos cos 2sin sin 2sin sin 2sin sin2 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

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

⎝ ⎠ ⎝ ⎠

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

22. 2 7 2 7

2 7 3 3 3 3sin sin 2sin cos3 3 2 2

3 5 3 52sin cos 2sin 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

+ −⎛ ⎞ ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − =

24.

( ) ( ) 0.3 0.5 0.3 0.5cos 0.3 cos 0.5 2sin sin2 2

2sin(0.4 )sin( 0.1 ) 2sin(0.4 )sin(0.1 )

x x x xx x

x x x x

+ −⎛ ⎞ ⎛ ⎞− = − ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − − =

Section 7.6

895

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

⎛ ⎞ ⎛ ⎞− +− = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= − = −

26.

( ) ( ) 3 7 2 7 3 7 2 7cos 3 7 cos 2 7 2sin sin2 2

7 5 7 7 5 72sin sin 2sin sin2 2 2 2

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

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

⎝ ⎠ ⎝ ⎠

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

28.

( ) ( )

3 5 3 53 5 4 4 4 4sin sin 2sin cos4 4 2 2

2sin cos 2sin cos4 4

x x x xx x

x x x x

π π π ππ π

π ππ 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

−⎛ ⎞= − = −⎜ ⎟− ⎝ ⎠⎛ ⎞

⎜ ⎟⎝ ⎠

30.

2sin 4 sin 2cos 4 cos 2

x xx x+

=−

4 2sin2

x x+⎛ ⎞⎜ ⎟⎝ ⎠

4 2cos2

2

x x−⎛ ⎞⎜ ⎟⎝ ⎠

−4 2sin

2x x+⎛ ⎞

⎜ ⎟⎝ ⎠

4 2cot cot24 2sin

2

x x xx x

−⎛ ⎞= − = −⎜ ⎟− ⎝ ⎠⎛ ⎞

⎜ ⎟⎝ ⎠

Chapter 7

896

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=

32.


x xx x+

=+

4 2 4 2sin cos2 2

x x x x+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 4 2 4 2cos cos2 2

x x x x+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

4 2tan tan 32

x x 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−⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

34.


x xx x−

=−

7 2 7 2cos sin2 2

x x x x+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2− 7 2 7 2sin sin2 2

x x x x+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

7 2 9cot 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+⎛ ⎞= ⎜ ⎟⎝ ⎠

36.

2sin sincos cos

A BA B−

=+

sin cos2 2

A B A B− +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 cos2

A B+⎛ ⎞⎜ ⎟⎝ ⎠

tan2

cos2

A BA B

−⎛ ⎞= ⎜ ⎟− ⎝ ⎠⎛ ⎞

⎜ ⎟⎝ ⎠

Section 7.6

897

37.

2cos cossin sin

A BA B

−−

=+

sin2

A B+⎛ ⎞⎜ ⎟⎝ ⎠

sin2

2

A B−⎛ ⎞⎜ ⎟⎝ ⎠

sin2

A B+⎛ ⎞⎜ ⎟⎝ ⎠

tan2

cos2

A BA B

−⎛ ⎞= − ⎜ ⎟− ⎝ ⎠⎛ ⎞

⎜ ⎟⎝ ⎠

38.

2cos cossin sin

A BA B

−−

=−

sin sin2 2

A B A B+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 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

+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= ⋅

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

+ −⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

40.

2cos coscos cos

A BA B

−−

=+

sin sin2 2

2

A B A B+ −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

sin sin2 2

cos cos cos cos2 2 2 2

tan tan2 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

= =

Chapter 7

898

42.

2sin2cos( ) cos( )

sin( ) sin( )

A B A BA B A BA B A B

− + +⎛ ⎞− ⎜ ⎟− − + ⎝ ⎠

=+ + −

sin2

2sin2

A B A B

A B A B

− − −⎛ ⎞⎜ ⎟⎝ ⎠

+ + −⎛ ⎞⎜ ⎟⎝ ⎠

cos2

sin( ) sin tancos cos

A B A B

B B BB B

+ − +⎛ ⎞⎜ ⎟⎝ ⎠

−= − = =

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+

= .

44. Description of F note: ( )cos 2 (349)tπ Description of A note: ( )cos 2 (440)tπ

Combining the two notes: ( ) ( )cos 698 cos 880t tπ π+ . Using the sum-to-product identity then yields:

( ) ( )

( ) ( )( ) ( )

698 880 698 880cos 698 cos 880 2cos cos2 2

2cos 789 cos 912cos 789 cos 91 (since cosine is even)

t t t tt t

t tt t

π π π ππ π

π ππ π

+ −⎛ ⎞ ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −=

The beat frequency is 440 – 349 = 91 Hz.

The average frequency is 440 349 394.5 Hz2+

= .

Section 7.6

899

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

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

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

46. Using the same model as in Exercise 33, we find that:

The beat frequency is 66 6

1 1 10 Hz1.55 10 0.63 10 1.55 0.63

c c c− −

⎡ ⎤− = −⎢ ⎥× × ⎣ ⎦

The average frequency is 6 6

61 11.55 10 0.63 10 10 Hz2 2 1.55 0.63

c cc− −+ ⎡ ⎤× × = +⎢ ⎥⎣ ⎦

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

+ −⎛ ⎞ ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − =

48.

( ) ( )

( ) ( ) ( ) ( )

1394 2954 1394 2954sin 2 (697) sin 2 (1477) 2sin cos2 2

2sin 2174 cos 780 2sin 2174 cos 780

t t t tt t

t t t t

π π π ππ π

π π π π

+ −⎛ ⎞ ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − =

Chapter 7

900

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

⎡ ⎤⋅ − − +⎣ ⎦=

⎡ ⎤−⎣ ⎦=

⎡ ⎤−⎢ ⎥ − −⎣ ⎦= = =

−= ≈

50. Note that 75 45 180A+ + = , so that 60A = . So, the area is

( ) ( ) ( )( )

( ) ( )( )

( ) ( )( )

( ) ( )

( )

2

2

2

2 2 2

2 2

1144 cos 75 45 cos 75 4512 in. sin 75 sin 45 2 in.2sin 60 2sin 60

36 cos 30 cos 120in.

sin 60

3 136 36 3 1 36 3 1 32 2in. in. in.

33 32

36 12 3 in. 56.78 in.

⎡ ⎤⋅ − − +⎣ ⎦=

⎡ ⎤−⎣ ⎦=

⎡ ⎤+⎢ ⎥ + +⎣ ⎦= = =

= + ≈

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. 52. In general, sin cos sin cos 0A B B A− ≠ . Should have used the product-to-sum identity to simplify this. 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. 54. False. From the sum-to-product identities, we have

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

A B A B A B= − − + ,

and the right-side is not, in general, expressible as the sine of a product.

Section 7.6

901

55. True. From the product-to-sum identities, we have

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

A B A B A B= + + − .

56. True. From the product-to-sum identities, we have

[ ]1sin sin 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. 58. Observe that

[ ]

( )( )

( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

cos cos cos cos cos cos

1 cos cos( ) cos2

1 cos cos cos( ) cos21 1 cos cos2 2

1 cos cos2

1 cos cos cos cos4

A B C A B C

A B A B C

A B C A B C

A B C A B C

A B C A B C

A B C A B C 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. 59.

[ ] [ ]1 12 2cos cos sin sin cos( ) cos( ) cos( ) cos( )cos( )

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

− = + + − − − + +

= +

Chapter 7

902

60. [ ] [ ]

( )( )

1 12 2

12

12

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

sin( ) sin ( )

sin( ) sin

sin( )

A B B A A B A B B A B A

A B A B

A B A B

A 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:

62. Observe that

[ ][ ]

4sin(2 1)cos(2 )2 sin(2 1 2 ) sin(2 1 2 )

2 sin( 1) sin(3 3)

y x xx x x x

x x

= − −

= − + − + − − +

= + + −


Section 7.6

903

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

= −

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


64. Observe that

[ ][ ] [ ]

12

1 12 2

cos(2 )sin(3 )sin(2 3 ) sin(2 3 )

sin(5 ) sin( ) sin(5 ) sin

y x x xx x x x x

x x x x x x

= −

= − + + −

= − + − = − −


Chapter 7

904

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=

= = = ⋅ =

66. Consider the graph of 1 tan tan 2y x x= + , as seen below:

From the graph, it seems as though this function is equivalent tosec 2x . We prove this identity below:

sin 2sin cossin sin 21 tan tan 2 1 1cos cos 2

x x xx xx xx x

+ = + ⋅ = +( )

cos x

( )2 2 22 2

2 2

cos 2

cos sin 2sin2sin cos 2 2sin1cos 2 cos 2 cos 2

cos sin 1 sec 2cos 2 cos 2

x

x x xx x xx x x

x x xx x

− ++= + = =

+= = =

Section 7.6

905

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.

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

[ ]

1

2

13 2

cos 4 cos 2 (solid)cos 6 (dashed)

cos 6 cos 2

y x xy xy x x

==

= +

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

Chapter 7

906

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 = .

2. The equation 2arccos2

θ⎛ ⎞− =⎜ ⎟⎜ ⎟⎝ ⎠

is

equivalent to 2cos2

θ = − . Since the range


34πθ = .

3. The equation 3arcsin2

θ⎛ ⎞− =⎜ ⎟⎜ ⎟⎝ ⎠

is

equivalent to 3sin2

θ = − . Since the

range of arcsine is ,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, we conclude

3πθ = − .

4. The equation 1arcsin2

θ⎛ ⎞ =⎜ ⎟⎝ ⎠

is equivalent

to 1sin2

θ = . Since the range of arcsine is

,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, we conclude that 6πθ = .

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 = .

6. The equation 1 3tan3

θ− ⎛ ⎞=⎜ ⎟⎜ ⎟

⎝ ⎠ is

equivalent to 13 sin2tan

3 cos32

θθθ

= = = .

Since the range of inverse tangent is

,2 2π π⎛ ⎞−⎜ ⎟

⎝ ⎠, we conclude that

6πθ = .

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 = .

8. The equation ( )arccsc 1 θ− = is equivalent to csc 1θ = − , which is further the same as sin 1θ = − . Since the range of inverse cosecant is ,0 0,

2 2π π⎡ ⎞ ⎛ ⎤− ∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦

, we

conclude that2πθ = − .

Section 7.7

907

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 = .

10. The equation ( )1sec 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 23πθ = .

11. The equation ( )arc tan 3 θ− = is

equivalent to 3 sin2tan 3 1 cos2

θθθ

= − = − = .


,2 2π π⎛ ⎞−⎜ ⎟


3πθ = − .

12. The equation ( )arccot 3 θ= is

equivalent to 3 cos2cot 3 1 sin2

θθθ

= = = .

Since the range of inverse cotangent is

( )0,π , we conclude that 6πθ = .

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

2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, we conclude 0θ = .

14. The equation ( )arc tan 1 θ= is equivalent

to 2 sin2tan 1

cos22

θθθ

= = = . Since the range

of inverse tangent is ,2 2π π⎛ ⎞−⎜ ⎟

⎝ ⎠, we conclude

that 4πθ = .

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= .

16. 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 = .

17. The equation 1 1cos2

θ− ⎛ ⎞ =⎜ ⎟⎝ ⎠

is

equivalent to 1cos2



3πθ = , which corresponds to 60θ = .

18. The equation 1 3cos2

θ− ⎛ ⎞− =⎜ ⎟⎜ ⎟⎝ ⎠

is

equivalent to 3cos2


of arccosine is [ ]0,π , we conclude that 56πθ = , which corresponds to 150θ = .

Chapter 7

908

19. The equation 1 2sin2

θ− ⎛ ⎞=⎜ ⎟⎜ ⎟

⎝ ⎠ is

equivalent to 2sin2


of arcsine is ,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, we conclude that

4πθ = , which corresponds to 45θ = .

20. The equation ( )1sin 0 θ− = is equivalent to sin 0θ = . Since the range of arcsine is

,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

, we conclude that 0θ = , which

corresponds to 0θ = .

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 = .

22. The equation ( )1tan 3 θ− = is

equivalent to 3 sin2tan 3 1 cos2

θθθ

= = = .


,2 2π π⎛ ⎞−⎜ ⎟


3πθ = , which


23. The equation 3arc tan3

θ⎛ ⎞

=⎜ ⎟⎜ ⎟⎝ ⎠

is

equivalent to 13 sin2tan

3 cos32

θθθ

= = = .


,2 2π π⎛ ⎞−⎜ ⎟


6πθ = , which


24. The equation ( )arccot 1 θ= is equivalent

to 2 cos2cot 1

sin22

θθθ

= = = . Since the range

of inverse cotangent is ( )0,π , we conclude

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

Section 7.7

909

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

the same as 1sin2


of inverse cosecant is ,0 0,2 2π π⎡ ⎞ ⎛ ⎤− ∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦

, we

conclude that6πθ = − , which corresponds

to 30θ = − .

26. The equation 1 2 3csc3

θ− ⎛ ⎞− =⎜ ⎟⎜ ⎟⎝ ⎠

is

equivalent to 2 3csc3

θ = − , which is further

the same as 3 3sin22 3

θ = − = − . Since the

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

,

we conclude that3πθ = − , which corresponds

to 60θ = − .

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 = .

28. 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 that 4πθ = − which corresponds to

45θ = − . 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 = − .

30. The equation ( )arc tan 1 θ− = is

equivalent to 2 sin2tan 1

cos22

θθθ

= − = − = .


,2 2π π⎛ ⎞−⎜ ⎟


4πθ = − , which

corresponds to 45θ = − . 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 = .

32. 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

thatθ π= , which corresponds to 180θ = .

33. ( )1cos 0.5432 57.10− ≈ 34. ( )1sin 0.7821 51.45− ≈

35. ( )1tan 1.895 62.18− ≈ 36. ( )1tan 3.2678 72.99− ≈

Chapter 7

910

37.

( )1 1 1sec 1.4973 cos 48.101.4973

− − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

38.

( )1 1 1sec 2.7864 cos 68.972.7864

− − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

39.

( )1 1 1csc 3.7893 sin 15.303.7893

− − ⎛ ⎞− = ≈ −⎜ ⎟−⎝ ⎠

40.

( )1 1 1csc 6.1324 sin 9.396.1324

− − ⎛ ⎞− = ≈ −⎜ ⎟−⎝ ⎠

41.

( )1 1 1cot 4.2319 tan4.2319

180 13.30 166.70

π− − ⎛ ⎞− = + ⎜ ⎟−⎝ ⎠

≈ − =

42.

( )1 1 1cot 0.8977 tan0.8977

180 48.09 131.91

π− − ⎛ ⎞− = + ⎜ ⎟−⎝ ⎠

≈ − =

43. ( )1sin 0.5878 0.63− − ≈ − 44. ( )1sin 0.8660 1.05− ≈

45. ( )1cos 0.1423 1.43− ≈ 46. ( )1tan 0.9279 0.75− − ≈ −

47. ( )1tan 1.3242 0.92− ≈ 48.

( )1 1 1cot 2.4142 tan 0.392.4142

− − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

49.

( )1 1 1cot 0.5774 tan 2.090.5774

π− − ⎛ ⎞− = + ≈⎜ ⎟−⎝ ⎠

50.

( )1 1 1sec 1.0422 cos 2.861.0422

− − ⎛ ⎞− = ≈⎜ ⎟−⎝ ⎠

51.

( )1 1 1csc 3.2361 sin 0.313.2361

− − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

52.

( )1 1 1csc 2.9238 sin 0.352.9238

− − ⎛ ⎞− = ≈ −⎜ ⎟−⎝ ⎠

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

since 52 12 2π π π

− ≤ ≤ .

54. 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.

56. ( )( )1sin sin 1.1− is undefined since 1.1 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− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

.

58. 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− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

.

Section 7.7

911

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

.

60. Note that we need to use the angle θ in [ ]0,π such that 5cos cos3πθ ⎛ ⎞= −⎜ ⎟

⎝ ⎠ . To this

end, observe that 1 15cos cos cos cos3 3 3π π π− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

.

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

62. Note that we need to use the angle θ in [ ]0,π such that 5cot cot4πθ ⎛ ⎞= ⎜ ⎟

⎝ ⎠ . To this

end, observe that 1 15cot cot cot cot4 4 4π π π− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

.

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

.

64. 1 1sec sec2

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

is undefined since 12

is not in the domain of inverse secant.

65. 1 1csc csc2

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

is undefined since 12

is not in the domain of inverse cosecant.

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

such that 7csc csc .6πθ ⎛ ⎞= ⎜ ⎟

⎝ ⎠

To this end, observe that 1 17csc csc csc csc6 6 6π π π− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

.

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

68. Note that we need to use the angle θ in [ ]0,π such that cot cot4πθ ⎛ ⎞= −⎜ ⎟

⎝ ⎠ . To this

end, observe that 1 1 3 3cot cot cot cot4 4 4π π π− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

.

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

xπ π− < < , we see that 1tan tan

4 4π π− ⎛ ⎞⎛ ⎞− = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

.

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

xπ π− < < , we see that 1tan tan

4 4π π− ⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

.

71. Not possible

Chapter 7

912

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

3 33cot cot cotπ π− −= − =

74. ( ) ( )1 1tan tan8 tan 0 0π− −= =

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

76. Not possible 77. Let 1 3sin

4θ − ⎛ ⎞= ⎜ ⎟

⎝ ⎠. Then, 3sin

4θ = , 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−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

78. Let 1 2cos3

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 2cos3

θ = , as


Using the Pythagorean Theorem, we see that

2 2 22 3z + = , so that 5z = .

Hence, 1 2 5sin cos sin3 3

θ−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

θ θ

Section 7.7

913

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−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

80. Let 1 7tan24

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 7tan24

θ = , as



2 2 27 24 z+ = , so that 25z = .

Hence, 1 7 24cos tan cos24 25

θ−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

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−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

82. Let 1 2cos5

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 2cos5

θ = , as



2 2 22 5z + = , so that 21z = .

Hence, 1 2 21tan cos tan5 2

θ−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

θθ

θθ

Chapter 7

914

83. Let 1 2sin5

θ − ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠. Then, 2sin

5θ = ,

as shown in the diagram:

Using the Pythagorean Theorem, we see

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

1 3 1 5 5 23sec sin sec4 cos 2323

θθ

−⎛ ⎞⎛ ⎞ = = = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

84. Let 1 7cos4

θ − ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠. Then, 7cos

4θ = , as



( )22 27 4z + = , so that 3z = . So,

1 7 1 4 4 7sec cos sec4 cos 77

θθ

−⎛ ⎞⎛ ⎞

= = = =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

Alternatively, you can use the formula 1 1 1sec cosx

x− − ⎛ ⎞= ⎜ ⎟

⎝ ⎠ here. Indeed,

1 1

1

7 1sec cos sec cos 447

4 4 4 7sec sec77 7

− −

−

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

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

θ2

θ

7

Section 7.7

915

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

−⎛ ⎞⎛ ⎞ = = = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

.

86. Let 1 1sin4

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 1sin4

θ = , as



2 2 21 4z + = , so that 15z = . Hence,

1 1 1csc sin csc 44 sin

θθ

−⎛ ⎞⎛ ⎞ = = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

Alternatively, you can use the formula 1 1 1csc sinx

x− − ⎛ ⎞= ⎜ ⎟

⎝ ⎠ here. Indeed, observe

that

( )( )

1 1

1

1 1csc sin csc sin 14 4

csc csc 4 4

− −

−

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

= =

θ θ

Chapter 7

916

87. Let 1 60sin61

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 60sin61

θ = ,

as shown in the diagram:

Using the Pythagorean Theorem, we see that 2 2 260 61z + = , so that 11z = . Hence,

1 60 11cot sin cot61 60

θ−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

88. Let 1 41sec9

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠

. Then, 41sec9

θ = , so

that 9cos41

θ = , as shown in the diagram:


2 2 29 41z + = , so that 40z = . Hence,

1 41 9cot sec cot9 40

θ−⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

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 ≈ = .

90. Use ( )sin 2i I f tπ= with f = 100 and I = 240. Find the smallest positive value of t for which i = 100. To this end, observe

( )

( )

1

240sin 2 100 100100sin 200240

100sin240 0.000684

200

t

t

t

π

π

π

−

⋅ =

=

⎛ ⎞⎜ ⎟⎝ ⎠= ≈

So, 0.000684 sec. 0.68 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.

θθ

Section 7.7

917

92. Given that ( ) 12 2.4sin(0.017 1.377)H t t= + − , we must find the value of t for which ( ) 9.6H t = . To this end, we have

1

12 2.4sin(0.017 1.377) 9.6sin(0.017 1.377) 1

0.017 1.377 sin ( 1) 21.3772 11.4

0.017

tt

t

t

π

π

−

+ − =− = −

− = − = −

− += ≈ −

So, counting backward into December 12 days implies this corresponds to Dec.19. 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.

94. We need to find the smallest value of t larger than 11.3 for which

( )12.5cos 0.157 2.5 15t + = . We approach this graphically, where the solid graph corresponds to the left-side of the equation. We have:

This occurs at approximately 40 years.

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+

= + =−

+= = =

− −⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

α

β

Chapter 7

918

96. Using Exercise 95, we see that 2

8tan7

xx

θ =−

, so that 12

8tan7

xx

θ − ⎛ ⎞= ⎜ ⎟−⎝ ⎠. Thus,

we have the following specific calculations:

x = 10: 1 80tan 0.71 radians 4193

θ − ⎛ ⎞= ≈ ≈⎜ ⎟⎝ ⎠

x = 20: 1 160tan 0.39 radians 22393

θ − ⎛ ⎞= ≈ ≈⎜ ⎟⎝ ⎠

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

−

−

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

⎢ ⎥⎣ ⎦

98. Use the formula 12 tan

12

kf dM

π

−⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥= −

⎢ ⎥⎢ ⎥⎣ ⎦

with 3m 0.003 km and 2.5 kmf d= = = .

We have the following specific calculation:

k = 5 km: ( )1

1

52 tan0.003 0.00152.51 2 tan 2 0.00044 km 0.44m

2M π

π π

−

−

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

⎢ ⎥⎣ ⎦

k = 10 km: ( )1

1

102 tan0.003 0.00152.51 2 tan 4 0.00023 km 0.23m

2M π

π π

−

−

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

⎢ ⎥⎣ ⎦

Section 7.7

919


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 − −⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠.


Since 280tan200 x

α =−

, we have

1 280tan200 x

α − ⎛ ⎞= ⎜ ⎟−⎝ ⎠.

Also, since 130tanx

β = , we have

1 130tanx

β − ⎛ ⎞= ⎜ ⎟⎝ ⎠

.

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

1 1280 130tan 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 .

102. The identity ( )1cos cos x x− = is valid only for x in the interval[ ]0,π , not ,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

.

103. In general, 11

1cottan

xx

−−≠ .

104. In general, 11

1cscsin

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.

θα β

θα β

Chapter 7

920

106. True. First, judging from the graph of 1 1 1csc siny xx

− − ⎛ ⎞= = ⎜ ⎟⎝ ⎠

, it seems as though the

inverse cosecant function is odd. To prove this, observe that since the inverse sine function is odd, we obtain

( ) ( )1 1 1 11 1csc sin sin cscx xx x

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

From the definition of odd, we are done. (Note: In general, if f and g are both odd functions, then the composition f g is odd on its domain.) 107. False. This holds only on a subset of the domain to which cosecant is restricted in order to define its inverse.

108. False. In general, 11

1cscsin

θθ

−−≠ . For instance, let 1

2x = . Observe that

( )1 12csc 2 0− ⋅ = , but 1 1sin (1) csc (1) 0 1− −⋅ = ≠ .

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).

110. 1 1csc2

− ⎛ ⎞⎜ ⎟⎝ ⎠

does not exist since 12

is not in the domain of the inverse cosecant

function (which coincides with the range of the cosecant 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− −⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − = − = −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦

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

112. Observe that ( ) ( )( ) ( )1 13 32 2sin 2sin cos sin sin(3 ) 3x x x x− −= = provided that

2 23xπ π− ≤ ≤ , so that 6 6xπ π− ≤ ≤ .

Section 7.7

921

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− ⎛ ⎞= ⋅ = =⎜ ⎟

⎝ ⎠.

114. Consider the function ( ) 2 4sin2

f x x π⎛ ⎞= − −⎜ ⎟⎝ ⎠

.

a. Note that this function has a phase shift of 2π units to the right. So, we take the

interval used to define 1sin x− , namely ,2 2π π⎡ ⎤−⎢ ⎥⎣ ⎦

and add 2π to both endpoints to

get the interval [ ]0,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

2 4sin2

2 sin4 22sin

4 22sin

2 4

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 2( ) sin2 4

xf x π− − −⎛ ⎞= + ⎜ ⎟⎝ ⎠

The domain of 1f − is equal to the range of f. Since the amplitude of f is 4 and there is a vertical shift up of 2 units, we see that the range of f is [ ]2,6− . Hence, the

domain of 1f − is [ ]2,6− .

Chapter 7

922

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 .

116. Consider the function ( )3( ) 1 tanf x x π= − + .

(a) We know that tany x= is 1-1 on ( )2 2,π π− . As such, since ( )3tany x π= + is simply a horizontal shift of tany x= to the left 3

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

( ) ( )52 3 2 3 6 6, ,π π π π π π− − − = − . Since reflecting over the x-axis 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

6 6,π π− , we determine the inverse as follows:

( )( )

( )( )

( )

3

3

3

13

13

1 tan

1 tan

tan 1

tan 1

tan 1

y x

x y

y x

y x

y x

π

π

π

π

π

−

−

= − +

= − +

+ = −

+ = −

= − + −

Hence, the inverse is ( )1 13( ) tan 1f x xπ− −= − + − with domain .

Section 7.7

923

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 .

118. Consider the function ( )4( ) csc 1f x xπ= − − .

(a) We know that cscy x= is 1-1 on ) (2 2,0 0,π π− ∪⎡ ⎤⎣ ⎦. As such, ( )4csc 1y xπ= − is 1-1 on the following interval:

( ) ( ) ( ) ( )2 4 4 2

4 4 4 42 2

4 4 4 4

1 0 or 0 1

1 1 or 1 1

2 or 2

x x

x x

x x

π π π π

π ππ π π π

π π π π

− ≤ − < < − ≤

− ≤ < < ≤ +

− ≤ < < ≤ +

Since multiplying by a constant does not affect whether or not it is 1-1, we conclude that f is 1-1 on this set as well. (b) Restricting our attention to x values listed above, we determine the inverse as follows:

( )( )

( )

4

4

4

14

14

csc 1

csc 1

csc 1

csc ( ) 1

1 csc ( )

y x

x y

x y

x y

x y

π

π

π

π

π

−

−

= − −

= − −

− = −

− = −

⎡ ⎤+ − =⎣ ⎦

Hence, the inverse is 1 14( ) 1 csc ( )f x xπ− −⎡ ⎤= + −⎣ ⎦ with domain ( ] [ ), 1 1,−∞ − ∪ ∞ .

Chapter 7

924

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− ≤ ≤ . 120. The graphs of the following two functions on the interval [ ]3,1− is below:

( )11 cos cosY x−= , 2Y x=

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

Section 7.7

925

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⎡ ⎞ ⎛ ⎤∈ − ∪⎟ ⎜⎢ ⎥⎣ ⎠ ⎝ ⎦.

122. The graphs of the following two functions on the interval [ ]0,π is below:

( )11 sec secY x−= , 2Y x=

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

( )1sec sec x x− = holds when 0 x π≤ ≤ .

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.

Chapter 7

926

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

a. ( )( )

1 23 3

22 8193

22 tan 3tan 21 tan 41

xxx

− −= = = = −

− − −

b. ( )1 110

sin 0.32175x −= − ≈ − . So,

( )tan 2 tan 0.64350 0.7500x = − = − 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 .

2. The values of θ in [ ]0, 2π that satisfy 2sin2

θ = − are θ = 5 7,4 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= + + = .

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

θ = − . The values of θ in [ ]0, 4π

that satisfy 1cos2

θ = − are

2 2 4 4 2 4 8 10, 2 , , 2 , , ,3 3 3 3 3 3 3 3π 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 . 6. The only way cot 0θ = is for cos 0θ = . The values of θ in that satisfy

cos 0θ = (and hence, the original equation) are θ =(2 1) , where is an integer

2n nπ+ .

10

Section 7.8

927

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 =

8. The values of θ in [ ]0, 2π that satisfy 3cos 22

θ = must satisfy

11 11 11 13 232 , 2 , , 2 , , ,6 6 6 6 6 6 6 6π π π π π π π πθ π π= + + = .


11 13 23, , ,12 12 12 12π π π πθ =

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

⎛ ⎞ ⎛ ⎞= + + = + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + +

10. The values of θ in that satisfy cos 12θ⎛ ⎞ = −⎜ ⎟⎝ ⎠

must satisfy

( )2 12

nθ π= + , where n is an integer.

So, multiplying by 2 yields the following values ofθ which satisfy the original equation: ( )2 2 1 , where is an integern nθ π= + .

Chapter 7

928

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

= − − − −

= − − − −

12. The values of θ in that satisfy tan 2 3θ = − must satisfy 2 52 2 , 23 3

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

This reduces to 223

nπθ π= + , where n is an integer.


( )2 3, where is an integer

3 2 6nn n

ππ πθ+

= + = .

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 = − − .

14. First, observe that 2 3csc3

θ = is equivalent to 3 3sin22 3

θ = = . The values of θ

in [ ],π π− that satisfy 3sin2

θ = are 2,3 3π πθ = .

Section 7.8

929

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




6 4 12nn n

ππ πθ+

= + = .

16. The values of θ in that satisfy tan 5 1θ = must satisfy 55 2 , 2

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

This reduces to

54




20 5 20nn n

ππ πθ+

= + = .

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 += .

18. First, observe that sec 4 2θ = is equivalent to 1cos 42

θ = . The values of θ in

for which this is true must satisfy 74 2 , 2


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

7 9 15, , ,16 16 16 16π π π πθ =

Chapter 7

930

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 = − − −

20. First, observe that 2 3csc63

θ = − is equivalent to 3sin 62

θ = − . The values of θ in

for which this is true must satisfy 4 5 (4 6 ) (5 6 )6 2 , 2 ,3 3 3 3

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

So, dividing all values by 6 yields the following values ofθ in [ ]0,π which satisfy the original equation:

2 5,9 18π πθ =

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


[ ]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 = = .

22. First, observe that 2cos 22θ⎛ ⎞ = −⎜ ⎟⎝ ⎠

is equivalent to 2cos2 2θ −⎛ ⎞ =⎜ ⎟⎝ ⎠

. The values of

θ in [ ]0,2π that satisfy 2cos2 2θ −⎛ ⎞ =⎜ ⎟⎝ ⎠

must satisfy 32 4θ π= , so that 3

2πθ = .

Section 7.8

931

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


[ ]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 = .

24. First, observe that 4 tan 4 02θ⎛ ⎞ − =⎜ ⎟⎝ ⎠

is equivalent to tan 12θ⎛ ⎞ =⎜ ⎟⎝ ⎠

. The values of θ in

[ ]0, 2π that satisfy tan 12θ⎛ ⎞ =⎜ ⎟⎝ ⎠

must satisfy 2 4θ π= , so that

2πθ = .

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


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 = =

26. First, observe that ( )4csc 2 8 0θ + = is equivalent to ( )csc 2 2θ = − , which can be

further simplified to ( ) 1sin 22

θ = − . The values of θ in [ ]0,2π that satisfy

( ) 1sin 22


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= + + = .


7 11 19 23, , ,12 12 12 12π π π πθ =

Chapter 7

932

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 = .

28. First, observe that ( )3 sec 2 2 0θ + = is equivalent to ( ) 2sec 23

θ = − , which can be

further simplified to ( ) 3cos 22

θ = − . The values of θ in [ ]0, 2π that satisfy

( ) 3cos 22


5 5 7 7 5 7 17 192 , 2 , , 2 , , ,6 6 6 6 6 6 6 6π π π π π π π πθ π π= + + = .


5 7 17 19, , ,12 12 12 12π π π πθ =

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 = .

30. Factoring the left-side of 2sin 2sin 1 0θ θ+ + = yields the equivalent equation ( )2sin 1 0θ + = which is satisfied when sin 1 0θ + = . The value of θ in [ ]0,2π that

satisfies sin 1θ = − is 32πθ = .

Section 7.8

933

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 = .

32. Factoring the left-side of 2tan 3 tan 0θ θ− = yields the equivalent equation

( )tan tan 3 0θ θ − = which is satisfied when either tan 0 or tan 3 0θ θ= − = . The

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

32tan 3 12

θ = = are 4,3 3π πθ = . Thus, the solutions to the original equation are

40, , 2 , ,3 3π πθ π π= .

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 = .

34. Observe that 2cot 1θ = is equivalent to 2cot 1 0θ − = . Factoring the left-side of this equation yields the equivalent equation ( )( )cot 1 cot 1 0θ θ− + = which is satisfied when

either cot 1 0 or cot 1 0θ θ− = + = . The values of θ in [ ]0,2π that satisfy cot 1θ = are 5,

4 4π πθ = and those which satisfy cot 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 = .

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 = .

Chapter 7

934

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

Note that the equation 1sec2

θ = (or equivalently cos 2θ = ) has no solution since 2 is not

in the range of cosine. The value of θ in [ ]0, 2π that satisfies sec 1θ = − is θ π= .

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= .

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

sin 1θ = ± . The values of θ in [ ]0,2π for which this is true are 3,2 2π πθ = .

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 =

40. Factoring the left-side of 2csc (2 ) 4 0θ − = yields the equivalent equation ( )( )csc 2 2 csc 2 2 0θ θ− + = , which is satisfied whenever csc 2 2θ = ± , or equivalently

1sin 22


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 =

Section 7.8

935

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 ≈

42. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy cos 2 0.5136θ = , we proceed as follows: Step 1: Find the values of 2θ whose cosine is 0.5136 . Indeed, observe that one solution is ( )1cos 0.5136 59.096− ≈ , which is in QI.

A second solution occurs in QIV, and has value 360 59.096 300.904− = . 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 59.096 , 59.096 360 , 300.904 ,300.904 36059.096 , 419.096 , 300.904 , 660.904

θ = + +

=

and so, the solutions to the original equation are approximately: 29.55 , 209.55 , 150.45 , 330.45θ ≈

Chapter 7

936

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 ≈ .

44. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy sec 1.42752θ⎛ ⎞ =⎜ ⎟⎝ ⎠

, we

proceed as follows:

Step 1: Find the values of 2θ whose secant is 1.4275 .

Indeed, observe that one solution is ( )1 1 1sec 1.4275 cos 45.5311.4275

− − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

, which

is in QI. A second solution occurs in QIV, and has value 360 45.531 314.469− ≈ . 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

45.531 , 314.4692θ= .


91.06θ ≈ .

Section 7.8

937

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 ≈ .

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

Step 1: First, observe that the equation is equivalent to 6sec5

θ = − .

Step 2: Find the values of θ whose secant is 65

− .

Indeed, note that one solution is 1 1 16 1 5sec cos cos 146.442765 65

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

,

which is in QII. A second solution occurs in QIII, and has value 360 146.4427 213.5573− = . Since the input of secant is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately 146.44 , 213.56θ ≈ .

Chapter 7

938

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 ≈ .

48. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 3cos 5 0θ − = , we proceed as follows:

Step 1: First, observe that the equation is equivalent to 5cos3

θ = .

Step 2: Find the values of θ whose cosine is 53

.

Indeed, observe that one solution is 1 5cos 41.81033

− ⎛ ⎞≈⎜ ⎟⎜ ⎟

⎝ ⎠, which is in QI. A

second solution occurs in QIV, and has value 360 41.8103 318.19− = . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately 41.81 , 318.19θ ≈ .

Section 7.8

939

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

.


− ⎛ ⎞ ≈⎜ ⎟⎝ ⎠

, 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 ≈ .

50. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 26sin 13sin 5 0θ θ− − = , we proceed as follows:


( )( )

No solution

3sin 1 2sin 5 03sin 1 0 or 2sin 5 0

1 5sin or sin3 2

θ θθ θ

θ θ

+ − =

+ = − =

= − =

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

− .


− ⎛ ⎞− ≈ −⎜ ⎟⎝ ⎠

, which is in QIV. Since

the angles we seek have positive measure, we use the representative 360 19.4712 340.53− ≈ . A second solution occurs in QIII, and has value 180 19.47 199.47+ = . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to the original equation are approximately 199.47 , 340.53θ ≈ .

Chapter 7

940

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:


( )( )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 ≈ .

Section 7.8

941

52. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 26sec 7sec 20 0θ θ− − = , we proceed as follows:


( )( )2sec 5 3sec 4 02sec 5 0 or 3sec 4 0

5 4sec or sec2 3

θ θθ θ

θ θ

− + =

− = + =

= = −

Step 2: Solve 5sec2

θ = , or equivalently 2cos5

θ = .

To do so, we must find the values of θ whose cosine is 25

.


− ⎛ ⎞ ≈⎜ ⎟⎝ ⎠


solution occurs in QIV, and has value 360 66.42 293.58− = . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 66.42 , 293.58θ ≈ .

Step 3: Solve 4sec3

θ = − , or equivalently, 3cos4

θ = − .

To do so, we must find the values of θ whose cosine is 34

− .


− ⎛ ⎞− ≈⎜ ⎟⎝ ⎠

, which is in QII. A second

solution occurs in QIII, and has value 360 138.59 221.41− = . Since the input of cosine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 138.59 , 221.41θ ≈ . Step 4: Conclude that the solutions to the original equation are

66.42 , 138.59 , 221.41 , 293.58θ ≈ .

Chapter 7

942

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

943

54. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy

212cos 13cos 3 02 2θ θ⎛ ⎞ ⎛ ⎞− + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

, we proceed as follows:


3cos 1 4cos 3 02 2

1 33cos 1 0 or 4cos 3 0 or equivalently cos or cos2 2 2 3 2 4

θ θ

θ θ θ θ

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

Step 2: Solve ( ) 1cos 2 3θ = .

Step a: Find the values of 2θ whose cosine is 1

3.


− ⎛ ⎞ ≈⎜ ⎟⎝ ⎠


solution occurs in QIV, and has value 360 70.5288 289.47− = . Step b: Use Step a to find all values of θ that satisfy the original equation, and exclude any value of θ that satisfies the equation, but lies outside 0 ,360⎡ ⎤⎣ ⎦ . The solutions obtained is Step a are

70.5288 , 289.472θ = .

When multiplied by 2, the solution corresponding to 289.47 will no longer be in the interval. So, the solution to the equation in 0 ,360⎡ ⎤⎣ ⎦ is approximately 141.06θ ≈ .

Step 3: Solve ( ) 3cos 2 4θ = .

Step a: Find the values of 2θ whose cosine is 3

4.


− ⎛ ⎞ ≈⎜ ⎟⎝ ⎠


solution occurs in QIV, and has value 360 41.4096 318.59− = . Step b: Use Step a to find all values of θ that satisfy the original equation, and exclude any value of θ that satisfies the equation, but lies outside 0 ,360⎡ ⎤⎣ ⎦ .

The solutions obtained is Step a are 41.4096 , 318.592θ = .


82.82θ ≈ .

Step 4: Conclude that the solutions to the original equation are 82.82 , 141.06θ ≈ .

Chapter 7

944

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 ≈ .


56. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 2sin 3sin 3 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): ( )( )( )

23 3 4 1 3 3 21sin2 1 2

θ− ± − − − ±

= =

So, θ is a solution to the original equation if

3 21No solution since 12

3 21 3 21sin or sin2 2

θ θ

− −<−

− − − += = .

Step 2: Find the values of θ whose sine is 3 212

− + .

Indeed, observe that one solution is 1 3 21sin 52.3062

− ⎛ ⎞− +≈⎜ ⎟⎜ ⎟


second solution occurs in QII, and has value 180 52.306 127.69− = . Since the input of sine is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 52.306 , 127.69θ ≈ .


Section 7.8

945

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

+ .


− ⎛ ⎞+≈⎜ ⎟⎜ ⎟


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 ≈ .

Chapter 7

946

58. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 23cot 2cot 4 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):

( )( )( )

22 2 4 3 4 2 52 1 13cot2 3 6 3

θ− ± − − − ± − ±

= = =

So, θ is a solution to the original equation if either 1 13 1 13cot or cot

3 3θ θ− − − += = .

Step 2: Solve 1 13cot3

θ − −= .

To do so, we must find the values of θ whose cotangent is 1 133

− − .

Indeed, observe that one solution is

1 11 13 3cot 180 tan 146.923 1 13

− −⎛ ⎞− − ⎛ ⎞= + ≈⎜ ⎟ ⎜ ⎟⎜ ⎟ − −⎝ ⎠⎝ ⎠, which is in QII.

A second solution occurs in QIV, and has value 180 146.92 326.92+ = . Since the input of cotangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 146.92 , 326.92θ ≈ .

Step 3: Solve 1 13cot3

θ − +=

To do so, we must find the values of θ whose cotangent is 1 133

− + .

Indeed, observe that one solution is 1 11 13 3cot tan 49.0253 1 13

− −⎛ ⎞− + ⎛ ⎞= ≈⎜ ⎟ ⎜ ⎟⎜ ⎟ − +⎝ ⎠⎝ ⎠,

which is in QI. A second solution occurs in QIII, and has value 180 49.025 229.025+ = . Since the input of cotangent is simply θ , and not some multiple thereof, we conclude that the solutions to this equation are approximately 49.03 , 229.03θ ≈ . Step 4: Conclude that the solutions to the original equation are

49.03 , 146.92 , 229.03 , 326.92θ ≈ .

Section 7.8

947

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


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 =

60. In order to find all of the values of θ in 0 ,360⎡ ⎤⎣ ⎦ that satisfy ( )22sec 2 0θ − = , we

proceed as follows: Step 1: Simplify the equation algebraically. Factoring the left-side of ( )2

2sec 2 0θ − = yields the equivalent equation

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

equivalently ( )22cos

2θ = ± . The values of θ for which this is true must satisfy

3 5 72 , 2 , 2 , 22 4 4 4 4

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

which reduces to 34 , 4


Step 2: Convert the appropriate values obtained in Step 1 to degrees. 90 , 270θ =

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

sin cosx x= are 5,4 4

x π π= .

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

3 7,4 4

x π π= .

Chapter 7

948

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.

64. Observe that

2

2

2

sin csc 21sin 2

sinsin 1 2sin

sin 2sin 1 0(sin 1) 0

sin 1 0sin 1

x x

xx

x xx x

xx

x

+ =

+ =

+ =

− + =

− =− =

=

The value of x in [ ]0, 2π that satisfies the

equation sin 1x = is 2x π= . Substituting

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

Section 7.8

949

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 .

66. Observe that

( )

( )( )

( )

2

2

2

sec tan 11 sin 1

cos cos1 sin 1

cos1 sin

1cos

1 sin 1 sin1

1 sin1 sin

x xx

x xx

xxx

x xx

x

+ =

+ =

+=

+=

+ +=

−+ ( )

( )1 sin

1 sin

x

x

+

+ ( )1

1 sin

1 sin 1 sin2sin 0

sin 0

x

x xxx

=−

− = +==


equation sin 0x = are 0, , 2x π π= . Substituting these values into the original equation shows that while 0, 2π are solutions, π is extraneous. Indeed, note that

( ) ( ) 1 0sec tan 1 11 1

π π+ = + = − ≠− −

So, the only solutions are 0, 2π .

Chapter 7

950

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

=−

− = +

− = +=

=


equation 1cos2

x = are 5,3 3

x π π= .

Substituting this value into the original

equation shows that while3π is a solution,


15 5 1 2csc cot3 3 3 3

2 23 3 33

π π⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ − −

= − = − ≠


68. Observe that

( )

( )( )

( )

2

2

2

3csc cot3

1 cos 3sin sin 3

1 cos 3sin 3

1 cos 1sin 3

1 cos 1 cos 11 cos 3

1 cos

x x

xx x

xxxx

x xx

x

− =

− =

−=

−=

− −=

−− ( )

( )1 cos

1 cos

x

x

−

− ( )( )

131 cos

3 1 cos 1 cos3 3cos 1 cos

4cos 21cos2

x

x xx xx

x

=+

− = +

− = +=

=


equation 1cos2

x = are 5,3 3

x π π= .

Substituting this value into the original

equation shows that while 53π is a solution,


11 2csc cot3 3 3 3

2 23 3

3 3

π π⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − ≠


Section 7.8

951

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.

70. Observe that

( )( )

2

2

2

2sin csc 312sin 3

sin2sin 1 3

sin2sin 1 3sin

2sin 3sin 1 02sin 1 sin 1 0

x x

xx

xxx x

x xx x

+ =

+ =

+=

+ =

− + =

− − =

2sin 1 0 or sin 1 01sin or sin 12

x x

x x

− = − =

= =


are x = 5, ,6 6 2π 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.

72. Observe that

( )

sin 2 3 sin

2sin cos 3 sin

sin 2cos 3 0

x x

x x x

x x

=

=

− =

sin 0 or 2cos 3 0

3sin 0 or cos2

x x

x x

= − =

= =


are x = 110, , 2 , ,6 6π ππ π . Substituting


Chapter 7

952

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

= − =

= =




74. Observe that

( )

( )( )

2 2

2 2

2

cos 2 sincos sin sin

1 sin sin sin

2sin sin 1 02sin 1 sin 1 0

x xx x x

x x x

x xx x

=

− =

− − =

+ − =

− + =

2sin 1 0 or sin 1 01sin or sin 12

x x

x x

− = + =

= = −


are x = 5 3, ,6 6 2π π π . Substituting these

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

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.

Section 7.8

953

76. Observe that

( )

( )

3cot 2 cot3cos 2 cossin 2 sin

3cos 2 sin cos sin 23cos 2 sin cos sin 2 0

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

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

sin 2cos 2 1 0

x xx x

x xx x x x

x x x xx x x x x x

x x x xx x x

x x xx x

=

=

=− =

+ − =

+ − =+ − =

− =

− =

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

x x

x x

= − =

= =

Note that the solutions of sin 0x = in [ ]0,2π are 0, , 2x π π= . However, substituting these values into the original equation show that NONE of them are solutions since the right-side is undefined at each of these values.

Next, the solutions of 1cos 22

x = are

5 52 , 2 , , 23 3 3 3

x π π π ππ π= + +

so that 5 7 11, , ,

6 6 6 6x π π π π= .

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

Chapter 7

954

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= .


78. Observe that

( )

3 tan 2sin

3 sin 2sincos3 sin 2sin cos

3 sin 2sin cos 0

sin 3 2cos 0

x x

x xx

x x x

x x x

x x

=

=

=

− =

− =

sin 0 or 3 2cos 0

3sin 0 or cos2

x x

x x

= − =

= =




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.

80. Observe that

( )

2sin 2sin 0sin sin 2 0

x xx x

− =

− =

No solution

sin 0 or sin 2 0sin 0 or sin 2

x xx x= − == =


are x = 0, , 2π π . Substituting these into the original equation shows that they are all, in fact, solutions to the original equation.

Section 7.8

955

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.

82. Observe that

( )

( )( )

2

2

2

2

2cos sin 1

2 1 sin sin 1

2 2sin sin 12sin sin 1 0

sin 1 2sin 1 0

x x

x x

x xx x

x x

= +

− = +

− = +

+ − =

+ − =

sin 1 0 or 2sin 1 01sin 1 or sin2

x x

x x

+ = − =

= − =


are x = 5 3, ,6 6 2π π π . Substituting these into

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


84. Observe that

( )

( )

2

2

2

2

2

4cos 4sin 5

4 1 sin 4sin 5

4 4sin 4sin 54sin 4sin 1 0

2sin 1 02sin 1 0

1sin2

x x

x x

x xx x

xx

x

− =

− − =

− − =

+ + =

+ =

+ =

= −

The solutions to this equation in [ ]0, 2π

are x = 7 11,6 6π π . Substituting these into


Chapter 7

956

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


86. Observe that

( )

2cot csc2cos 1sin sin

2cos sin sinsin 2cos 1 0

x xx

x xx x x

x x

=

=

=

− =

sin 0 or 2cos 1 01sin 0 or cos2

x x

x x

= − =

= =

Note that the solutions of sin 0x = in [ ]0, 2π are 0, , 2x π π= . However, substituting these values into the original equation show that NONE of them are solutions since both the left- and right-sides are undefined at each of these values.

Next, the solutions of 1cos2

x = are

5,3 3

x π π=


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=

88. Observe that ( ) ( )

( ) ( )

( ) ( )( )

14 2 2

22

2 2

2

12

csc cos1 cos

4sin

1 4sin cos

1 2sin 2

sin

x x

xx

x x

x

x

− =

− =

− =

− = ⋅

− =

The solutions of this equation are 7 11,6 6

x π π= .

Section 7.8

957

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

+ .


− ⎛ ⎞+≈⎜ ⎟⎜ ⎟


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

− .


− ⎛ ⎞−≈ −⎜ ⎟⎜ ⎟


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.

Chapter 7

958

90. Observe that

( )

2

2

sec tan 11 tan tan 1

tan tan 1 0

x xx x

x x

= +

+ = +

− =

tan 0 or tan 1 0 so that tan 0 or tan 1x x x x= − = = =

The solutions to these equations in 0 ,360⎡ ⎤⎣ ⎦ are x = 0 , 180 , 360 , 45 , 225 . Substituting these into the original equation shows that they are all solutions. 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.

Section 7.8

959

92. In order to find all of the values of x in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 2sec 2 tan 4x x= + , we proceed as follows: Step 1: Simplify the equation algebraically.

( )( )

2

2

2

sec 2 tan 41 tan 2 tan 4

tan 2 tan 3 0tan 3 tan 1 0

x xx x

x xx x

= +

+ = +

− − =

− + =

tan 3 0 or tan 1 0tan 3 or tan 1

x xx x− = + == = −

Step 2: Solve tan 3x = . To do so, we must find the values of x whose tangent is 3. Indeed, observe that one solution is ( )1tan 3 71.57− ≈ , which is in QI.

A second solution occurs in QIII, and has value180 71.57 251.57+ = . Since the input of tangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately 71.57 , 251.57x ≈ . Step 3: Solve tan 1x = − . To do so, we must find the values of x whose tangent is 1.− Indeed, observe that one solution is ( )1tan 1 135− − = , which is in QII.

A second solution occurs in QIV, and has value180 135 315+ = . Since the input of tangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately 135 , 315x = . Step 4: Conclude that the solutions to the original equation are

135 , 315 , 71.57 , 251.57x ≈ . Substituting these into the original equation shows that they are, in fact, solutions to the original equation.

Chapter 7

960

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

961

94. In order to find all of the values of x in 0 ,360⎡ ⎤⎣ ⎦ that satisfy 2csc cot 7x x+ = , we proceed as follows: Step 1: Simplify the equation algebraically:

( )

( )( )

2

2

2

csc cot 7

1 cot cot 7

cot cot 6 0cot 3 cot 2 0

x x

x x

x xx x

+ =

+ + =

+ − =

+ − =

cot 3 0 or cot 2 0cot 3 or cot 2

x xx x+ = − == − =

Step 2: Solve cot 3x = − . To do so, we must find the values of x whose cotangent is 3− . Indeed, observe that one solution is

( )1 1 1cot 3 180 tan 161.573

− − ⎛ ⎞− = + − ≈⎜ ⎟⎝ ⎠

, which is in QII. A second solution occurs

in QIV, and has value 180 161.57 341.57+ = . Since the input of cotangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately 161.57 , 341.57x ≈ . Step 3: Solve cot 2x = To do so, we must find the values of x whose cotangent is 2 .

Indeed, observe that one solution is ( )1 1 1cot 2 tan 26.572

− − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

,

which is in QI. A second solution occurs in QIII, and has value 180 26.57 206.57+ = . Since the input of cotangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately 26.57 , 206.57x ≈ . Step 4: Conclude that the solutions to the original equation are

26.57 , 206.57 , 161.57 , 341.57x ≈ .

Chapter 7

962

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.

Section 7.8

963

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

( )2

2

2

2

sec tan 2 0

1 tan tan 2 0

tan tan 1 0

1 1 4(1)( 1) 1 5tan2(1) 2

x x

x x

x x

x

+ − =

+ + − =

+ − =

− ± − − − ±= =

Step 2: Solve 1 5tan

2x − −= .

To do so, we must find the values of x whose tangent is 1 52

− − .


− ⎛ ⎞− −≈ −⎜ ⎟⎜ ⎟


Since the angles we seek have positive measure, we use the representative 360 58.28 301.72− = A second solution occurs in QII, and has value 301.72 180 121.72− = . Since the input of tangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately 121.72 , 301.72x ≈ .

Step 3: Solve 1 5tan

2x − += .

To do so, we must find the values of x whose tangent is 1 5

2− +

.


− ⎛ ⎞− +≈⎜ ⎟⎜ ⎟

⎝ ⎠, which is in QI.

A second solution occurs in QIII, and has value180 31.72 211.72+ = . Since the input of tangent is simply x, and not some multiple thereof, we conclude that the solutions to this equation are approximately 31.72 , 211.72x ≈ . Step 4: Conclude that the solutions to the original equation are

31.72 , 211.72 , 121.72 , 301.72x ≈ . Substituting these into the original equation shows that they are, in fact, solutions to the original equation.

Chapter 7

964


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

( ) ( )( )

( )

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= .


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

( ) ( )( )

( )

2 214 8 8

1 18 2 8 8 2 8

18 2 8

12 8 8

1 12 2 8

4

sec sin 0

sin sec sin sec 0

sin sec

sin cos

sin 2

1 sin

x x

x x x x

x x

x x

x

x

− + =

− + =

= ±

=±

=± ⋅

± =

The values of x that satisfy these equation must satisfy 3,4 2 2x π π= . So, the desired

solution is 2x 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

965

100. Solving for x yields:

1,800 400sin 2,0006

200 400sin6

1 sin2 6

x

x

x

π

π

π

⎛ ⎞= +⎜ ⎟⎝ ⎠⎛ ⎞− = ⎜ ⎟⎝ ⎠

⎛ ⎞− = ⎜ ⎟⎝ ⎠

This equation is satisfied when 7 11,6 6 6

xπ π π= , so that 7, 11x = . So, the sales reach

1,800 during in July and November.

Chapter 7

966


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

Section 7.8

967

102. Observe that

( )

( )( )

2 2

2 2

2

cos sin cos 0

cos 1 cos cos 0

2cos cos 1 02cos 1 cos 1 0

θ θ θ

θ θ θ

θ θθ θ

− + =

− − + =

+ − =

− + =

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

Observe that 2cos 1 0θ − = is equivalent to 1cos2

θ = , which is satisfied when 5,3 3π πθ = .

Also, cos 1 0θ + = is equivalent to cos 1θ = − , which is satisfied when θ π= .

So, we conclude that the solutions to the original equation are 5, ,3 3π πθ π= .

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 = + = .

104. Solving the equation 200 100sin 1506

xπ⎛ ⎞+ =⎜ ⎟⎝ ⎠

yields:

200 100sin 1506

100sin 506

1sin6 2

x

x

x

π

π

π

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

Observe that this equation is satisfied when 7 11,6 6 6

xπ π π= , so that

6 7 6 11, 7, 116 6

x π ππ π⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

. So, the first year after 2000 in which this occurs is 2007.

Chapter 7

968

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−

=

⎛ ⎞⎜ ⎟≈ =⎜ ⎟⎝ ⎠

106. Use ( ) ( )sin sini i r rn nθ θ= with the given information to obtain

( ) ( )2.417sin 15 1.00sin rθ= so that

( ) ( )

( )1

2.417sin 15sin

1.002.417sin 15

39 sin1.00

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.

108. Observe that using the identity sin 2 2sin cosA A A= with

3A xπ= yields

2sin cos 3 23 3

sin 2 13

x x

x

π π

π

⎛ ⎞ ⎛ ⎞ + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⋅ = −⎜ ⎟⎝ ⎠

This equation is satisfied when 2 33 2

xπ π= , so

that 94

x = . So, it takes 9 sec.4

for the

volume of air to equal 2 liters. 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 .

Section 7.8

969

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

[ ]( )

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. 112. The value 2

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

3sin 2 3 2 12π⎛ ⎞ − = − =⎜ ⎟⎝ ⎠

, while sin 12π⎛ ⎞− = −⎜ ⎟⎝ ⎠

. 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…

114. Forgot to check for extraneous solutions. Note that for x π= , we have

1 sin 1 1π+ = = , while cos 1π = − . Hence, x π= is not a solution to the equation. The remaining values ARE solutions.

115. False. For instance, 3sin2

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

116. False. For instance, 2 3sin2

θ = has four solutions on [ ]0, 2π , namely

2 4 5, , ,3 3 3 3π π π πθ = .

Chapter 7

970

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

118. False. This is not sufficient. For instance, the equation sin 1x = has infinitely many solutions, but there are values of x in the domain for which it is not true (for example, 0x = ).

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 = .

120. Solving the equation 3cos4 2πθ⎛ ⎞+ =⎜ ⎟

⎝ ⎠ on is equivalent to

3cos4 2πθ⎛ ⎞+ =⎜ ⎟

⎝ ⎠ or 3cos

4 2πθ⎛ ⎞+ = −⎜ ⎟

⎝ ⎠.

Observe that 3cos4 2πθ⎛ ⎞+ =⎜ ⎟

⎝ ⎠ is satisfied when 112 , 2

4 6 6n nπ π πθ π π+ = + + , so that

11 192 , 2 2 , 26 4 6 4 12 12

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

Also, 3cos4 2πθ⎛ ⎞+ = −⎜ ⎟

⎝ ⎠ is satisfied when 5 72 , 2

4 6 6n nπ π πθ π π+ = + + , so that

5 7 7 112 , 2 2 , 26 4 6 4 12 6

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

So, we conclude that the solutions to the original equation are 19 7 112 , 2 , 2 , 2

12 12 12 6n n n nπ π π πθ π π π π= − + + + + , where n is an integer,

which can be further simplified as: 7, 2

12 12n nπ πθ π π= − + + , where n is an integer.

Section 7.8

971

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= .

122. First, observe that using the addition formula cos( ) cos cos sin sinA B A B A B− = + yields

( ) ( )( )

cos cos 15 sin sin 15 0.7

cos 15 0.7

x x

x

+ =

− =

We now find the solutions to this equation in 0 ,360⎡ ⎤⎣ ⎦ :

One solution is ( )115 cos 0.7 15 45.57 60.57x −= + ≈ + = , which is in QI. The second

solution is in QIV and has value 15 314.43 329.43+ = . Observe that the smallest positive solution is approximately 60.57x ≈ .

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.

124. Factoring the left-side of ( )43sec 1 0θ − = yields the equivalent equation

( )( ) ( )( )2 23 3sec 1 sec 1 0θ θ− + = , which is satisfied whenever either ( )2

3sec 1 0θ − = or

( )23sec 1 0θ + = . The second equation has no solution since the left-side is always greater

than or equal to 1. The first equation is equivalent to ( )23cos 1θ = , which holds whenever

( )3cos 1θ = ± . The values of θ for which this is true must satisfy

3nθ π= , where n is an integer.

So, dividing by 3 then yields the values of θ in for which this is true: 3 , where is an integer .n nθ π=

Chapter 7

972

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 = +

126. Observe that 2

2

3212

2 tan 3 3 3 tan (3 )2 tan 3 3

1 tan (3 )

tan 6 3

x xxx

x

= −

=−

= =

The values of x for which this equation holds are

63

x nπ π= + , where n is an integer,

which is equivalent to (1 3 )

18 6 18n nx π π π+

= + = , where n is an integer.

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 .

128. Consider the graphs below of 1 2csc , secy yθ θ= = .

Observe that the approximate solution to this

equation is 0.785. The exact solution is 4π .

Section 7.8

973

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 .

130. Consider the graphs below of 1 2cos , cscy yθ θ= = .

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

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.

132. Consider the graphs below of 1 2cos ,y y eθθ= = .

First, note that while there are no positive solutions of the equation cos eθθ = , the two curves do intersect at 0θ = .

Chapter 7

974

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 .

134. To determine the smallest positive solution (approximately) of the equation cot 5 tan 2 3x x+ = − graphically, we search for the intersection points of the graphs of the following two functions:

1 2cot 5 tan 2 , 3y x x y= + = −

The x-coordinate of the intersection point is in radians. Observe that the smallest positive solution, in degrees, is approximately 33.92 .

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Section 7.6 Solutionsfaculty.oprfhs.org/cavalos/Book Chapters and Book Solutions/CATA...cos 3 7 cos 2 7 2sin sin()() 37 2 7 37 2 7 22 757 7 57 2sin sin 2sin sin 22 2 2 xx x x xx x