1. θ is in Quadrant III. - Washington State University

1. is in Quadrant III.

r

1

6

12 + 62 = r2 r2 = 1 + 36 r = 37

cos

=

- 1 37

=-

37 37

sin

=

- 6 37

=

6 37 - 37

tan = 6

sec = - 37

csc = -

37 6

cot

=

1 6

2. cos2 + sin2 = 1 cos2 = 1 - sin2 cos = ? 1 - sin2

Since is in QII, cos = - 1 - sin2

3.

sin x tan x + sin x sin x(tan x + 1) (a) tan x + tan2 x = tan x(1 + tan x)

= sin x tan x

= sin x - cot x

= sin x

cos x sin x

= cos x

(b)

sec x sin x

-

csc x sec x

=

sec2 x - csc x sin x sin x sec x

=

sec2 x - 1

sin x

cos x

= tan2 x tan x

= tan x

(c)

cos x - sec x sec x

=

cos x sec x

-1

= cos2 x - 1

= - sin2 x

(d) cos2 x tan2 x = cos2 x

sin2 x cos2 x

= sin2 x

= 1 - cos2 x

4. x = -1, x = 2 , x = 0

5.

Note that the

period

of tan

is

so

tan(

+

)

=

tan

=

-

4 3

or

use the

sum identity:

tan(

+ )

=

tan + tan 1 - tan tan

=

-

4 3

+

0

1-

-

4 3

(0)

=

4 -3

6. (a)

cos - 12

= cos

6-4

= cos cos + sin sin

64

64

=

3 2

2 2

+

1 2

2 2

=

6+ 4

2

(b)

sin

13 12

= sin

5 6

+

4

=

sin

5 6

cos

4

+

cos

5 6

sin

4

=

1 2

2 2

+

3 -2

2 2

=

2- 4

6

(c)

cos2 (22.5) - sin2 (22.5) = cos (2 ? 22.5) = cos 45

=

2 2

3

3

3

3

2

(d) 2 sin

cos

= sin 2 ? = sin

=

8

8

8

4

2

11

5

11

5

11 5

2

3

(e) sin

cos

+ cos

sin

= sin

+ = sin

=

24

24

24

24

24 24

3

2

7. Since and are acute angles, they are in QI

13

b

12

r

3

4

122 + b2 = 132 b2 = 169 - 144 b = 25 = 5

42 + 32 = r2 r2 = 16 + 9 r = 25 = 5

(a) sin( - ) = sin cos - cos sin =

5 13

(b) cos( + ) = cos cos - sin sin =

12 13

(c)

tan( + )

=

tan + tan 1 - tan tan

=

5 12

+

1-

5 12

3 4

3 4

8. Since and are obtuse angles, they are in QII

4

12

5 - 13

3 5

=

20 - 36 65

=

16 - 65

4

5

5 - 13

3 5

=

48 - 15 65

=

33 65

=

14

12

1

-

15 48

=

14 12

?

48 33

=

56 33

17 b 8

25

7

b

82 + b2 = 172 b2 = 289 - 64 b = 225 = 15

72 + b2 = 252 b2 = 625 - 49 b = 576 = 24

(a) sin( - ) = sin cos - cos sin =

15 17

7

8

- 25 - - 17

24 25

=

-105 + 192 425

=

87 425

(b) cos( + ) = cos cos - sin sin =

-

8 17

-

7 25

-

15 17

(c)

tan( + )

=

tan + tan 1 - tan tan

=

-

15 8

-

1-

-

15 8

24

7

-

24 7

-105-192

=

56

1

-

360 56

=

24 25

=

56 - 360 425

=

-

304 425

297 - 56

56 - 304

=

297 304

9. Let a be the unknown side length.

a2 + 12 = 52

a2 = 25 - 1

a = 24 = 2 6

sin

=

1 5

sin 2 = 2 sin cos = 2

1 5

26 5

=

46 25

cos

=

2

6 5

cos 2 = cos2 - sin2 =

2

26

5

-

1 2 24 1 23 5 = 25 = 25 = 25

10.

45

r7

4 72 + 42 = r2

r2 = 49 + 16

r = 65

11. Domain: (-, )

Range:

-

2

,

2

12.

(a)

6

(b)

-

4

(c)

-3

6 (d) 7

(e)

3

Since

-1

6 7

1

Since

6 5

>

1

13. (a) Let = sin-1 5 . 13

cos = cos( + 45)

= cos cos 45 - sin sin 45

= 7 65

2 2

-

4 65

2 2

= 7 65 65

2

4 65

2 - 65

2 2

=

7 130 - 4 130

130

= 3 130

130

(f)

2 3

(g)

2 3

(h) Undefined

(i)

12 15

(b) Let = tan-1

9 -2

.

13

5

a

a2 + 52 = 132 a2 = 169 - 25 a = 144 = 12

tan sin-1 5 13

=

tan

=

5 12

r

2

9

22 + 92 = r2

r2 = 4 + 81

r = 85

sec

tan-1

9 -

2

= sec =

85 2

(c) Let = cos-1

x .

9

(d) Let = sec-1

1. 2x

9

b

x

x2 + b2 = 92 b2 = 81 - x2

b = 81 - x2

cot cos-1 x = cot = x

9

81 - x2

1

b

2x

(2x)2 + b2 = 12 b2 = 1 - 4x2

b = 1 - 4x2

sin sec-1 1 2x

= sin = 1 - 4x2

14. i. the principal root, ii. all solutions in [0, 2), and iii. all real roots

(a) 8 cos x = -4 2

cos x = -

2 2

i. x = 3 4

ii.

x=

3 4

,

5 4

iii.

x=

3 4

+

2k,

5 4

+ 2k;

k

Z

(b) 3 tan x + 2 = 3

tan x = 1 = 3

3 3

i.

x=

6

ii.

x=

6

,

7 6

iii. x = 6 + k; k Z

15. (a) 3 cos x tan(2x) - cos x = 0 cos x( 3 tan(2x) - 1) = 0

From cos x = 0, we have

From 3 tan(2x) - 1 = 0, we have

x

=

2,

3 2

tan(2x) = 1 = 3

3 3

2x

=

6

+

k;

k

Z

x

=

12

+

k 2

;

k

Z

and so, in the interval [0, 2),

x

=

12 ,

7 12 ,

13 12 ,

19 12

Gathering all of these solutions gives the final answer:

x

=

,

2

3 ,

2

,

12

7 ,

12

13 ,

12

19 12

(b)

4 sin x cos x - 2 3 sin x - 2 cos x + 3 = 0

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

(2 cos x - 3)(2 sin x - 1) = 0

From 2 cos x - 3 = 0, we have

cos x =

3 2

x

=

6

,

11 6

From 2 sin x - 1 = 0, we have

sin x

=

1 2

x

=

, 6

5 6

Gathering all of these solutions gives the final answer:

x = , 11 , 5 66 6

(c)

cos x - sin x =

2 2

2

(cos x - sin x)2 = 2

2

cos2 x - 2 sin x cos x + sin2 x = 1 2

1 1 - sin(2x) = 2

sin(2x)

=

1 2

2x

=

6

+

2k,

5 6

+

2k;

k

Z

x

=

12

+

k,

5 12

+

k;

k

Z

and so, in the interval [0, 2),

x

=

12 ,

5 12

,

13 12

,

17 12

After checking all of these solutions in the original equation, we have

x = , 17 12 12

5 and 13 are extraneous

12

12

(d)

2 cos 3

2x - 6

+3=5 26

2 3

cos

2x - 6

=

-

2 3

cos 2x - 6 = -1

2x

-

6

=

3 2

+

2k;

k

Z

2x

=

5 3

+

2k;

k

Z

x

=

5 6

+

k;

k

Z

and so, in the interval [0, 2),

x = 5 , 11 66

1

2

16. (a) csc x = - 2 sin x = - 2 sin x = - 2

x

=

5 4

+

2k,

7 4

+

2k;

k

Z

(b) -8 sin

1 2

x

=4 3

sin

1 2

x

=-

3 2

1 2

x

=

4 3

+

2k,

5 3

+

2k;

k

Z

x

=

8 3

+

4k,

10 3

+

4k;

k

Z

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