1. x

Solutions to Homework Problems from Section 7.3 of Stewart

1. Given that sin x 5/13 and that x is in quadrant I, we have

cos2x 1 sin2x 1

5 13

2 144 169

so

cos x

144 169

12 13

.

Since x is in quadrant I we know that cos x 0 so cos x 12/13. From this we obtain

sin 2x 2 sin x cos x 2

5 13

12 13

120 169

cos 2x cos2x sin2x

12 2 13

5 2 119

13

169

tan 2x

sin 2x cos 2x

120 119

.

3. Given that tan x 4/3 and that x is in quadrant II we have

sec2x tan2x 1

4 3

21

25 9

which means that sec x 5/3 and consequently cos x 3/5. Since x is in quadrant II we

know that cos x 0. Thus cos x 3/5. This gives us sin x 4/5 (using the main

Pythagorean identity and the fact that x is in quadrant II). Thus

sin 2x 2 sin x cos x 2

4 5

3 5

24 25

cos 2x cos2x sin2x

3 5

2

4 5

2 7 25

tan 2x

sin 2x cos 2x

24 7

.

5. Given that sin x 3/5 and that x is in quadrant III, we obtain

cos x 1 sin2x 1

3 5

2 4 5

which gives us

sin 2x 2 sin x cos x 2

3 5

4 5

24 25

cos 2x cos2x sin2x

4 5

2

3 5

2 7 25

tan 2x

sin 2x cos 2x

24 7

.

7. We want to write sin4x in terms of first powers of cosine. First note that

sin2x

1 cos2x 2

which means that sin4x sin2x 2

1 cos2x 2

2

1

2

cos2x 4

cos22x

.

Now note that

cos22x

1

cos4x 2

.

This gives us

sin4x

1 2 cos2x

1cos4x 2

4

2 2

2 4 cos2x 1 cos4x 8

3 4 cos2x cos4x 8

or

sin4x

3 8

1 2

cos2x

1 8

cos4x.

9.

cos4x sin4x

1 16

2

sin

x

cos

x4

1 16

sin2x4

1 16

sin42x.

Using the result of problem 7, we know that

sin42x

3 8

1 2

cos4x

1 8

cos8x.

Thus

cos4x sin4x

1 16

3 8

1 2

cos4x

1 8

cos8x

3 128

1 32

cos4x

1 128

cos8x.

11. Using the result from problem 7 we have

cos2x sin4x cos2x

3 4 cos2x cos4x 8

1 cos2x 2

3 4 cos2x cos4x 8

3 4 cos2x cos4x 3 cos2x 4 cos22x cos4x cos2x 16

3

cos2x

cos4x

4 cos22x 16

cos4x

cos2x

.

We must now deal with the 4 cos22x and the cos4x cos2x. First note that

4 cos22x 4

1 cos4x 2

2 2 cos4x

To deal with cos4x cos2x we use the "product to sum" formula on page 480 of the

textbook:

cos u cos v

cosu

v

2

cosu

v

.

This gives us

cos4x cos2x

cos6x

2

cos2x

.

Finally,

cos2x sin4x

3 cos2x cos4x 4 cos22x cos4x cos2x 16

3 cos2x cos4x 2 2 cos4x

cos6xcos2x 2

2

16

2

6 2 cos2x 2 cos4x 4 4 cos4x cos6x cos2x 32

2

cos2x

2

cos4x 32

cos6x

.

In conclusion,

cos2x sin4x

1 16

1 32

cos2x

1 16

cos4x

1 32

cos6x.

13.

sin215

1 cos30 2

Thus

1

3 2

2

2

2

2 4

3

.

sin15

2 3 4

2 3 2.

15.

cos222. 5

1 cos45 2

Thus

1

2 2

2

22

2 4

2

.

cos22. 5

2 2 2.

17. This is really the same as problem 13 because /12 radians is the same as 15. Thus

sin

12

2 3 2.

19.

a. 2 sin18 cos18 sin2 18 sin36 .

b. 2 sin3 cos3 sin6.

21.

a. cos234 sin234 cos2 34 cos68 .

b. cos25 sin25 cos2 5 cos10.

23.

a. First note that

cos24

1 cos8 2

and thus

1 cos8 2 cos24 .

Also sin8 2 sin4 cos4 .

Thus

sin8 1 cos8

2 sin4 cos4 2 cos24

tan4 .

b. Using the same kind of reasoning as in part a we have 1 cos4 2 sin22

and sin4 2 sin2 cos2.

Thus

1 cos4 sin4

2 sin22 2 sin2 cos2

tan2.

25. Given that sin x 3/5 and that 0 x 90 we can use the main Pythagorean identity to deduce that cos x 4/5. This gives us

sin2

x 2

1 cos x

1

4 5

2

2

and since 0

x 2

45 we conclude that

1 10

sin

x 2

1 10

10 10

.

Likewise and we obtain

cos2

x 2

1 cos x

1

4 5

9

2

2

10

cos

x 2

3 10 10

.

This gives us

tan

x 2

1 3

.

27. Given that csc x 3 and that 90 x 180, we have sin x 1/3 and we can use the main Pythagorean identity to deduce that cos x 2 2 /3. This gives us

sin2

x 2

1 cos x

1

22 3

2

2

and since 45 x 90 we conclude that

1 2

2 3

32 6

2

sin

x 2

32 6

2

.

Likewise

cos2

x 2

and we obtain

1 cos x

1

22 3

1

2 32 2

2

2

23

6

cos

x 2

32 6

2

.

This gives us

tan

x 2

32 32

2 2

.

29. Given that sec x 3/2 and that 270 x 360, we have cos x 2/3 and

sin2

x 2

1 cos x

1

2 3

1

2

2

6

and since 135

x 2

180, we conclude that

sin

x 2

1 6

6 6

.

Likewise and we obtain

cos2

x 2

1 cos x

1

2 3

5

2

2

6

cos

x 2

5 6

30 6

.

This gives us

tan

x 2

5 5

.

31. We want to write the product sin2x cos3x as a sum. Note that

sin2x cos3x cos2x sin3x sin2x 3x

and that

sin2x cos3x cos2x sin3x sin2x 3x

Adding the above two equations we obtain

2 sin2x cos3x sin5x sinx.

Using the fact that sinx sin x, we obtain

sin2x cos3x

1 2

sin5x

sinx.

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