1. x

[Pages:12]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.

33. We want to write the product 3 cos4x cos7x as a sum. Note that cos4x cos7x sin4x sin7x cos4x 7x

and that

cos4x cos7x sin4x sin7x cos4x 7x

Adding the above two equations we obtain

2 cos4x cos7x cos3x cos11x.

Using the fact that cosx cos x, we obtain

cos4x cos7x

1 2

cos3x

cos11x.

This gives us

3 cos4x cos7x

3 2

cos3x

cos11x.

35. We want to express the sum sin5x sin3x as a product. This is done as follows:

sin5x sin4x x sin4x cosx cos4x sinx

and sin3x sin4x x sin4x cosx cos4x sinx.

Adding the above equations gives us sin5x sin3x 2 sin4x cosx.

37. We want to express the sum cos4x cos6x as a product. This is done as follows: cos4x cos5x x cos5x cosx sin5x sinx

and cos6x cos5x x cos5x cosx sin5x sinx.

Subtracting the second equation from the first gives us cos4x cos6x 2 sin5x sinx.

39. We want to express the sum sin2x sin7x as a product. This is done as follows: sin2x sin4. 5x 2. 5x sin4. 5x cos2. 5x cos4. 5x sin2. 5x

and sin7x sin4. 5x 2. 5x sin4. 5x cos2. 5x cos4. 5x sin2. 5x.

Subtracting the second equation from the first gives us sin2x sin7x 2 cos4. 5x sin2. 5x.

41. We want to find the value of 2 sin52. 5 sin97. 5 . We proceed as follows:

cos52. 5 cos97. 5 sin52. 5 sin97. 5 cos52. 5 97. 5 cos45 cos45

2 2

and

cos52. 5 cos97. 5 sin52. 5 sin97. 5

cos52. 5 97. 5

cos150

3 2

.

By subtracting the second result from the first, we obtain

2 sin52. 5 sin97. 5

2 2

3 2

2 2

3.

Thus

2 sin52. 5 sin97. 5

2 2

3.

43. We want to find the value of cos37. 5 sin7. 5 . We proceed as follows:

sin37. 5 cos7. 5 cos37. 5 sin7. 5 sin37. 5 7. 5 sin45

2 2

and sin37. 5 cos7. 5 cos37. 5 sin7. 5

sin37. 5 7. 5

sin30

1 2

.

By subtracting the second result from the first, we obtain

2 cos37. 5 sin7. 5

2 1 22

2 2

1

.

Thus

cos37. 5 sin7. 5

2 4

1

.

45. We want to find the value of cos255 cos195 . We proceed as follows:

cos255 cos225 30 cos225 cos30 sin225 sin30

and cos195 cos225 30 cos225 cos30 sin225 sin30

Subtraction of the second result from the first gives us cos255 cos195 2 sin225 sin30

2

2 2

1 2

2 2

.

47. Using the identity cos2 sin2 cos2 and letting 5x, we immediately obtain the identity cos25x sin25x cos10x.

49.

sin x cos x2 sin2x 2 sin x cos x cos2x

sin2x cos2x 2 sin x cos x

1 sin2x.

51.

sin4x sinx

2 sin2x cos2x sinx

22 sinx cosx cos2x sinx

4 cosx cos2x.

53.

2tan x cot x tan2x cot2x

2tan x cot x tan x cot xtan x cot x

tan x 2 cot x

2

sin x cos x

cos x sin x

sin x cos x sin x cos x

2 sin x cos x sin2x cos2x

2 sin x cos x 1

2 sin x cos x.

55. First note that

tan3x tan2x x

tan2x tanx 1 tan2x tanx

and then note that

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