1. x - Kennesaw State University
[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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