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

2

? 144

cos 2 x ? 1 ? sin 2 x ? 1 ? 5

169

13

so

144 ? ? 12 .

13

169

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

12 ? 120

sin 2x ? 2 sin x cos x ? 2 5

169

13

13

2

2

? 119

? 5

cos 2x ? cos 2 x ? sin 2 x ? 12

169

13

13

tan 2x ? sin 2x ? 120 .

119

cos 2x

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

2

? 1 ? 25

sec 2 x ? tan 2 x ? 1 ? ? 4

9

3

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

? 3 ? ? 24

sin 2x ? 2 sin x cos x ? 2 4

5

5

25

2

2

?? 7

? 4

cos 2x ? cos 2 x ? sin 2 x ? ? 3

5

5

25

tan 2x ? sin 2x ? 24 .

7

cos 2x

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

cos x ? ?

cos x ? ? 1 ? sin 2 x ? ? 1 ? ? 3

5

2

? ?4

5

which gives us

? 4 ? 24

sin 2x ? 2 sin x cos x ? 2 ? 3

5

5

25

2

2

? 7

cos 2x ? cos 2 x ? sin 2 x ? ? 4

? ?3

5

5

25

tan 2x ? sin 2x ? 24 .

7

cos 2x

4

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

1 ? cos?2x?

sin 2 x ?

2

which means that

1 ? cos?2x? 2

1 ? 2 cos?2x? ? cos 2 ?2x?

2

?

.

sin 4 x ? sin 2 x ?

2

4

Now note that

cos 2 ?2x? ?

1 ? cos?4x?

.

2

This gives us

1?cos?4x?

1 ? 2 cos?2x? ?

2

sin x ?

? 2

4

2

2 ? 4 cos?2x? ? 1 ? cos?4x?

?

8

3 ? 4 cos?2x? ? cos?4x?

?

8

4

or

sin 4 x ? 3 ? 1 cos?2x? ? 1 cos?4x?.

8

8

2

9.

cos 4 x sin 4 x ? 1 ?2 sin x cos x? 4

16

? 1 ?sin?2x?? 4

16

? 1 sin 4 ?2x?.

16

Using the result of problem 7, we know that

sin 4 ?2x? ? 3 ? 1 cos?4x? ? 1 cos?8x?.

8

8

2

Thus

cos 4 x sin 4 x ? 1 3 ? 1 cos?4x? ? 1 cos?8x?

8

2

16 8

? 3 ? 1 cos?4x? ? 1 cos?8x?.

128

128

32

11. Using the result from problem 7 we have

3 ? 4 cos?2x? ? cos?4x?

cos 2 x sin 4 x ? cos 2 x

8

1 ? cos?2x?

3 ? 4 cos?2x? ? cos?4x?

?

8

2

3 ? 4 cos?2x? ? cos?4x? ? 3 cos?2x? ? 4 cos 2 ?2x? ? cos?4x? cos?2x?

?

16

2

3 ? cos?2x? ? cos?4x? ? 4 cos ?2x? ? cos?4x? cos?2x?

?

.

16

We must now deal with the 4 cos 2 ?2x? and the cos?4x? cos?2x?. First note that

1 ? cos?4x?

? 2 ? 2 cos?4x?

4 cos 2 ?2x? ? 4 ?

2

To deal with cos?4x? cos?2x? we use the ��product to sum�� formula on page 480 of the

textbook:

cos?u ? v? ? cos?u ? v?

cos u cos v ?

.

2

This gives us

cos?4x? cos?2x? ?

cos?6x? ? cos?2x?

.

2

Finally,

3 ? cos?2x? ? cos?4x? ? 4 cos 2 ?2x? ? cos?4x? cos?2x?

16

cos?6x??cos?2x?

3 ? cos?2x? ? cos?4x? ? ?2 ? 2 cos?4x?? ?

2

?

? 2

16

2

6 ? 2 cos?2x? ? 2 cos?4x? ? 4 ? 4 cos?4x? ? cos?6x? ? cos?2x?

?

32

2 ? cos?2x? ? 2 cos?4x? ? cos?6x?

?

.

32

cos 2 x sin 4 x ?

In conclusion,

cos 2 x sin 4 x ? 1 ? 1 cos?2x? ? 1 cos?4x? ? 1 cos?6x?.

32

32

16

16

13.

1 ? cos?30 ? ?

2

sin 2 ?15 ? ? ?

1 ? 23

?

? 2

2

2

2? 3

?

.

4

Thus

2? 3

4

sin?15 ? ? ?

?

2? 3

.

2

15.

cos 2 ?22. 5 ? ? ?

1 ? cos?45 ? ?

2

1 ? 22

?

? 2

2

2

2? 2

?

.

4

Thus

2? 2

.

2

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

cos?22. 5 ? ? ?

sin

?

12

?

2? 3

.

2

19.

a. 2 sin?18 ? ? cos?18 ? ? ? sin?2 ? 18 ? ? ? sin?36 ? ?.

b. 2 sin?3?? cos?3?? ? sin?6??.

21.

a. cos 2 ?34 ? ? ? sin 2 ?34 ? ? ? cos?2 ? 34 ? ? ? cos?68 ? ?.

b. cos 2 ?5?? ? sin 2 ?5?? ? cos?2 ? 5?? ? cos?10??.

23.

a. First note that

cos 2 ?4 ? ? ?

1 ? cos?8 ? ?

2

and thus

1 ? cos?8 ? ? ? 2 cos 2 ?4 ? ?.

Also

sin?8 ? ? ? 2 sin?4 ? ? cos?4 ? ?.

Thus

sin?8 ? ?

2 sin?4 ? ? cos?4 ? ?

?

? tan?4 ? ?.

1 ? cos?8 ? ?

2 cos 2 ?4 ? ?

b. Using the same kind of reasoning as in part a we have

1 ? cos?4?? ? 2 sin 2 ?2??

and

sin?4?? ? 2 sin?2?? cos?2??.

Thus

2 sin 2 ?2??

1 ? cos?4??

?

? tan?2??.

2 sin?2?? cos?2??

sin?4??

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

1 ? 45

sin 2 x ? 1 ? cos x ?

? 1

2

10

2

2

x

?

?

and since 0 ? 2 ? 45 we conclude that

sin x

2

?

1 ?

10

10

.

10

Likewise

cos

2

x

2

1?

? 1 ? cos x ?

2

2

4

5

? 9

10

and we obtain

cos x

2

?

3 10

.

10

This gives us

tan x ? 1 .

3

2

?

?

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

1 ? 2 32

1

?

cos

x

x

?

?

sin

2

2

2

?

?

and since 45 ? x ? 90 we conclude that

2

sin x

2

?

2

3?2 2

? 1 ?

?

2

3

6

3?2 2

.

6

Likewise

cos

2

x

2

1 ? 2 32

1

?

cos

x

?

?

2

2

2

3?2 2

? 1 ?

?

2

3

6

and we obtain

cos x

2

?

3?2 2

.

6

tan x

2

?

3?2 2

.

3?2 2

This gives us

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

1 ? 23

? 1

sin 2 x ? 1 ? cos x ?

2

6

2

2

x

?

?

and since 135 ? 2 ? 180 , we conclude that

sin x

2

?

1 ?

6

6

.

6

Likewise

cos

2

x

2

1?

? 1 ? cos x ?

2

2

2

3

? 5

6

and we obtain

cos x

2

??

5 ? ? 30 .

6

6

This gives us

5

.

tan x ? ?

5

2

31. We want to write the product sin?2x? cos?3x? as a sum. Note that

sin?2x? cos?3x? ? cos?2x? sin?3x? ? sin?2x ? 3x?

and that

sin?2x? cos?3x? ? cos?2x? sin?3x? ? sin?2x ? 3x?

Adding the above two equations we obtain

2 sin?2x? cos?3x? ? sin?5x? ? sin??x?.

Using the fact that sin??x? ? ? sin x, we obtain

sin?2x? cos?3x? ? 1 ?sin?5x? ? sin?x??.

2

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