chapter 7 additional topics in trigonometry

Chapter 7 Additional Topics in Trigonometry

894 Copyright © 2014 Pearson Education, Inc.

Mid-Chapter 7 Check Point

1. 180 32 41 107C Use the Law of Sines to find b.

sin sin20

sin 32 sin 4120sin 41

sin 3224.8

a b

A Bb

b

b

Use the Law of Sines to find c.

sin sin20

sin 32 sin10720sin107

sin 3236.1

a c

A Cc

c

c

The solution is 107 , 24.8,and 36.1C b c .

2. Use the Law of Sines to find B.

sin sin63 57

sin 42 sin57sin 42

sin63

sin 0.6054

a b

A B

B

B

B

There are two angles possible:

1 237 , 180 37 143B B

2B is impossible, since 42 143 185 .

1180 180 37 42 101C B A

Use the Law of Sines to find c.

sin sin63

sin101 sin 4263sin101

sin 4292.4

c a

C Ac

c

c

There is one triangle and the solution is

1 (or ) 37 , 101 , and 92.4B B C c .

3. Use the Law of Sines to find angle B.

sin sin6 7

sin 65 sin7sin 65

sin6

sin 1.0574

a b

A B

B

B

B

The sine can never exceed 1. There is no triangle with the given measurements.

4. Use the Law of Cosines to find b. 2 2 2

2 2 2

2

2 cos

10 16 2(10)(16)cos110

465.4464

21.6

b a c ac B

b

b

b

Use the Law of Sines to find A.

sin sin21.6 10

sin110 sin10sin110

sin21.6

sin 0.4350

26

b a

B A

A

A

A

A

Find the third angle. 180 180 26 110 44C A B

The solution is 26 , 44 , and 21.6A C b .

5. Use the Law of Sines to find angle A.

sin sin16 13

sin sin 4216sin 42

sin13

sin 0.8235

a c

A C

A

A

A

There are two angles possible:

1 255 , 180 55 125A A

There are two triangles:

1 1

2 2

180 180 42 55 83

180 180 42 125 13

B C A

B C A

Use the Law of Sines to find 1 2andb b .

Mid-Chapter Check Point

Copyright © 2014 Pearson Education, Inc. 895

1

1

1

1

2

2

2

2

sin sin

13

sin83 sin 4213sin83

19.3sin 42

sin sin

13

sin13 sin 4213sin13

4.4sin 42

b c

B C

b

b

b c

B C

b

b

In one triangle, the solution is

1 1 155 , 83 , 19.3A B b .

In the other triangle, 2 2 2125 , 13 , 4.4A B b .

6. Use the Law of Cosines to find the angle opposite the longest side. Thus, find angle C.

2 2 2

2 2 2

2 2 2

2 cos

cos2

5 7.2 10.1cos

2 5 7.2cos 0.3496

110

c a b ab C

a b cC

ab

C

C

C

Use the Law of Sines to find angle A.

sin sin5 10.1

sin sin1105sin110

sin10.1

sin 0.4652

28

a c

A C

A

A

A

A

Find the third angle. 180 180 28 110 42B A C

The solution is 28 , 42 , andA B 110C

7. The area of the triangle is half the product of the lengths of the two sides times the sine of the included angle.

1Area (5)(7)(sin 36 ) 10

2

The area of the triangle is approximately 10 square feet.

8. Begin by calculating one-half the perimeter: 1 1

( ) (7 9 12) 142 2

s a b c

Use Heron’s formula to find the area.

Area ( )( )( )

14(14 7)(14 9)(14 12)

980 31

s s a s b s c

The area of the triangle is approximately 31 square meters.

9. The first train traveled 100 miles, the second train traveled 80 miles. Use the Law of Cosines to find the distance.

2 2 2

2 2 2

2

2 cos

100 80 2(100)(80)cos110

21872.32229

147.9

c a b ab C

c

c

c

The two trains are 147.9 miles apart.

10. Let the fire be at point C. 90 56 34

90 23 67

180 34 67 79

A

B

C

Use the law of sines to find b.

sin sin16


sin 7915.0

b c

B Cb

b

b

The fire is 15.0 miles from station A



11. Let point A be where the angle of elevation is 66 .

Let point B be where the angle of elevation is 50 .

Let point C be at the top of the tree. 180 180 66 50 64C A B

Use the law of sines to find a.

sin sin420


sin 64426.9

a c

A Ca

a

a

The height of the tree, h, is given by sin

426.9sin50

327.0

h a B

h

h

The tree is 327.0 feet tall.

12. 5 3 2

cos 3cos4 2

x rπθ

5 3 2sin 3sin

4 2y r

πθ

Ordered pair: 3 2 3 2

,2 2

13. cos 6cos 02

x rπθ

sin 6sin 62

y rπθ

Ordered pair: 0, 6

14. 22 2 22 2 3 4r x y

2 3tan 3

2

y

xθ

Because θ lies in quadrant IV, 5

23 3

π πθ π

Polar coordinates: 5

4,3

π

15. 2 2 2 2( 6) 0 6r x y

0tan 0

6

y

xθ

θ π

Polar coordinates: 6,π

16.

a. 11

4,4

π

b. 7

4,4

π

c. 5

4,4

π

17.

a. 5 5

,2 2

π

b. 5 3

,2 2

π

c. 5 3

,2 2

π

18.

5 7

5 cos sin 7

5cos sin 7

7

5cos sin

x y

r r

r

r

θ θθ θ

θ θ

Mid-Chapter Check Point


19. 7

sin 7

7

sin7csc

y

r

r

r

θ

θθ

20. 2 2

2 2

2 2 2 2 2

2

2

( 1) 1

( cos 1) ( sin ) 1

cos 2 cos 1 sin 1

2 cos 0

2 cos

2cos

x y

r r

r r r

r r

r r

r

θ θ

θ θ θθ

θθ

21. 2

2 2

6

36

36

r

r

x y

22. 3

tan tan3

tan 3

3

3

y

x

y x

πθ

πθ

θ

23. 3csc

3

sinsin 3

3

r

r

r

y

θ

θθ

24. 2

2 2

2 2

2 2

2 2

10cos

10 cos

10

10 0

10 25 25

( 5) 25

r

r r

x y x

x x y

x x y

x y

θθ



25. 2

2

2

2 2

2

2

4sin sec

4sin

cos

cos 4sin

cos 4 sin

4

1

4

r

r

r

r r

x y

y x

θ θθθ

θ θθ θ

26. 1 4cosr θ

a. Replace θ with .θ

1 4cos( )

1 4cos

r

r

θθ

The graph is symmetric with respect to the polar axis.

b. Replace ,r θ with , .r θ

1 4cos( )

1 4cos

r

r

θθ

The graph may or may not be symmetric with

respect to the line 2

πθ .

c. Replace r with –r. 1 4cos

1 4cos

r

r

θθ

The graph may or may not be symmetric with respect to the polar axis.

27. 2 4cos 2r θ

a. Replace θ with .θ

2

2

4cos 2

4cos 2

r

r

θ

θ


b. Replace ,r θ with , .r θ

2

2

4cos 2

4cos 2

r

r

θ

θ

The graph is symmetric with respect to the line

2

πθ .

c. Replace r with –r.

22

4cos 2

4cos 2

r

r

θ

θ


28.

29.

30.

Section 7.5 Complex Numbers in Polar Form; Demoivre’s Theorem


31.

32.

Section 7.5

Check Point Exercises

1. a. z = 2 + 3i corresponds to the point (2, 3). Plot the complex number by moving two units to the right on the real axis and 3 units up parallel to the imaginary axis.

b. z = –3 – 5i corresponds to the point (–3, –5). Plot the complex number by moving three units to the left on the real axis and five units down parallel to the imaginary axis.

c. Because 4 4 0 ,z i this complex number corresponds to the point (–4, 0). Plot the complex number by moving four units to the left on the real axis.

d. Because 0 ,z i i this complex number corresponds to the point (0, –1). Plot the complex number by moving one unit down on the imaginary axis.

2 a.

2 2

5 12

5, 12

5 12 25 144 169 13

z i

a b

z

b.

2 2

2 3

2, 3

2 ( 3) 4 9 13

z i

a b

z



3. 1 3z i corresponds to the point 1, 3 .

Use 2 2r a b with a = –1 and 3b to find r.

22 2 2( 1) 3

1 3 4 2

r a b

Use tan with 1 and 3 to find .b

a ba

θ θ

3tan 3

1

b

aθ

Because tan 33

π and θ lies in

quadrant III, 3 4

.3 3 3 3

π π π πθ π

The polar form of 1 3z i is

4 4(cos sin ) 2 cos sin .

3 3z r i i

π πθ θ

4. The complex number 4(cos30 sin 30 )z i is in polar form, with r = 4 and 30 .θ We use exact values for

cos30 and sin 30 to write the number in rectangular form.

3 14(cos30 sin 30 ) 4 2 3 2

2 2i i i

The rectangular form of 4(cos30 sin 30 )z i is 2 3 2z i .

5. 1 2 [6(cos 40 sin 40 )][5(cos 20 sin 20 )] (6 5)[(cos(40 20 ) sin(40 20 )]

30(cos60 sin 60 )

z z i i i

i

6. 1

2

4 450 cos sin

50 4 43 3cos sin 10(cos sin )

5 3 3 3 35 cos sin

3 3

iz

i iz

i

π ππ π π π π π

π π

7. 5 5 3 12(cos30 sin 30 ) 2 cos(5 30 ) sin(5 30 ) 32(cos150 sin150 ) 32

2 2

16 3 16

i i i i

i



8. Write 1 in (cos sin )i r iθ θ form.

2 2 2 21 1 2

1tan 1 and

1 4

1 (cos sin ) 2 cos sin4 4

r a b

b

a

i r i i

πθ θ

π πθ θ

Use DeMoivre’s Theorem to raise 1 + i to the fourth power.

4

44(1 ) 2 cos sin 2 cos 4 sin 4 4(cos sin ) 4( 1 0 ) 44 4 4 4

i i i i iπ π π π π π

9. From DeMoivre’s Theorem for finding complex roots, the fourth roots of 16(cos60 sin 60 )i are

4 60 360 60 36016 cos sin , 0,1,2,3.

4 4kk k

z i k

Substitute 0, 1, 2, and 3 for k in the above expression for .kz

4 40

4 41

60 360 0 60 360 0 60 6016 cos sin 16 cos sin 2(cos15 sin15 )

4 4 4 4


4 4 4 4

z i i i

z i i i

4 42

4 43


4 4 4 4

60 360 3 60 360 3 1140 114016 cos sin 16 cos sin 2(cos 285 si

4 4 4 4

z i i i

z i i i

n 285 )

10. First, write 27 in polar form. 27 (cos sin ) 27(cos0 sin 0).r iθ θ From DeMoivre’s theorem for finding complex

roots, the cube roots of 27 are

3

30

31

0 2 0 227 cos sin , 0,1,2.

3 3

0 2 0 0 2 027 cos sin 3(cos0 sin 0) 3(1 0) 3

3 3

0 2 1 0 2 1 2 2 127 cos sin 3 cos sin 3

3 3 3 3 2

kk k

z i k

z i i i

z i i i

π π

π π

π π π π

32

3 3 3 3

2 2 2

0 2 2 0 2 2 4 4 1 3 3 3 327 cos sin 3 cos sin 3

3 3 3 3 2 2 2 2

i

z i i i iπ π π π



Concept and Vocabulary Check 7.5

1. real; imaginary

2. absolute value

3. modulus; argument

4. 2 2a b ; b

a

5. 1 2r r ; 1 2θ θ ; 1 2θ θ ; multiplying; adding

6. 1

2

r

r; 1 2θ θ ; 1 2θ θ

7. dividing; subtracting

8. nr ; nθ ; nθ ; n

Exercise Set 7.5 1. Because 4 0 4 ,z i i this complex number

corresponds to the point (0, 4).

With a = 0 and b = 4, 2 20 4 16 4.z

2. Because 3 0 3 ,z i i this complex number corresponds to the point (0, 3).

With a = 0 and b = 3, 2 20 3 9 3.z

3. Because 3 3 0 ,z i this complex number corresponds to the point (3, 0).

With a = 3 and b = 0, 2 23 0 9 3.z

4. Because 4 4 0 ,z i this complex number corresponds to the point (4, 0).

With a = 4 and b = 0, 2 24 0 16 4.z

5. z = 3 + 2i corresponds to the point (3, 2).

With a = 3 and b = 2, 2 23 2 9 4 13 .z


With a = 2 and b = 5,

2 22 5 4 25 29 .z



7. z = 3 – i corresponds to the point (3, –1).

With a = 3 and b = –1, 2 23 ( 1) 9 1 10 .z

8. z = 4 – i corresponds to the point (4, –1).

With a = 4 and b = –1,

2 24 ( 1) 16 1 17 .z

9. z = –3 + 4i corresponds to the point (–3, 4).

With a = –3 and b = 4, 2 2( 3) 4 9 16 25 5.z

10. z = –3 – 4i corresponds to the point (–3, –4).

With a = –3 and b = –4,

2 2( 3) ( 4) 9 16 25 5.z


Use 2 2 and tan ,b

r a ba

θ with a = 2 and b =

2, to find r and .θ 2 22 2 4 4 8 2 2

2tan 1

2

r

θ

Because tan 14

π and θ lies in quadrant I,

4

πθ .

2 2 (cos sin )

2 2 cos sin4 4

or 2 2(cos 45 sin 45 )

z i r i

i

i

θ θπ π


Use 2 2 and tan ,b

r a ba

θ with a = 1

and 3,b to find r and .θ

221 3 1 3 4 2

3tan 3

1

r

θ

Because tan 33

π and θ lies in quadrant I, 3

πθ .

1 3 (cos sin ) 2 cos sin3 3

z i r i iπ πθ θ

or

2(cos60 sin 60 )i



13. z = –1 – i corresponds to the point (–1, –1).

Use 2 2 and tan ,b

r a ba

θ with

a = –1 and b = –1, to find r and .θ 2 2( 1) ( 1) 1 1 2

1tan 1

1

r

θ

Because tan 14

π and θ lies in

quadrant III,5

4 4

π πθ π .

1 (cos sin )

5 52 cos sin

4 4

or 2(cos 225 sin 225 )

z i r i

i

i

θ θπ π

14. z = 2 – 2i corresponds to the point (2, –2).

Use 2 2 and tan ,b

r a ba


–2, to find r and .θ 2 22 ( 2) 4 4 8 2 2

2tan 1

2

r

θ

Because tan 14

π and θ lies in quadrant IV,

72

4 4

π πθ π .

2 2 (cos sin )

7 72 2 cos sin

4 4

z i r i

i

θ θπ π

or 2 2(cos315 sin 315 )i

15. z = –4i corresponds to the point (0, –4).

Use 2 2 and tan ,b

r a ba

θ with

a = 0 and b = –4, to find r and .θ 2 20 ( 4) 16 4

4tan is undefined.

0

r

θ

Because tan2

π is undefined and θ lies on the

negative y-axis, 3

2 2

π πθ π .

4 (cos sin )

3 34 cos sin

2 2

or 4(cos270 sin 270 )

z i r i

i

i

θ θπ π

16. z = –3i corresponds to the point (0, –3).

Use 2 2 and tan ,b

r a ba


–3, to find r and .θ 2 20 ( 3) 9 3

3tan is undefined.

0

r

θ

Because tan2

π is undefined and θ lies on the

negative y-axis, 3

2 2

π πθ π .

3 33 (cos sin ) 3 cos sin

2 2z i r i i

π πθ θ

or 3(cos270 sin 270 )i



17. 2 3 2z i corresponds to the point 2 3, 2 .

Use 2 2 and tan ,b

r a ba

θ with

a = 2 3 and b = –2, to find r and .θ

2 22 3 ( 2) 12 4 16 4

2 1tan

2 3 3

r

θ

Because 1

tan6 3

π and θ lies in

quadrant IV,11

26 6

π πθ π .

2 3 2 (cos sin )

11 114 cos sin

6 6

or 4(cos330 sin 330 )

z i r i

i

i

θ θπ π

18. 2 2 3z i corresponds to the point 2, 2 3 .

Use 2 2 and tan ,b

r a ba

θ with a = –2 and

2 3b , to find r and .θ

22( 2) 2 3 4 12 16 4

2 3tan 3

2

r

θ

Because tan 33

π and θ lies in

quadrant II,2

3 3

π πθ π .

2 2 3 (cos sin )

2 24 cos sin

3 3

z i r i

i

θ θπ π

or 4(cos120 sin120 )i

19. z = –3 corresponds to the point (–3, 0).

Use 2 2 and tan ,b

r a ba

θ with

a = –3 and b = 0, to find r and .θ 2 2( 3) 0 9 3

0tan 0

3

r

θ

Because tan 0 0 and θ lies on the negative x-axis, 0θ π π .

3 (cos sin )

3 cos sin

or 3(cos180 sin180 )

z r i

i

i

θ θπ π

20. z = –4 corresponds to the point (–4, 0).

Use 2 2 and tan ,b

r a ba

θ with a = –4 and b

= 0, to find r and .θ 2 2( 4) 0 16 4

0tan 0

4

r

θ

Because tan 0 0 and θ lies on the negative x-axis, 0θ π π .

4 (cos sin ) 4 cos sinz r i iθ θ π π

or 4(cos180 sin180 )i



21. 3 2 3 3z i corresponds to the point

3 2, 3 3 .

Use 2 2 and tan ,b

r a ba

θ with

a = 3 2 and b = 3 3 , to find r and .θ

2 23 2 3 3 18 27

45 3 5

3 3 3 6tan

23 2 2

r

θ

Because θ lies in quadrant III,

1 6180 tan 180 50.8

2

230.8

θ

3 2 3 3 (cos sin )

3 5(cos 230.8 sin 230.8 )

z i r i

i

θ θ

22. 3 2 3 2z i corresponds to the point

3 2, 3 2 .

Use 2 2 and tan ,b

r a ba

θ with 3 2a and

3 2,b to find r and .θ

2 23 2 3 2 18 18 36 6

3 2tan 1

3 2

r

θ

Because tan 14


72

4 4

π πθ π .

3 2 3 2 (cos sin )

7 76 cos sin

4 4

z i r i

i

θ θπ π

or 6(cos315 sin 315 )i


Use 2 2 and tan ,b

r a ba

θ with

a = –3 and b = 4, to find r and .θ 2 2( 3) (4) 9 16 25 5

4 4tan

3 3

r

θ

Because θ lies in quadrant II,

1 4180 tan 180 53.1 126.9 .

3θ

3 4 (cos sin )

5(cos126.9 sin126.9 )

z i r i

i

θ θ


Use 2 2 and tan ,b

r a ba

θ with a = –2 and b

= 3, to find r and .θ 2 2( 2) (3) 4 9 13

3 3tan

2 2

r

θ



Because θ lies in quadrant II,

1 3180 tan 180 56.3 123.7

2θ

.

2 3 (cos sin )

13(cos123.7 sin123.7 )

z i r i

i

θ θ


Use 2 2 and tan ,b

r a ba

θ with

a = 2 and b = 3 , to find r and .θ

222 3 4 3 7

3 3tan

2 2

r

θ

Because θ lies in quadrant IV,

1 3360 tan 360 40.9 319.1

2θ

2 3 (cos sin )

7(cos319.1 sin 319.1 )

z i r i

i

θ θ

26. 1 5z i corresponds to the point 1, 5

Use 2 2 and tan ,b

r a ba

θ with a = 1 and

5b , to find r and .θ

221 5 1 5 6

5tan 5

1

r

θ

Because θ lies in quadrant IV,

1360 tan 5 360 65.9 294.1θ

1 5 (cos sin )

6(cos294.1 sin 294.1 )

z i r i

i

θ θ

27. 3 1

6(cos30 sin 30 ) 62 2

3 3 3

i i

i

The rectangular form of

6(cos30 sin 30 ) is 3 3 3 .z i z i

28. 1 3

12(cos60 sin 60 ) 12 6 6 32 2

i i i


12(cos60 sin 60 ) is 6 6 3z i z i .

29. 1 3

4(cos240 sin 240 ) 42 2

2 2 3

i i

i


4(cos 240 sin 240 ) is 2 2 3.z i z i

30. 3 1

10(cos 210 sin 210 ) 102 2

5 3 5

i i

i


10(cos 210 sin 210 ) is = 5 3 5 .z i z i

31. 7 7 2 2

8 cos sin 84 4 2 2

4 2 4 2

i i

i

π π


7 78 cos sin is 4 2 4 2 .

4 4i z i

π π

32. 5 5 3 1

4 cos sin 46 6 2 2

2 3 2

i i

i

π π


5 54 cos sin is 2 3 2 .

6 6z i z i

π π



33. 5 cos sin 5 0 (1)2 2

5

i i

i

π π


5 cos sin is 5 .2 2

z i z iπ π

34. 3 37 cos sin 7 0 ( 1) 7

2 2i i i

π π


3 37 cos sin is 7 .

2 2z i z i

π π

35.

20 cos 205 sin 205

20 0.91 ( 0.42) 18.2 8.4

i

i i


20 cos 205 sin 205z i is 18.2 8.5z i .

36. 30 cos2.3 sin 2.3 20.0 22.4i i


30 cos2.3 sin 2.3 is 20.0 22.4z i z i

37.

1 2

6(cos 20 sin 20 ) 5(cos50 sin50 )

(6 5) cos(20 50 ) sin(20 50 )

30(cos70 sin 70 )

z z

i i

i

i

38.

1 2

4(cos15 sin15 ) 7(cos 25 sin 25 )

(4 7) cos(15 25 ) sin(15 25 )

28(cos 40 sin 40 )

z z

i i

i

i

39. 1 2 3 cos sin 4 cos sin (3 4) cos sin5 5 10 10 5 10 5 10

3 312 cos sin

10 10

z z i i i

i

π π π π π π π π

π π

40. 1 25 5 5 5

3 cos sin 10 cos sin (3 10) cos sin8 8 16 16 8 16 8 16

11 1130 cos sin

16 16

z z i i i

i


π π

41. 1 23 4 3 4

cos sin cos sin cos sin cos sin4 4 3 3 4 3 4 3 12 12 12 12

7 7cos sin

12 12

z z i i i i

i

π π π π π π π π π π π π

π π



42. 1 22 3 2 3

cos sin cos sin cos sin cos sin6 6 4 4 6 4 6 4 12 12 12 12

5 5cos sin

12 12

z z i i i i

i

π π π π π π π π π π π π

π π

43. Begin by converting 1 21 and 1z i z i to polar form.

For 1 :z a = 1 and b = 1

2 2 2 2

1

2

2 2 2 2

1 1 2

1tan 1 and .

1 4

(cos sin ) 2 cos sin4 4

For : 1 and 1

( 1) 1 2

1tan 1

1

r a b

b

a

z r i i

z a b

r a b

b

a

πθ θ

π πθ θ

θ

Because tan 1 and 4

π θ lies in quadrant II, 3

.4 4

π πθ π

23 3


z r i iπ πθ θ

Now, find the product.

1 2 (1 )( 1 )

3 3 3 32 cos sin 2 cos sin 2 2 cos sin

4 4 4 4 4 4 4 4

2 cos sin

z z i i

i i i

i


π π

44. Begin by converting 1 21 and 2 2z i z i to polar form.

1

2 2 2 2

For : 1 and 1

1 1 2

1tan 1 and

1 4

z a b

r a b

b

a

πθ θ


z r i iπ πθ θ

2

2 2 2 2

For : 2 and 2

2 2 8 2 2

2tan 1 and

2 4

z a b

r a b

b

a

πθ θ

2 (cos sin ) 2 2 cos sin4 4

z r i iπ πθ θ

Now, find the product.

1 2 (1 )(2 2 ) 2 cos sin 2 2 cos sin 2 2 2 cos sin4 4 4 4 4 4 4 4

4 cos sin2 2

z z i i i i i

i


π π



45. 1

2

20(cos75 sin 75 ) 20cos(75 25 ) sin(75 25 )

4(cos 25 sin 25 ) 4

5(cos50 sin50 )

z ii

z i

i

46. 1

2

50(cos80 sin80 ) 50cos(80 20 ) sin(80 20 )

10(cos 20 sin 20 ) 10

5(cos60 sin 60 )

z ii

z i

i

47. 1

2

3 cos sin3 35 5

cos sin cos sin4 5 10 5 10 4 10 10

4 cos sin10 10

iz

i iz

i

π ππ π π π π π

π π

48. 1

2

5 53 cos sin

3 5 5 3 40 9 40 918 18cos sin cos sin

10 18 16 18 16 10 144 144 144 14410 cos sin

16 16

3 31 31cos sin

10 144 144

iz

i iz

i

i

π ππ π π π π π π π

π π

π π

49. 1

2

cos80 sin80cos(80 200 ) sin(80 200 ) cos( 120 ) sin( 120 )

cos 200 sin 200

cos240 sin 240

z ii i

z i

i

50. 1

2

cos70 sin 70cos(70 230 ) sin(70 230 ) cos( 160 ) sin( 160 )

cos 230 sin 230

cos200 sin 200

z ii i

z i

i

51. Begin by converting 1 22 2 and 1z i z i to polar form.

For 1 :z a = 2 and b = 2

2 2 2 2

1

2

2 2 2 2

2 2 8 2 2

2tan 1 and

2 4

(cos sin ) 2 2 cos sin4 4

For : 1 and 1

1 1 2

1tan 1 and

1 4

r a b

b

a

z r i i

z a b

r a b

b

a

πθ θ

π πθ θ

πθ θ

2 (cos sin ) 2 cos sin .4 4

z r i iπ πθ θ

Now, find the quotient.

1

2

2 2 cos sin2 2 4 4

2 cos sin 2(cos0 sin 0)1 4 4 4 4

2 cos sin4 4

iz i

i iz i

i

π ππ π π π

π π



52. Begin by converting 1 22 2 and 1z i z i to polar form. 1For : 2 and 2z a b

2 2 2 22 ( 2) 8 2 2

2tan 1

2

r a b

b

aθ

Because tan 1 and 4

π θ lies in quadrant IV,

72 .

4 4

π πθ π

17 7

(cos sin ) 2 2 cos sin4 4

z r i iπ πθ θ

2

2 2 2 2

For : 1 and 1

1 ( 1) 2

1tan 1

1

z a b

r a b

b

aθ

Because tan 1 and 4

π θ lies in quadrant IV,

72 .

4 4

π πθ π

27 7


z r i iπ πθ θ

Now, find the quotient.

1

2

7 72 2 cos sin

2 2 7 7 7 74 42 cos sin

7 71 4 4 4 42 cos sin

4 4

2(cos0 sin 0)

iz i

iz i

i

i

π ππ π π π

π π

53. 3 3 2 24(cos15 sin15 ) (4) cos(3 15 ) sin(3 15 ) 64(cos 45 sin 45 ) 64

2 2

32 2 32 2

i i i i

i

54. 3 3 3 12(cos10 sin10 ) (2) cos(3 10 ) sin(3 10 ) 8(cos30 sin 30 ) 8 4 3 4

2 2i i i i i

55. 3 32(cos80 sin80 ) (2) cos(3 80 ) sin(3 80 ) 8(cos 240 sin 240 )

1 38 4 4 3

2 2

i i i

i i

56. 3 32(cos 40 sin 40 ) (2) cos(3 40 ) sin(3 40 ) 8(cos120 sin120 )

1 38 4 4 3

2 2

i i i

i i



57. 6 6

1 1 1 1 1cos sin cos 6 sin 6 cos sin (0 )

2 12 12 2 12 12 64 2 2 64 64i i i i i

π π π π π π

58. 5 5

1 1 1cos sin cos 5 sin 5 cos sin

2 10 10 2 10 10 32 2 2

1 1(0 )

32 32

i i i

i i

π π π π π π

59. 4

45 5 5 52 cos sin 2 cos 4 sin 4

6 6 6 6

20 20 4 44 cos sin 4 cos sin

6 6 3 3

1 34 2 2 3

2 2

i i

i i

i i

π π π π

π π π π

60. 6

65 5 5 5 30 303 cos sin 3 cos 6 sin 6 27 cos sin

18 18 18 18 18 18

5 527 cos sin

3 3

1 3 27 27 327

2 2 2 2

i i i

i

i i

π π π π π π

π π

61. Write 1 i in (cos sin )r iθ θ form.

2 2 2 21 1 2

1tan 1 and

1 4


r a b

b

a

i r i i

πθ θ

π πθ θ

Use DeMoivre’s Theorem to raise 1 i to the fifth power.

5

55(1 ) 2 cos sin 2 cos 5 sin 54 4 4 4

5 5 2 24 2 cos sin 4 2 4 4

4 4 2 2

i i i

i i i

π π π π

π π



62. Write 1 – i in (cos sin )r iθ θ form.

2 2 2 21 ( 1)

2

1tan 1

1

r a b

b

aθ

Because tan 14


7

24 4

π πθ π .

1 (cos sin )

7 72 cos sin

4 4

i r i

i

θ θπ π

Use DeMoivre’s Theorem to raise 1 – i to the fifth power.

5

55 7 7 7 7 35 35(1 ) 2 cos sin 2 cos 5 sin 5 4 2 cos sin

4 4 4 4 4 4

2 24 2

2 2

4 4

i i i i

i

i

π π π π π π

63. Write 3 i in (cos sin )r iθ θ form.

2 22 2 3 1 4 2

1 1tan

3 3

r a b

b

aθ

Because 1

tan 30 and 3

θ lies in quadrant IV, 360 30 330 .θ

3 (cos sin ) 2(cos330 sin 330 )i r i iθ θ

Use DeMoivre’s Theorem to raise 3 i to the sixth power.

66 6( 3 ) 2(cos330 sin 330 ) (2) cos(6 330 ) sin(6 330 )

64(cos1980 sin1980 ) 64(cos180 sin180 )

64( 1 0 ) 64

i i i

i i

i

64. Write 2 in (cos sin )i r iθ θ form.

22 2 22 ( 1) 3

1 2tan

22

r a b

b

aθ

Because θ lies in quadrant IV, 1 2360 tan 360 35.3 324.7 .

2θ

2 (cos sin ) 3(cos324.7 sin 324.7 )i r i iθ θ



Use DeMoivre’s Theorem to raise 2 i to the fourth power.

44

4

2 3(cos324.7 sin 324.7 )

3 cos(4 324.7 ) sin(4 324.7 )

9(cos1298.8 sin1298.8 )

7 5.7

i i

i

i

i

Exact answer is 7 4 2.i

65. 9(cos30 sin 30 )i

2

0

1

30 360 30 3609 cos sin , 0,1

2 2


2 2 2 2

30 360 1 30 3609 cos sin

2

kk k

z i k

z i i i

z i

1 390 390

9 cos sin 3(cos195 sin195 )2 2 2

i i

66. 25(cos 210 sin 210 )i

2

0

1

210 360 210 36025 cos sin , 0,1

2 2

210 360 0 210 360 0 210 21025 cos sin 25 cos sin

2 2 2 2

5(cos105 sin105 )

210 360 125 cos

2

kk k

z i k

z i i

i

z

210 360 1 570 570

sin 25 cos sin2 2 2

5(cos 285 sin 285 )

i i

i

67. 8(cos210 sin 210 )i

3

3 30

31

210 360 210 3608 cos sin , 0,1,2

3 3

210 360 0 210 360 0 210 2108 cos sin 8 cos sin

3 3 3 3

2(cos70 sin 70 )

210 360 18 cos

3

kk k

z i k

z i i

i

z

3

3 32

210 360 1 570 570sin 8 cos sin

3 3 3

2(cos190 sin190 )

210 360 2 210 360 2 930 9308 cos sin 8 cos sin

3 3 3 3

2(cos310 sin 310

i i

i

z i i

i

)



68. 27(cos306 sin 306 )i

3

3 30

31

306 360 306 36027 cos sin , 0,1,2

3 3

306 360 0 306 360 0 306 30627 cos sin 27 cos sin

3 3 3 3

3(cos102 sin102 )

306 360 127 cos

3

kk k

z i k

z i i

i

z

3

3 32

306 360 1 666 666sin 27 cos sin

3 3 3

3(cos 222 sin 222 )

306 360 2 306 360 2 1026 102627 cos sin 27 cos sin

3 3 3 3

3(cos34

i i

i

z i i

2 sin 342 )i

69. 4 4

81 cos sin3 3

iπ π

4 43 34

4 43 34 4

0

4 43 34

1

2 281 cos sin , 0,1,2,3

4 4

2 0 2 0 1 3 3 3 381 cos sin 81 cos sin 3

4 4 3 3 2 2 2 2

2 1 2 181 cos sin

4 4

k

k kz i k

z i i i i

z i

π π

π π

π π

π π

π π π π

π π

4

4 43 34 4

2

4 43 34

3

5 5 3 1 3 3 381 cos sin 3

6 6 2 2 2 2

2 2 2 2 4 481 cos sin 81 cos sin

4 4 3 3

1 3 3 3 33

2 2 2 2

2 3 2 381 cos sin

4

i i i

z i i

i i

z i

π π

π π

π π

π π π π

π π

4 11 11

81 cos sin4 6 6

3 1 3 3 33

2 2 2 2

i

i i

π π



70. 5 5

32 cos sin3 3

iπ π

5 53 35

5 53 35 5

0

5 53 35

1

2 232 cos sin , 0,1,2,3,4

5 5

2 0 2 0 1 332 cos sin 32 cos sin 2 1 3

5 5 3 3 2 2

2 1 2 132 cos sin

5 5

k

k kz i k

z i i i i

z i

π π

π π

π π

π π

π π π π

π π

5

5 53 35 5

2

5 53 35

3

11 1132 cos sin

15 15

2 0.67 (0.74) 1.3 1.5

2 2 2 2 17 1732 cos sin 32 cos sin

5 5 15 15

2( 0.91 ( 0.41)) 1.8 0.8

2 332 cos sin

5

i

i i

z i i

i i

z i

π π

π π

π π

π π π π

π

5

5 53 35 5

4

2 3 23 2332 cos sin

5 15 15

2(0.10 ( 0.99)) 0.2 2.0

2 4 2 4 29 2932 cos sin 32 cos sin

5 5 15 15

2(0.98 ( 0.21)) 2.0 0.4

i

i i

z i i

i i

π π

π π π

π π π π

71. 32 = 32(cos0°+ isin0°)

5

5 50

5 51

0 360 0 36032 cos sin , 0,1,2,3,4

5 5

0 360 0 0 360 032 cos sin 32(cos0 sin 0 ) 2(1 0 ) 2

5 5

0 360 1 0 360 132 cos sin 32(c

5 5

kk k

z i k

z i i i

z i

5 52

5 53

os72 sin 72 ) 2(0.31 (0.95))

0.6 1.9

0 360 2 0 360 232 cos sin 32(cos144 sin144 ) 2( 0.81 (0.59))

5 5

1.6 1.2

0 360 3 0 360 332 cos sin 32(cos

5 5

i i

i

z i i i

i

z i

5 54

216 sin 216 ) 2( 0.81 ( 0.59))

1.6 1.2

0 360 4 0 360 432 cos sin 32(cos 288 sin 288 ) 2(0.31 ( 0.95))

5 5

0.6 1.9

i i

i

z i i i

i



72. 64 64 cos0 sin 0i

6

6 60

6 61

0 360 0 36064 cos sin , 0,1,2,3,4,5

6 6

0 360 0 0 360 064 cos sin 64 cos0 sin 0 2(1 0 ) 2

6 6

0 360 1 0 360 164 cos sin 64

6 6

kk k

z i k

z i i i

z i

6 62

6 63

1 3cos60 sin 60 2 1 3

2 2

0 360 2 0 360 2 1 364 cos sin 64 cos120 sin120 2 1 3

6 6 2 2

0 360 3 0 360 364 cos sin 64 cos180 sin180 2

6 6

i i i

z i i i i

z i i

6 64

6 65

1 3cos 240 sin 240

2 2

( 1 0 ) 2

0 360 4 0 360 464 cos sin 64 2 1 3

6 6

0 360 5 0 360 5 1 364 cos sin 64 cos300 sin 300 2

6 6 2 2

i i

i

z i i

z i i i

1 3i

73. 1 1(cos0 sin 0 )i

3

3 30

3 31

0 360 0 3601 cos sin , 0,1,2

3 3

0 360 0 0 360 01 cos sin 1 cos0 sin 0 1(1 0 ) 1

3 3

0 360 1 0 360 11 cos sin 1 cos120 sin1

3 3

kk k

z i k

z i i i

z i i

3 32

1 3 1 320 1

2 2 2 2

0 360 2 0 360 2 1 31 cos sin 1 cos240 sin 240 1

3 3 2 2

1 3

2 2

i i

z i i i

i

74. 1(cos90 sin 90 )i i

3

3 30

31

90 360 90 3601 cos sin , 0,1,2

3 3

90 360 0 90 360 0 3 1 3 11 cos sin 1 cos30 sin 30 1

3 3 2 2 2 2

90 360 1 90 360 11 cos sin

3 3

kk k

z i k

z i i i i

z i

3

3 32

3 1 3 11 cos150 sin150 1

2 2 2 2

90 360 2 90 360 21 cos sin 1 cos 270 sin 270 1(0 ( 1))

3 3

i i i

z i i i i



75. 1 2 cos45 sin 45i i

4

4 40

41

45 360 45 3602 cos sin , 0,1,2,3

4 4

45 360 0 45 360 02 cos sin 2 cos11.25 sin11.25 1.1 0.2

4 4

45 360 1 45 360 12 cos sin

4 4

kk k

z i k

z i i i

z i

4

4 42

4 43

2 cos101.25 sin101.25 0.2 1.1

45 360 2 45 360 22 cos sin 2 cos191.25 sin191.25 1.1 0.2

4 4

45 360 3 45 360 32 cos sin 2 cos281.25 sin 281.25

4 4

i i

z i i i

z i i

0.2 1.1i

76. 1 2 cos135 sin135i i

5

5 50

51

135 360 135 3602 cos sin , 0,1,2,3,4

5 5

135 360 0 135 360 02 cos sin 2 cos 27 sin 27 0.95 0.49

5 5

135 360 1 135 360 12 cos sin

5 5

kk k

z i k

z i i i

z i

5

5 52

5 53

2 cos99 sin 99 0.17 1.06

135 360 2 135 360 22 cos sin 2 cos171 sin171 1.06 0.17

5 5

135 360 3 135 360 32 cos sin 2 cos243 sin 243 0.49 0

5 5

i i

z i i i

z i i

5 54

.95

135 360 4 135 360 42 cos sin 2 cos315 sin 315 0.76 0.76

5 5

i

z i i i

77.

2 2 3

1 cos90 sin 90 2 2 cos 45 sin 45 2 cos150 sin150

4 2 cos285 sin 285

1.4641 5.4641

i i i

i i i

i

i

78.

1 1 3 3

2 cos 45 sin 45 2 cos300 sin 300 2 cos150 sin150

4 2 cos 495 sin 495

4 2 cos135 sin135

4 4

i i i

i i i

i

i

i



79.

1 3 1

2 3 2

2 cos60 sin 60 2 cos315 sin 315

4 cos330 sin 330

2cos45 sin 45

21 1

2 2

i i

i

i i

i

i

i

80.

1 3 2 2 3

4 3 4

2 cos120 sin120 4 cos300 sin 300

8 cos330 sin 330

1 cos90 sin 90

cos90 sin 90

i i

i

i i

i

i

i

i

81. 6

6

6

6

6

1 0

1

1

1 0

cos0 sin 0

x

x

x

x i

x i

6

60

61

0 360 0 3601 cos sin , 0,1,2,3,4,5

6 6

0 360 0 0 360 01 cos sin cos0 sin 0 1 0 1

6 6

0 360 1 0 360 11 cos sin cos60 sin 6

6 6

kk k

z i k

z i i i

z i i

62

63

64

1 30

2 2

0 360 2 0 360 2 1 31 cos sin cos120 sin120

6 6 2 2

0 360 3 0 360 31 cos sin cos180 sin180 1 0 1

6 6

0 360 4 0 360 41 cos sin

6

i

z i i i

z i i i

z i

65

1 3 1 3cos 240 sin 240

6 2 2 2 2

0 360 5 0 360 5 1 3 1 31 cos sin cos300 sin 300

6 6 2 2 2 2

i i i

z i i i i



82.

6

6

6

6

6

1 0

1

1

1 0

1 cos180 sin180

x

x

x

x i

x i

6

60

61

180 360 180 3601 cos sin , 0,1,2,3,4,5

6 6

180 360 0 180 360 0 3 11 cos sin cos30 sin 30

6 6 2 2

180 360 1 180 360 11 cos sin

6 6

kk k

z i k

z i i i

z i

62

63

64

cos90 sin 90 0

180 360 2 180 360 2 3 11 cos sin cos150 sin150

6 6 2 2

180 360 3 180 360 3 3 11 cos sin cos 210 sin 210

6 6 2 2

180 3601 cos

i i i

z i i i

z i i i

z

65

4 180 360 4sin cos 270 sin 270 0

6 6

180 360 5 180 360 5 3 11 cos sin cos330 sin 330

6 6 2 2

i i i i

z i i i

83.

4

4

4

4

4

16 0

16

16

0 16

16 cos 270 sin 270

x i

x i

x i

x i

x i

4

40

41

270 360 270 36016 cos sin , 0,1,2,3

4 4

270 360 0 270 360 016 cos sin 2 cos67.5 sin 67.5 0.7654 1.8478

4 4

270 360 1 270 360 116 cos sin

4 4

kk k

z i k

z i i i

z i

42

43

2 cos157.5 sin157.5 1.8478 0.7654

270 360 2 270 360 216 cos sin 2 cos 247.5 sin 247.5 0.7654 1.8478

4 4

270 360 3 270 360 316 cos sin

4 4

i i

z i i i

z i

2 cos337.5 sin 337.5 1.8478 0.7654i i



84.

5

5

5

5

5

32 0

32

32

0 32

32 cos90 sin 90

x i

x i

x i

x i

x i

5

50

51

90 360 90 36032 cos sin , 0,1,2,3,4

5 5

90 360 0 90 360 032 cos sin 2 cos18 sin18 1.9021 0.6180

5 5

90 360 1 90 360 132 cos sin

5 5

kk k

z i k

z i i i

z i

52

53

2 cos90 sin 90 0 2 2

90 360 2 90 360 232 cos sin 2 cos162 sin162 1.9021 0.6180

5 5

90 360 3 90 360 332 cos sin 2 cos234 sin 234 1.1756 1.61

5 5

i i i

z i i i

z i i

54

80

90 360 4 90 360 432 cos sin 2 cos306 sin 306 1.1756 1.6180

5 5

i

z i i i

85. 3

3

3

3

1 3 0

1 3

1 3

2(cos60 sin 60 )

x i

x i

x i

x i

3

3 30

3 31

60 360 60 3602 cos sin , 0,1,2

3 3

60 360 0 60 360 02 cos sin 2 cos 20 sin 20 1.1839 0.4309

3 3

60 360 1 60 360 12 cos sin 2

3 3

kk k

z i k

z i i i

z i

3 32

cos140 sin140 0.9652 0.8099

60 360 2 60 360 22 cos sin 2 cos 260 sin 260 0.2188 1.2408

3 3

i i

z i i i



86. 3

3

3

3

1 3 0

1 3

1 3

2(cos300 sin 300 )

x i

x i

x i

x i

3

3 30

31

300 360 300 3602 cos sin , 0,1,2

3 3

300 360 0 300 360 02 cos sin 2 cos100 sin100 0.2188 1.2408

3 3

300 360 1 300 360 12 cos sin

3 3

kk k

z i k

z i i i

z i

3

3 32

2 cos220 sin 220 0.9652 0.8099

300 360 2 300 360 22 cos sin 2 cos340 sin 340 1.1839 0.4309

3 3

i i

z i i i

87. 4 cos sin4 4

2 2

2 2

i

e i

i

π π π

88. 6 3 1cos sin

6 6 2 2

i

e i iπ π π

89. 1 cos sin

( 1) (0)

1 0

ie i

i

i

π π π



90.

22 2 cos 2 sin 2

2 cos0 sin 0

2 0

ie i

i

i

π π π

91. z i

a.

1

2 22

2 22 23 2

222 2 24 3

2 25 4

2 26 5

( ) 1

1

( ) 1

( 1 )

( ) 1

z z i

z z z i i i

z z z z z z i i i

z z z z z z z i i i

z z z i i i

z z z i i i

b. 2 2

2 2

1 ( 1) 1 2

0 1 1

i

i

The absolute values of the terms in the sequence are 1 and 2 .

Choose a complex number with absolute value less than 1, and another with absolute value greater than 2 . Complex numbers may vary.

92. z i

a.

1

2 22

2 22 23 2

222 2 24 3

2 25 4

2 26 5

( ) ( ) 1

1 ( )

( ) 1

( 1 ) ( )

( ) 1

z z i

z z z i i i

z z z z z z i i i

z z z z z z z i i i

z z z i i i

z z z i i i

b. 2 2

2 2

1 ( 1) ( 1) 2

0 1 1

i

i

The absolute values of the terms of the sequence are 1 and 2 .

Choose a complex number with absolute value less than 1, and another with absolute value greater than 2 . Complex numbers may vary.



93. – 105. Answers may vary.

106. makes sense

107. does not make sense; Explanations will vary. Sample explanation: This process involves four multiplications.

108. makes sense

109. does not make sense; Explanations will vary. Sample explanation: 1 3i and 1 3i are the other 2 cube roots of 8.

110.

1 1 1 1 1 1 2 2

2 2 2 2 2 2 2 2

1 1 2 1 2 1 2 1 2

2 2 2 2 2 2 2 2 2

1 1 2

cos sin cos sin cos sin

cos sin cos sin cos sin

cos cos sin cos cos sin sin sin

cos cos sin cos sin cos sin sin

cos cos sin

r i r i i

r i r i i

r i

r i

r

θ θ θ θ θ θθ θ θ θ θ θ

θ θ θ θ θ θ θ θ

θ θ θ θ θ θ θ θ

θ θ

1 2 1 2 1 2

2 22 2 2

1 1 2 1 2

2

11 2 1 2

2

sin sin cos cos sin

cos sin (0)

cos sin

1 0

cos sin

i

r i

r i

r i

ri

r

θ θ θ θ θ θ

θ θ

θ θ θ θ

θ θ θ θ

111. 1 1 cos0 sin 0i

4

4 40

4 41

0 360 0 3601 cos sin , 0,1,2,3

4 4

0 360 0 0 360 01 cos sin 1 cos0 sin 0 1(1 0 ) 1

4 4

0 360 1 0 360 11 cos sin 1 cos90 sin 9

4 4

kk k

z i k

z i i i

z i i

4 42

4 43

0 1(0 (1))

0 360 2 0 360 21 cos sin 1 cos180 sin180 1( 1 0 ) 1

4 4

0 360 3 0 360 31 cos sin 1 cos270 sin 270 1(0 ( 1))

4 4

i i

z i i i

z i i i i

112. Answers may vary.



113. Find the distance from ( 3, 3) and (0,3).

2 21 2 1 2

2 2

( ) ( )

(3 3) (0 3)

45

3 5

d x x y y

Find the distance from (0,0) and (3,6).

2 21 2 1 2

2 2

( ) ( )

(6 0) (3 0)

45

3 5

d x x y y

The line segments have the same length.

114. ( 3, 3) and (0,3)

2 1

2 1

3 ( 3)

0 ( 3)

2

y ym

x x

(0,3) and (3,6)

2 1

2 1

6 0

3 02

y ym

x x

The lines have the same slope. Thus, the lines are parallel.

115. 4(5 4 ) 2(6 9 )

20 16 12 18

8 34

x y x y

x y x y

x y

chapter 7 additional topics in trigonometry

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