cat 2e ism chapter 8 part 3 - oak park and river forest ...faculty.oprfhs.org/cavalos/book chapters...

Section 8.8

1171

Section 8.8 Solutions ------------------------------------------------------------------------------------ 1. – 10. The following graph shows the points in Exercises 1 – 10:

11. In order to convert the point ( )2,2 3 (expressed in rectangular coordinates) to polar

coordinates, first observe that 2, 2 3x y= = , so that the point is in QI. Hence,

( ) ( )222 2 3 4r = + =

2 3tan 32

yx

θ = = = so that ( )1tan 33πθ −= = .

So, the given point can be expressed in polar coordinates as 4,3π⎛ ⎞

⎜ ⎟⎝ ⎠

.

12. In order to convert the point ( )3, 3− (expressed in rectangular coordinates) to polar coordinates, first observe that 3, 3x y= = − , so that the point is in QIV. Hence,

( ) ( )2 23 3 3 2r = + − = 3tan 1

3yx

θ −= = = − so that ( )1tan 1

4πθ −= − = − .

Since 0 2θ π≤ < , we use the reference angle 74πθ = .

So, the given point can be expressed in polar coordinates as 73 2,4π⎛ ⎞

⎜ ⎟⎝ ⎠

.

Chapter 8

1172

13. In order to convert the point ( )1, 3− − (expressed in rectangular coordinates) to polar

coordinates, first observe that 1, 3x y= − = − , so that the point is in QIII. Hence,

( ) ( )221 3 2r = − + − =

3tan 31

yx

θ −= = =

− so that ( )1 4tan 3

3πθ π −= + = .

(Remember to add π to ( )1tan 3− since the point is in QIII.)


⎜ ⎟⎝ ⎠

.

14. In order to convert the point ( )6,6 3 (expressed in rectangular coordinates) to polar

coordinates, first observe that 6, 6 3x y= = , so that the point is in QI. Hence,

( ) ( )226 6 3 12r = + =

6 3tan 36

yx

θ = = = so that ( )1tan 33πθ −= = .


⎜ ⎟⎝ ⎠

.

15. In order to convert the point ( )4, 4− (expressed in rectangular coordinates) to polar coordinates, first observe that 4, 4x y= − = , so that the point is in QII. Hence,

( ) ( )2 24 4 4 2r = − + = 4tan 1

4yx

θ −= = = − so that ( )1 3tan 1

4πθ π −= + − = .

(Remember to add π to ( )1tan 1− − since the point is in QII.)


⎜ ⎟⎝ ⎠

.

Section 8.8

1173

16. In order to convert the point ( )0, 2 (expressed in rectangular coordinates) to polar

coordinates, first observe that 0, 2x y= = , so that the point is on the positive y-axis. Hence,

( ) ( )220 2 2,2

r πθ= + = =


⎜ ⎟⎝ ⎠

.

17. In order to convert the point ( )3,0 (expressed in rectangular coordinates) to polar coordinates, first observe that 3, 0x y= = , so that the point is on the positive x-axis. Hence,

( ) ( )2 23 0 3, 0r θ= + = =

So, the given point can be expressed in polar coordinates as ( )3,0 .

18. In order to convert the point ( )7, 7− − (expressed in rectangular coordinates) to polar coordinates, first observe that 7, 7x y= − = − , so that the point is in QIII. Hence,

( ) ( )2 27 7 7 2r = − + − = 7tan 17

yx

θ −= = =

− so that ( )1 5tan 1

4πθ π −= + = .

(Remember to add π to ( )1tan 1− since the point is in QIII.)


⎜ ⎟⎝ ⎠

.

19. In order to convert the point ( )3, 1− − (expressed in rectangular coordinates) to polar

coordinates, first observe that 3, 1x y= − = − , so that the point is in QIII. Hence,

( ) ( )2 23 1 2r = − + − =

1 1tan3 3

yx

θ −= = =

− so that 1 1 7tan

63πθ π − ⎛ ⎞= + =⎜ ⎟

⎝ ⎠.

(Remember to add π to 1 1tan3

− ⎛ ⎞⎜ ⎟⎝ ⎠

since the point is in QIII.)


⎜ ⎟⎝ ⎠

.

Chapter 8

1174

20. In order to convert the point ( )2 3, 2− (expressed in rectangular coordinates) to polar

coordinates, first observe that 2 3, 2x y= = − , so that the point is in QIV. Hence,

( ) ( )2 22 3 2 4r = + − =

2 1tan2 3 3

yx

θ −= = = − so that 1 1tan

63πθ − ⎛ ⎞= − = −⎜ ⎟

⎝ ⎠.



⎜ ⎟⎝ ⎠

.

21. In order to convert the point 54,3π⎛ ⎞

⎜ ⎟⎝ ⎠

(expressed in polar coordinates) to rectangular

coordinates, first observe that 54,3

r πθ= = . Hence,

5 1cos 4cos 4 23 2

5 3sin 4sin 4 2 33 2

x r

y r

πθ

πθ

⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞= = = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as ( )2, 2 3− .


⎜ ⎟⎝ ⎠



r πθ= = . Hence,

3 2cos 2cos 2 24 2

3 2sin 2sin 2 24 2

x r

y r

πθ

πθ

⎛ ⎞⎛ ⎞= = = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞= = = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as ( )2, 2− .

Section 8.8

1175

23. In order to convert the point 51,6π⎛ ⎞−⎜ ⎟

⎝ ⎠ (expressed in polar coordinates) to rectangular


r πθ= − = . Hence,

5 3 3cos cos6 2 2

5 1 1sin sin6 2 2

x r

y r

πθ

πθ

⎛ ⎞⎛ ⎞= = − = − − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞= = − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as 3 1,2 2

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠.


⎝ ⎠ (expressed in polar coordinates) to rectangular



7 2cos 2cos 2 24 2

7 2sin 2sin 2 24 2

x r

y r

πθ

πθ

⎛ ⎞⎛ ⎞= = − = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞= = − = − − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as ( )2, 2− .


⎜ ⎟⎝ ⎠



r πθ= = . Hence,

11cos 0cos 06

11sin 0sin 06

x r

y r

πθ

πθ

⎛ ⎞= = =⎜ ⎟⎝ ⎠⎛ ⎞= = =⎜ ⎟⎝ ⎠

So, the given point can be expressed in rectangular coordinates as ( )0,0 .

Chapter 8

1176

26. In order to convert the point ( )6,0 (expressed in polar coordinates) to rectangular coordinates, first observe that 6, 0r θ= = . Hence,

( ) ( )( ) ( )

cos 6cos 0 6 1 6sin 6sin 0 6 0 0

x ry r

θθ

= = = == = = =

So, the given point can be expressed in rectangular coordinates as ( )6,0 .

27. In order to convert the point ( )2, 240 (expressed in polar coordinates) to rectangular

coordinates, first observe that 2, 240r θ= = . Hence,

( )

( )

1cos 2cos 240 2 123sin 2sin 240 2 3

2

x r

y r

θ

θ

⎛ ⎞= = = − = −⎜ ⎟⎝ ⎠⎛ ⎞

= = = − = −⎜ ⎟⎜ ⎟⎝ ⎠

So, the given point can be expressed in rectangular coordinates as ( )1, 3− − .

28. In order to convert the point ( )3, 150− (expressed in polar coordinates) to

rectangular coordinates, first observe that 3, 150r θ= − = . Hence,

( )

( )

3 3 3cos 3cos 150 32 2

1 3sin 3sin 150 32 2

x r

y r

θ

θ

⎛ ⎞= = − = − − =⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞= = − = − = −⎜ ⎟⎝ ⎠

So, the given point can be expressed in rectangular coordinates as 3 3 3,2 2

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠.

29. In order to convert the point ( )1, 135− (expressed in polar coordinates) to

rectangular coordinates, first observe that 1, 135r θ= − = . Hence,

( )

( )

2 2cos cos 1352 2

2 2sin sin 1352 2

x r

y r

θ

θ

⎛ ⎞= = − = − − =⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞

= = − = − = −⎜ ⎟⎜ ⎟⎝ ⎠


⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠.

Section 8.8

1177



( ) ( )2 5 2 2 5 2cos 5cos 315 5 sin 5sin 315 52 2 2 2

x r y rθ θ⎛ ⎞ ⎛ ⎞

= = = = = = = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as 5 2 5 2,2 2

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠.

31. d 32. b 33. a 34. c 35. The graph of 5r = is a circle centered at the origin with radius 5 since for any given angle, the same distance from the origin is used to plot the point.

36. The graph 3πθ = − is a line through

the origin which makes an angle of 3π−

with the positive x-axis.

37. In order to graph 2cosr θ= , consider the following table of points:

θ 2cosr θ= ( ),r θ 0 2 ( )2,0

2π 0 ( )0, 2

π

π -2 ( )2,π− 3

2π 0 ( )30, 2

π

2π 2 ( )2, 2π

The graph is as follows:

Chapter 8

1178

38. In order to graph 3sinr θ= , consider the following table of points:

θ 3sinr θ= ( ),r θ 0 0 ( )0,0

2π 3 ( )3, 2

π

π 0 ( )0,π 3

2π -3 ( )33, 2

π−

2π 0 ( )0, 2π


39. In order to graph 4sin 2r θ= , consider the following table of points:

θ 4sin 2r θ= ( ),r θ 0 0 ( )0,0

4π 4 ( )4, 4

π

2π 0 ( )0, 2

π

34

π -4 ( )34, 4π−

π 0 ( )0,π 5

4π 4 ( )54, 4

π

32

π 0 ( )30, 2π

74

π -4 ( )74, 4π−

2π 0 ( )0, 2π


Section 8.8

1179

40. In order to graph 5cos 2r θ= , consider the following table of points:

θ 5cos 2r θ= ( ),r θ 0 5 ( )5,0

4π 0 ( )0, 4

π

2π -5 ( )5, 2

π−

34

π 0 ( )30, 4π

π 5 ( )5,π 5

4π 0 ( )50, 4

π

32

π -5 ( )35, 2π−

74

π 0 ( )70, 4π

2π 5 ( )5, 2π


Chapter 8

1180


θ 3sin 3r θ= ( ),r θ 0 0 ( )0,0

6π 3 ( )3, 6

π

26

π 0 ( )20, 6π

2π -3 ( )3, 2

π−

46

π 0 ( )40, 6π

56

π 3 ( )53, 6π

π 0 ( )0,π 7

6π -3 ( )73, 6

π−

86

π 0 ( )80, 6π

32

π 3 ( )33, 2π

106

π 0 ( )100, 6π

116

π -3 ( )113, 6π−

2π 0 ( )0, 2π


Section 8.8

1181

42. In order to graph 4cos3r θ= , consider the following table of points:

θ 4cos3r θ= ( ),r θ 0 4 ( )4,0

6π 0 ( )0, 6

π

26

π -4 ( )24, 6π−

2π 0 ( )0, 2

π

46

π 4 ( )44, 6π

56

π 0 ( )50, 6π

π -4 ( )4,π− 7

6π 0 ( )70, 6

π

86

π 4 ( )84, 6π

32

π 0 ( )30, 2π

106

π -4 ( )104, 6π−

116

π 0 ( )110, 6π

2π 4 ( )4, 2π


Chapter 8

1182

43. In order to graph 2 9cos 2r θ= , consider the following table of points:

θ 2 9cos 2r θ= r ( ),r θ 0 9 3± ( )3,0±

4π 0 0 ( )0, 4

π

2π -9 No points

34

π 0 0 ( )30, 4π

π 9 3± ( )3,π±

54

π 0 0 ( )50, 4π

3

2π -9 No points

74

π

0 0 ( )70, 4π

2π 9 3± ( )3,2π±


Section 8.8

1183

44. In order to graph 2 16sin 2r θ= , consider the following table of points:

θ 2 16sin 2r θ= r ( ),r θ 0 0 0 ( )0,0

4π 16 4± ( )4, 4

π±

2π 0 0 ( )0, 2

π

34

π -16 No points

π 0 0 ( )0,π 5

4π 16 4± ( )54, 4

π±

3

2π 0 0 ( )30, 2

π

7

4π

-16 No points

2π 0 0 ( )0, 2π

45. In order to graph 2cosr θ= − , consider the following table of points:

θ 2cosr θ= − ( ),r θ 0 -2 ( )2,0−

2π 0 ( )0, 2

π

π 2 ( )2,π 3

2π 0 ( )30, 2

π

2π -2 ( )2, 2π−


Chapter 8

1184

46. In order to graph 3sin 3r θ= − , consider the following table of points:

θ 3sin 3r θ= −

( ),r θ

0 0 ( )0,0

6π -3 ( )3, 6

π−

26

π 0 ( )20, 6π

2π 3 ( )3, 2

π

46

π 0 ( )40, 6π

56

π -3 ( )53, 6π−

π 0 ( )0,π 7

6π 3 ( )73, 6

π

86

π 0 ( )80, 6π

32

π -3 ( )33, 2π−

106

π 0 ( )100, 6π

116

π 3 ( )113, 6π

2π 0 ( )0, 2π


47. In order to graph 4r θ= , consider the following table of points:

θ 4r θ= ( ),r θ 0 0 ( )0,0

2π 2π ( )2 , 2

ππ

π 4π ( )4 ,π π 3

2π 6π ( )36 , 2

ππ

2π 8π ( )8 , 2π π


Section 8.8

1185

48. In order to graph 2r θ= − , consider the following table of points:

θ 2r θ= − ( ),r θ 0 0 ( )0,0

2π π− ( ), 2

ππ−

π 2π− ( )2 ,π π− 3

2π 3π− ( )33 , 2

ππ−

2π 4π− ( )4 , 2π π−


49. In order to graph 3 2cosr θ= − + , consider the following table of points:

θ 3 2cosr θ= − + ( ),r θ 0 -1 ( )1,0−

2π -3 ( )3, 2

π−

π -5 ( )5,π− 3

2π -3 ( )33, 2

π−

2π -1 ( )1, 2π−


50. In order to graph 2 3sinr θ= + , consider the following table of points:

θ 2 3sinr θ= + ( ),r θ 0 2 ( )2,0

2π 5 ( )5, 2

π

π 2 ( )2,π 3

2π -1 ( )31, 2

π−

2π 2 ( )2, 2π


Chapter 8

1186

51. Using cos , sinx r y rθ θ= = , we see that the equation becomes

( )sin 2cos 1sin 2 cos 1

2 12 1

rr r

y xy x

θ θθ θ+ =

+ =+ =

= − +

The graph of this curve is a line.


( )sin 3cos 2sin 3 cos 2

3 23 2

rr r

y xy x

θ θθ θ− =

− =− =

= +

The graph of this curve is a line. 53. Using cos , sinx r y rθ θ= = , we see that the equation becomes

2 2 2 2

2 2

2 2

2 2

cos 2 cos sin 82 8

2 1 8 1( 1) 9

r r rx x y

x x yx y

θ θ θ− + =

− + =

− + + = +

− + =

The graph of this curve is a circle.


2 2

2

2

cos sin 22

2

r rx y

y x

θ θ− = −

− = −

= +

The graph of this curve is a parabola.

55. The graph is as follows:


Section 8.8

1187





61. The graph of ( )0.587 1 0.9671 0.967cos

rθ

+=

− is as

follows:

62. The graph of ( )29.62 1 0.2491 0.249cos

rθ

+=

− is as

follows:

Chapter 8

1188

63. Consider the following table of values for the graphs of r θ= and r θ= :

θ r θ= r θ= 0 0 0

2π 2

π 2

π

π π π 3

2π 3

2π 3

2π

2π 2π 2π Notice that the graph of r θ= (dotted) is more tightly wound than is the graph of r θ= (solid).

64. Consider the following table of values for the graphs of r θ= and 4

3r θ= : θ r θ= 4

3r θ= 0 0 0

2π 2

π ( )4

3

2π

π π ( )4

3π 3

2π 3

2π ( )

433

2π

2π 2π ( )4

32π Notice that the graph of 4

3r θ= (dotted) is more loosely wound than is the graph of r θ= (solid).

Section 8.8

1189

65. Consider the following table of values for the graphs of 2 4cos 2r θ= and

2 14 cos 2r θ= : θ 2 4cos 2r θ= 2 1

4 cos 2r θ=0 4 1

4

2π -4 1

4−

π 4 14

32

π -4 14−

2π 4 14

Notice that the graph of 2 1

4 cos 2r θ= (dotted) is much closer to the origin than is the graph of 2 4cos 2r θ= (solid).

66. The graphs of 2 4cos 2r θ= and ( )2 4cos 2 2r θ= + are given to the right.

Notice that the graph of ( )2 4cos 2 2r θ= + (dotted) is simply a rotation of the graph of

2 4cos 2r θ= (solid). The same set of points is generated, but starting/ending at different positions on the graph.

Chapter 8

1190

67. In order to graph 2 2sinr θ= + , consider the following table of points:

θ 2 2sinr θ= + ( ),r θ 0 2 ( )2,0

2π 4 ( )4, 2

π

π 2 ( )2,π 3

2π 0 ( )30, 2

π

2π 2 ( )2, 2π


68. In order to graph 4 4sinr θ= − − , consider the following table of points:

θ 4 4sinr θ=− − ( ),r θ 0 -4 ( )4,0−

2π -8 ( )8, 2

π−

π -4 ( )4,π− 3

2π 0 ( )30, 2

π

2π -4 ( )4, 2π−


69. a) – c) All three graphs generate the same set of points, as seen below:

Note that all three graphs are figure eights. Extending the domain in b results in twice as fast movement, while doing so in c results in movement that is four times as fast.

Section 8.8

1191

70. We must choose L = 12, and choose a such that 2 cos( )r aθ= makes 8 complete cycles in the given interval. Using the observations in Exercise 69, we see that a must be 8, so that the equation is ( )2 12cos 8r θ= .

71. You need to start at 0 and hit x = 5 at 2θ π= , and then x = 10 at 4θ π= , and so on until x = 30 is hit at 12θ π= . Thus, we use the function 5

2 , 0 12r π θ θ π= ≤ ≤ , which is pictured below:

72. Similar to problem #71, the walkway can now be 7 feet wide. The path of the spiral equation 10

2 , 0 6r π θ θ π= ≤ ≤ is as follows:

73. (i) The equation of the polar curve that describes this scenario is

8sin 3 , 0 2r θ θ π= ≤ ≤ . (ii) It retraces the path 50 times in the interval [ ]0,100π since each complete trace occurs in an interval of length 2π .

Chapter 8

1192

74. In order to create an 8-petaled rose of the sort described in #73, we can use the polar equation 8cos 4 , 0 2r θ θ π= ≤ ≤ , whose graph is below:

75. The point is in QIII, so needed to add π to the angle.

76. The point is in QII, so needed to add π to the angle.

77. True. By definition. 78. False. The equation of a cardioid is either cosr a a θ= + or sinr a a θ= + , while the more general limacon has equation either cosr a b θ= + or sinr a b θ= + . So, simply choose a b≠ to get a limacon that is not a cardioid. 79. Observe that if x a= , then cosr aθ = ,

so that cos

arθ

= , which represents a

vertical line.

80. Observe that if y b= , then sinr bθ = ,

so that sin

brθ

= , which represents a

horizontal line. 81. The point ( ),a θ can also be represented as ( ), 180a θ− ± . This is because the point

( ),a θ− is diametrically opposed (through the origin) to the given point. As such, we need

to rotate from it 180 clockwise or counterclockwise to arrive at the given point. 82. In order to convert the point ( ),a b− (expressed in rectangular coordinates) to polar coordinates, first observe that ,x a y b= − = , so that the point is in QII. Hence,

( ) ( )2 2 2 2r a b a b= − + = +

tan y b bx a a

θ = = = −−

so that 1180 tan ba

θ − ⎛ ⎞= + −⎜ ⎟⎝ ⎠

.

So, the given point can be expressed in polar coordinates as

2 2 1,180 tan ba ba

−⎛ ⎞⎛ ⎞+ + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠.

Section 8.8

1193

83. The points of intersection of 4cosr θ= and cos 1r θ = are found by substituting the first equation in for r in the second, and solving for θ :

2 512 3 34cos 1 cos ,π πθ θ θ= ⇒ = ⇒ = (since r >0)

Hence, the points of intersection are ( ) ( )53 32, , 2,π π . The graphs of these two curves are

displayed below:

84. We convert sin cosr a bθ θ= + to a Cartesian equation as follows:

( ) ( )

2

2 2

2 2

2 22 2

2 2

sin cossin cos

0

(upon completing the square)4

b a

r a br ar br

x y ay bxx bx y ay

a bx y

θ θθ θ

= +

= +

+ = +

− + − =

+− + − =

This is the equation of a circle.

85. We convert 3 3

sin 2cos sin

ar θθ θ

=−

to a Cartesian equation as follows:

( )( )

( ) ( ) ( )( ) ( ) ( )( )

3 3

3 3

3 3 3 2

3 3 2

3 3

3 3

sin 2cos sin

cos sin sin 2

cos sin sin 2

cos sin 2sin cos

cos sin 2 sin cos

2 0

ar

r a

r ar

r r ar

r r a r r

x y ax y

θθ θ

θ θ θ

θ θ θ

θ θ θ θ

θ θ θ θ

=−

− =

− =

− =

− =

− − =

86. The equation for this graph would be 10sin(9 ), 0 2r θ θ π= ≤ ≤ .

Chapter 8

1194

87. Graphing the equation 2 cos( )r a bθ= − for different values of a and b always yields a circle with radius a centered at (a, b). 88. Note that the period of sin bθ (and of sina bθ ) is 2

bπ . As such, the smallest value of

M for which the graph starts to repeat is 2bπ .

89. The graph of cos2

r θ⎛ ⎞= ⎜ ⎟⎝ ⎠

is as follows:

The inner loop is generated beginning with

2πθ = and ending with 3

2π .

90. The graph of 32cos2

r θ⎛ ⎞= ⎜ ⎟⎝ ⎠

is as

follows:

The petal in the 1st quadrant is generated

beginning with 0θ = and ending with 3π .

Chapter 8 Review

1195

91. The graph of 1 3cosr θ= + is as follows:

The very tip of the inner loop begins with

2πθ = , then it crosses the origin (the first

time) at ( )1 13cosθ −= − , winds around, and

eventually ends with 32πθ = .

92. The graphs of 1 sin(2 )r θ= + (solid) and 1 cos(2 )r θ= − (dashed) are as follows:

Solving 1 sin(2 ) 1 cos(2 )θ θ+ = − , we see that sin(2 ) cos(2 ) 0θ θ+ = , which is satisfied when 2θ is an integer multiple of 4π that has terminal side in QII or QIV. As such, we see that

3 7 11 152 , , ,4 4 4 4π π π πθ =

and hence, 3 7 11 15, , ,8 8 8 8π π π πθ = .

93.

2

8 2

1 2 sin 41 sin 4

4 2

, where is an integern

n

n

π

π π

θθ

θ π

θ

= +== +

= +

94.

53 3

2 5 29 3 9 3

2 cos3 1.5cos3 0.5

3 2 , 2

, , an integern n

n n

n

π π

π π π π

θθθ π π

θ

− === + +

= + +

Chapter 8 Review Solutions --------------------------------------------------------------------------- 1. Observe that:

( )180 10 20 150γ = − + =

( )( )4 sin 20sin sin sin10 sin 20 84 sin10

ba b bα β= ⇒ = ⇒ = ≈

( )( )4 sin150sin sin sin10 sin150 124 sin10

ca c cα γ= ⇒ = ⇒ = ≈

Chapter 8

1196

2. Observe that: ( )180 40 60 80α = − + =

( )( )10 sin80sin sin sin80 sin 40 1510 sin 40

aa b aα β= ⇒ = ⇒ = ≈

( )( )10 sin 60sin sin sin 40 sin 60 1310 sin 40

cb c cβ γ= ⇒ = ⇒ = ≈

3. Observe that: ( )180 45 5 130γ = − + =

( )( )10 sin 5sin sin sin 5 sin130 1.14 110 sin130

aa c aα γ= ⇒ = ⇒ = ≈ ≈

( )( )10 sin 45sin sin sin 45 sin130 9.23 910 sin130

bb c bβ γ= ⇒ = ⇒ = ≈ ≈

4. Observe that: ( )180 70 60 50α = − + =


ba b bα β= ⇒ = ⇒ = ≈


ca c cα γ= ⇒ = ⇒ = ≈

5. Observe that: ( )180 11 11 158β = − + =

sin sin sin11 sin11 1111

aa c aα γ= ⇒ = ⇒ =


bb c bβ γ= ⇒ = ⇒ = ≈

6. Observe that: ( )180 20 50 110α = − + =

( )( )8 sin110sin sin sin110 sin 20 228 sin 20

aa b aα β= ⇒ = ⇒ = ≈


cb c cβ γ= ⇒ = ⇒ = ≈

Chapter 8 Review

1197

7. Observe that: ( )180 45 45 90β = − + =


aa b aα β= ⇒ = ⇒ = ≈


cb c cβ γ= ⇒ = ⇒ = ≈

8. Observe that: ( )180 60 20 100γ = − + =

( )( )17 sin 60sin sin sin 60 sin100 1517 sin100

aa c aα γ= ⇒ = ⇒ = ≈

( )( )17 sin 20sin sin sin 20 sin100 617 sin100

bb c bβ γ= ⇒ = ⇒ = ≈

9. Observe that: ( )180 12 22 146β = − + =


ba b bα β= ⇒ = ⇒ = ≈

( )( )99 sin 22sin sin sin12 sin 22 17899 sin12

ca c cα γ= ⇒ = ⇒ = ≈

10. Observe that: ( )180 102 27 51α = − + =

( )( )24 sin102sin sin sin 51 sin102 3024 sin 51

ba b bα β= ⇒ = ⇒ = ≈


cb c cβ γ= ⇒ = ⇒ = ≈

Chapter 8

1198

11. Step 1: Determineβ .

1sin sin sin 20 sin 9sin 20 9sin 20sin sin 267 9 7 7a b

α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

This is the solution in QI – label it as 1β . The second solution in QII is given by

2 1180 154β β= − ≈ . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: 1 26β ≈

( )1 180 26 20 134γ ≈ − + =

1 11

1 1

sin sin sin 26 sin134 9sin134 159 sin 26

cb c cβ γ

= ⇒ = ⇒ = ≈

QII triangle: 2 154β ≈

( )2 180 154 20 6γ ≈ − + =

2 22

2 2

sin sin sin154 sin 6 9sin 6 29 sin154

cb c cβ γ

= ⇒ = ⇒ = ≈

12. Step 1: Determineγ .

1sin sin sin sin16 30sin16 30sin16sin sin 20.154 2030 24 24 24c b

γ β γ γ γ − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈ ≈⎜ ⎟

⎝ ⎠

This is the solution in QI – label it as 1γ . The second solution in QII is given by

2 1180 160γ γ= − ≈ . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: 1 20.154 20γ ≈ ≈

( )1 180 20.154 16 143.84587 144α ≈ − + = ≈

1 11

1 1

sin sin sin143.84587 sin16 24sin143.84587 51.36 5124 sin16

aa c aα γ

= ⇒ = ⇒ = ≈ ≈

QII triangle: 2 160γ ≈

( )2 180 160 16 4α ≈ − + =

2 22

2 2

sin sin sin 4 sin160 30sin 4 630 sin160

aa c aα γ

= ⇒ = ⇒ = ≈

Chapter 8 Review

1199


1sin sin sin sin 24 12sin 24 12sin 24sin sin 2912 10 10 10c a

γ α γ γ γ − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠


2 1180 151γ γ= − ≈ . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: 1 29γ ≈

( )1 180 29 24 127α ≈ − + =

11

1 1

sin sin sin127 sin 24 10sin127 2010 sin 24

bb a bβ α

= ⇒ = ⇒ = ≈


( )2 180 151 24 5β ≈ − + =

22

2 2

sin sin sin 5 sin 24 10sin 5 210 sin 24

bb a bβ α

= ⇒ = ⇒ = ≈


1sin sin sin sin12 116sin12 116sin12sin sin 14116 100 100 100c b

γ β γ γ γ − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠


2 1180 166γ γ= − ≈ . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: 1 14γ ≈

( )1 180 12 14 154α ≈ − + =

11

1 1

sin sin sin154 sin12 100sin154 211100 sin12

aa b aα β

= ⇒ = ⇒ = ≈


( )2 180 166 12 2α ≈ − + =

22

2 2

sin sin sin 2 sin12 100sin 2 17100 sin12

aa b aα β

= ⇒ = ⇒ = ≈

Chapter 8

1200

15. First, determine α : 1sin sin sin sin150 40sin150 40sin150sin sin 42

40 30 30 30a bα β α α α − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Hence, there is no triangle in this case since 180α β+ > . 16. Step 1: Determineβ .

1sin sin sin165 sin 2sin165 2sin165sin sin 103 2 3 3c b

γ β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Note that there is only one triangle in this case since the angle in QII with the same sine as this value of β is 180 10 170− = . In such case, note that 180β γ+ > , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and γ , must sum to180 ). Step 2: Solve for the triangle. ( )180 10 165 5α ≈ − + =


ac a aγ α= ⇒ = ⇒ = ≈

17. Step 1: Determineβ .

1sin sin sin10 sin 6sin10 6sin10sin sin 154 6 4 4a b

α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠



( )1 180 15 10 155γ ≈ − + =

1 11

1 1

sin sin sin15 sin155 6sin155 106 sin15

cb c cβ γ

= ⇒ = ⇒ = ≈


( )2 180 165 10 5γ ≈ − + =

2 22

2 2


cb c cβ γ

= ⇒ = ⇒ = ≈

Chapter 8 Review

1201

18. Step 1: Determineα . 1sin sin sin sin 4 37sin 4 37sin 4sin sin 6

37 25 25 25a cα γ α α α − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

This is the solution in QI – label it as 1α . The second solution in QII is given by

2 1180 174α α= − ≈ . Both are tenable solutions, so we need to solve for two triangles. Step 2: Solve for both triangles. QI triangle: 1 6α ≈

( )1 180 6 4 170β ≈ − + =

11

1 1

sin sin sin170 sin 4 25sin170 6225 sin 4

bb c bβ γ

= ⇒ = ⇒ = ≈

QII triangle: 2 174α ≈

( )2 180 174 4 2β ≈ − + =

22

2 2

sin sin sin 2 sin 4 25sin 2 1325 sin 4

bb c bβ γ

= ⇒ = ⇒ = ≈

19. Consider the following diagram:

The given information yields a SAS triangle, so we use Law of Cosines:

2 2 28 6 2(8)(6)cos50100 96cos506.2 mi.

x = + −= −

≈


The given information yields a SAS triangle, so we use Law of Cosines:

2 2 210 3 2(10)(3)cos125109 60cos12512 mi.

x = + −= −

≈

50 125

Chapter 8

1202

21. This is SAS, so begin by using Law of Cosines: Step 1: Find c.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 40 60 2 40 60 cos 50 5200 4800cos 5045.9849 46

c a b abc

γ= + − = + − = −

≈ ≈

Step 2: Find α . 1sin sin sin sin 50 40sin 50 40sin 50sin sin 42

40 45.9849 45.9849 45.9849a cα γ α α α − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find β . ( )180 42 50 88β ≈ − + = 22. This is SAS, so begin by using Law of Cosines: Step 1: Find a.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 15 12 2 15 12 cos 140 369 360cos 14025.392 25

a b c bca

α= + − = + − = −

≈ ≈

Step 2: Find β .

1sin sin sin 40 sin 15sin140 15sin140sin sin 2225.392 15 25.392 25.392a b

α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find γ . ( )180 22 140 18γ ≈ − + = 23. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ .

( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 30 24 25 2 24 25 cos900 1201 1200cos

c a b ab γ γγ

= + − ⇒ = + −⇒ = −

Thus, 1301 301cos so that cos 75.47 751200 1200

γ γ − ⎛ ⎞= = ≈ ≈⎜ ⎟⎝ ⎠

.

Step 2: Find either of the remaining two angles using the Law of Sines. 1sin sin sin 75.47 sin 25sin 75.47 25sin 75.47sin sin 54

30 25 30 30c bγ β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find the third angle. ( )180 54 75 51α ≈ − + =

Chapter 8 Review

1203

24. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ .

( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 8 6 6 2 6 6 cos64 72 72cos

c a b ab γ γγ

= + − ⇒ = + −⇒ = −


γ γ − ⎛ ⎞= = ≈ ≈⎜ ⎟⎝ ⎠

.


8 6 8 8c bγ β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find the third angle. ( )180 48 84 48α ≈ − + = 25. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ .

( ) ( ) ( ) ( )( )2 222 2 2 2 cos 5 11 14 2 11 14 cos

25 25 2 11 14 cos

c a b ab γ γ

γ

= + − ⇒ = + −

⇒ = −

Thus, ( )1cos 0 so that cos 0 90γ γ −= = = . Step 2: Find either of the remaining two angles using the Law of Sines.

1sin sin sin 90 sin 14 sin 90 14 sin 90sin sin 485 5 514c b

γ β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟⎜ ⎟

⎝ ⎠


( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 122 120 22 2 120 22 cos14,884 14,884 5280cos

c a b ab γ γγ

= + − ⇒ = + −⇒ = −

Thus, ( )1cos 0 so that cos 0 90γ γ −= = = . Step 2: Find either of the remaining two angles using the Law of Sines.

1sin sin sin 90 sin 120sin 90 120sin 90sin sin 80122 120 122 122c b

γ β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠


Chapter 8

1204

27. This is SAS, so begin by using Law of Cosines: Step 1: Find a.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 7 10 2 7 10 cos 14 149 140cos 144

a b c bca

α= + − = + − = −

≈

Step 2: Find β .

1sin sin sin14 sin 7sin14 7sin14sin sin 283.627 7 3.627 3.627a b

α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find γ . ( )180 28 14 138γ ≈ − + = 28. This is SAS, so begin by using Law of Cosines: Step 1: Find c.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 6 12 2 6 12 cos 80 180 144cos 8012.449 12

c a b abc

γ= + − = + − = −

≈ ≈

Step 2: Find β .

1sin sin sin sin80 12sin80 12sin80sin sin 7212 12.449 12.449 12.449b c

β γ β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find α . ( )180 72 80 28α ≈ − + = 29. This is SAS, so begin by using Law of Cosines: Step 1: Find a.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 10 4 2 10 4 cos 90 116 80cos 9010.770 11

a b c bca

α= + − = + − = −

≈ ≈

Step 2: Find β .

1sin sin sin 90 sin 10sin 90 10sin 90sin sin 6810.770 10 10.770 10.770a b

α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find γ . ( )180 68 90 22γ ≈ − + =

Chapter 8 Review

1205

30. This is SAS, so begin by using Law of Cosines: Step 1: Find c.

( ) ( ) ( ) ( ) ( ) ( )2 22 2 2 2 cos 4 5 2 4 5 cos 75 41 40cos 755.536 6

c a b abc

γ= + − = + − = −

≈ ≈

Step 2: Find β .

1sin sin sin sin 75 5sin 75 5sin 75sin sin 615 5.536 5.536 5.536b c

β γ β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find α . ( )180 61 75 44α ≈ − + = 31. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ .

( ) ( ) ( ) ( ) ( )2 2 22 2 2 2 cos 12 10 11 2 10 11 cos144 221 220cos

c a b ab γ γγ

= + − ⇒ = + −⇒ = −


γ γ − ⎛ ⎞= = ≈ ≈⎜ ⎟⎝ ⎠

.


12 11 12 12c bγ β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠


( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 25 22 24 2 22 24 cos625 1060 1056cos

c a b ab γ γγ

= + − ⇒ = + −⇒ = −


γ γ − ⎛ ⎞= = ≈ ≈⎜ ⎟⎝ ⎠

.


25 24 25 25c bγ β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠


Chapter 8

1206

33. This is SAS, so begin by using Law of Cosines: Step 1: Find a.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 16 18 2 16 18 cos 100 580 576cos 10026.077 26

a b c bca

α= + − = + − = −

≈ ≈

Step 2: Find β .


α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find γ . ( )180 37 100 43γ ≈ − + = 34. This is SAS, so begin by using Law of Cosines: Step 1: Find b.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 25 25 2 25 25 cos 9 1250 1250cos 93.923 4

b a c acb

β= + − = + − = −

≈ ≈

Step 2: Find α .

1sin sin sin 9 sin 25sin 9 25sin 9sin sin 85.53.923 25 3.923 3.923b c

β γ γ γ γ − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find γ . ( )180 9 85.5 85.5γ ≈ − + = 35. This is SAS, so begin by using Law of Cosines: Step 1: Find a.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 12 40 2 12 40 cos 10 1744 960cos 1028.259 28

a b c bca

α= + − = + − = −

≈ ≈

Step 2: Find β .


α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find γ . ( )180 4 10 166γ ≈ − + =

Chapter 8 Review

1207

36. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is α .

( ) ( ) ( ) ( ) ( )2 2 22 2 2 2 cos 26 20 10 2 20 10 cos 676 500 400cosa b c bc α α α= + − ⇒ = + − ⇒ = −


α α − ⎛ ⎞= − = − ≈ ≈⎜ ⎟⎝ ⎠

.

Step 2: Find either of the remaining two angles using the Law of Sines. 1sin sin sin116.103 sin 20sin116.103 20sin116.103sin sin 44

26 20 26 26a bα β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find the third angle - ( )180 116 44 20γ ≈ − + =

37. There is no triangle in this case since 39a c+ = > 40 b= . 38. There is no triangle in this case since 3a b+ = > 3 c= . 39. This is SSA, so use Law of Sines: Step 1: Determineβ .

1sin sin sin15 sin 4.2sin15 4.2sin15sin sin 106.3 4.2 6.3 6.3a b

α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Note that there is only one triangle in this case since the angle in QII with the same sine as this value of β is 180 10 170− = . In such case, note that 180β α+ > , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and α , must sum to180 ). Step 2: Solve for the triangle. ( )180 10 15 155γ ≈ − + =

sin sin sin155 sin15 6.3sin155 10.36.3 sin15

cc a cγ α= ⇒ = ⇒ = ≈

40. This is SSA, so use Law of Sines: Step 1: Determineγ .

1sin sin sin sin 35 6sin 35 6sin 35sin sin 436 5 5 5c b

γ β γ γ γ − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Note that there is only one triangle in this case since the angle in QII with the same sine as this value of γ is 180 43 137− = . In such case, note that 180β γ+ > , therefore preventing the formation of a triangle (since the three interior angles, two of which are β and γ , must sum to 180 ). Step 2: Solve for the triangle.

( )180 35 43 102α ≈ − + = sin sin sin102 sin 35 5sin102 95 sin 35

aa b aα β= ⇒ = ⇒ = ≈

Chapter 8

1208

41. ( )( )1 16 18 sin100 141.82

A = ≈ 42. ( )( )1 25 25 sin 9 48.892

A = ≈

43. First, observe that 33 16.52 2

a b cs + += = = . So, the area is given by

( )( )( ) ( )( )( )( )16.5 6.5 5.5 4.5 51.5A s s a s b s c= − − − = ≈

44. First, observe that 71 35.52 2


( )( )( ) ( )( )( )( )35.5 13.5 11.5 10.5 240.6A s s a s b s c= − − − = ≈

45. First, observe that 56 282 2


( )( )( ) ( )( )( )( )28 2 8 18 89.8A s s a s b s c= − − − = ≈



( )( )( ) ( )( )( )( )48 24 16 8 384A s s a s b s c= − − − = ≈

47. ( )( )1 12 40 sin10 41.72

A = ≈ 48. ( )( )1 21 75 sin 60 682.02

A = ≈

49. Use the formula 4abcA

r= with 9 in.a b c= = = and 235 in.A = to find r:

( )( )( )9 9 935

4r= so that 140 729r = and hence, 729 5.2 in.

140r = ≈

50. Use the formula 4abcA

r= with 9 in., 12in., 15 in.a b c= = = and 254 in.A = to find

r: ( )( )( )9 12 15

544r

= so that 216 1620r = and hence, 7.5 in.r =

51.

( ) ( )2 2

8 4,2 ( 3) 12,5

12 5 13

AB

AB

= − − − − = −

= − + =

52.

( ) ( )2 2

2 ( 2),8 11 4, 3

4 3 5

AB

AB

= − − − = −

= + − =

53.

( ) ( )2 2

5 0,9 ( 3) 5,12

5 12 13

AB

AB

= − − − =

= + =

54.

( ) ( )2 2

9 3, 3 ( 11) 6,8

6 8 10

AB

AB

= − − − − =

= + =

Chapter 8 Review

1209

55. Given that 10, 24u = − , we have

( ) ( )2 210 24 26u = − + = 24tan10

θ =−

so that since the head is in

QII, 1 24180 tan 112.610

θ − ⎛ ⎞= + ≈⎜ ⎟−⎝ ⎠

56. Given that 5, 12u = − − , we have

( ) ( )2 25 12 13u = − + − =

12tan5

θ = so that 1 12tan 67.45

θ − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

57. Given that 16, 12u = − , we have

( ) ( )2 216 12 20u = + − = 12tan

16θ −= so that 1 12tan 323.1

16θ − −⎛ ⎞= ≈⎜ ⎟

⎝ ⎠

58. Given that 0,3u = , we have

( ) ( )2 20 3 3u = + = Since the vector is on the y-axis pointing in the positive direction, 90θ = .

59. 2 3 2 7, 2 3 4,5 14, 4 12,15 14 12, 4 15 2,11u v+ = − + − = − + − = − − + =

60. 7, 2 4,5 7 ( 4), 2 5 11, 7u v− = − − − = − − − − = −

61. 6 6 7, 2 4,5 42, 12 4,5 42 4, 12 5 38, 7u v+ = − + − = − + − = − − + = −

62. ( ) ( ) ( )3 2 3 7, 2 2 4,5 3 7, 2 8,10 3 1,8 3, 24u v− + = − − + − = − − + − = − − = −

63. cos , sin 10cos75 , 10sin 75 2.6, 9.7u u uθ θ= = ≈

64. cos , sin 8cos 225 , 8sin 225 5.7, 5.7u u uθ θ= = ≈ − −

65. cos , sin 12cos105 , 12sin105 3.1, 11.6u u uθ θ= = ≈ −

66. cos , sin 20cos15 , 20sin15 19.3, 5.2u u uθ θ= = ≈

67. ( ) ( )2 2

6, 6 6, 6 2 2,2 22 36 6

vv

− −= = = −

+ −

68. ( ) ( )2 2

11,60 11,60 11 60,61 61 6111 60

vv

− − −= = =

− +

69. 5i j+ 70. 15 2i j− + 71.

( )( ) ( )( )6, 3 1,4 6 1 3 4 6− ⋅ = + − = − 72.

( )( ) ( )( )6,5 4,2 6 4 5 2 34− ⋅ − = − − + =

Chapter 8

1210

73. ( )( ) ( )( )3,3 3, 6 3 3 3 6 9⋅ − = + − = −

74. ( )( ) ( )( )2, 8 1,1 2 1 8 1

6

− − ⋅ − = − − + −

= −

75. ( )( ) ( )( )0,8 1,2 0 1 8 2 16⋅ = + =

76. ( )( ) ( )( )4, 3 1,0 4 1 3 0 4− ⋅ − = − + − = −

77. Use the formula 1cos u vu v

θ − ⎛ ⎞⋅= ⎜ ⎟⎜ ⎟

⎝ ⎠ with the following computations:

( ) ( ) ( ) ( )2 2 2 23, 4 5,12 33, 3 4 5, 5 12 13u v u v⋅ = ⋅ − = = + = = − + =

So, ( )( )

1 33cos 595 13

θ − ⎛ ⎞= ≈⎜ ⎟⎜ ⎟

⎝ ⎠.


θ − ⎛ ⎞⋅= ⎜ ⎟⎜ ⎟


( ) ( ) ( ) ( )2 2 2 24,5 5, 4 40, 4 5 41, 5 4 41u v u v⋅ = − ⋅ − = − = − + = = + − =

So, ( )( )

1 40cos 16741 41

θ −⎛ ⎞−⎜ ⎟= ≈⎜ ⎟⎝ ⎠

.


θ − ⎛ ⎞⋅= ⎜ ⎟⎜ ⎟


( ) ( ) ( ) ( )2 22 21, 2 1,3 2 5, 1 2 3, 1 3 2 19u v u v⋅ = ⋅ − = = + = = − + =

So, ( )( )

1 5cos 493 19

θ −⎛ ⎞⎜ ⎟= ≈⎜ ⎟⎝ ⎠

.


θ − ⎛ ⎞⋅= ⎜ ⎟⎜ ⎟


( ) ( ) ( ) ( )2 2 2 27, 24 6,8 234, 7 24 25, 6 8 10u v u v⋅ = − ⋅ − = − = + − = = − + =

So, ( )( )

1 234cos 15925 10

θ − ⎛ ⎞−= ≈⎜ ⎟⎜ ⎟

⎝ ⎠.

Chapter 8 Review

1211


θ − ⎛ ⎞⋅= ⎜ ⎟⎜ ⎟


( ) ( ) ( ) ( )2 2 2 23,5 4, 4 32, 3 5 34, 4 4 32u v u v⋅ = ⋅ − − = − = + = = − + − =

So, ( )( )

1 32cos 16634 32

θ −⎛ ⎞−⎜ ⎟= ≈⎜ ⎟⎝ ⎠

.


θ − ⎛ ⎞⋅= ⎜ ⎟⎜ ⎟


( ) ( ) ( ) ( )2 2 2 21,6 2, 2 14, 1 6 37, 2 2 8u v u v⋅ = − ⋅ − = − = − + = = + − =

So, ( )( )

1 14cos 14437 8

θ −⎛ ⎞−⎜ ⎟= ≈⎜ ⎟⎝ ⎠

.

83. Since 8,3 3,12 12 0⋅ − = ≠ , these two vectors are not orthogonal.

84. Since 6, 2 4,12 0− ⋅ = , these two vectors are orthogonal.

85. Since 5, 6 12, 10 0− ⋅ − − = , these two vectors are orthogonal.

86. Since 1,1 4, 4 0⋅ − = , these two vectors are orthogonal.

87. Since 0, 4 0, 4 16 0⋅ − = − ≠ , these two vectors are not orthogonal.

88. Since 1 17,2 , 2 07 2

− ⋅ − = − ≠ , these two vectors are not orthogonal.

89. Since 6 , , 6 6 6 6 6 12 0z a b a b z za zb za bz zb− ⋅ + − = + − + = ≠ in general, these two vectors are not orthogonal. 90. Since ( )2 2 2 2, 1 , ( )( ) 0a b a b a b a b a b a b− − ⋅ + − = − + − − = , these two vectors are orthogonal.

Chapter 8

1212

91. & 92. The complex numbers –6 + 2i and 5i are plotted below:

93. In order to express 2 2i− in polar form, observe that 2, 2x y= = − , so that the point is in QIV. Now,

( ) ( )2 22 2 2 2 2r x y= + = + − =

2tan 12

yx

θ −= = = − , so that ( )1tan 1 45 or

4πθ −= − = − − .

Since 0 2θ π≤ < , we must use the reference angle 7315 or 4π .

Hence, ( ) ( )7 72 2 2 cos sin 2 cos 315 sin 3154 4

i i iπ π⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤− = + = +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎣ ⎦⎝ ⎠ ⎝ ⎠⎣ ⎦.

94. In order to express 3 i+ in polar form, observe that 3, 1x y= = , so that the point is in QI. Now,

( ) ( )2 22 2 3 1 2r x y= + = + =

3 1tan3 3

yx

θ = = = , so that 1 1tan or 3063πθ − ⎛ ⎞= =⎜ ⎟

⎝ ⎠.

Hence, ( ) ( )3 2 cos sin 2 cos 30 sin 306 6

i i iπ π⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤+ = + = +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎣ ⎦⎝ ⎠ ⎝ ⎠⎣ ⎦.

Chapter 8 Review

1213

95. In order to express 0 8i− in polar form, observe that 0, 8x y= = − , so that the point is on the negative y-axis. Now,

( ) ( )2 22 2 0 8 8r x y= + = + − = and 270θ = .

Hence, ( ) ( )0 8 8 cos 270 sin 270i i⎡ ⎤− = +⎣ ⎦ .

96. In order to express 8 8i− − in polar form, observe that 8, 8x y= − = − , so that the point is in QIII. Now,

( ) ( )2 22 2 8 8 8 2r x y= + = − + − = 8tan 18

yx

θ −= = =

−, so that ( )1180 tan 1 225θ −= + = .

(Remember to add 180 to 1tan (1)− since the point is in QIII.)

Hence, ( ) ( )8 8 8 2 cos 225 sin 225i i⎡ ⎤− − = +⎣ ⎦ .

97. In order to express 60 11i− + in polar form, observe that 60, 11x y= − = , so that the point is in QII. Now,

( ) ( )2 22 2 60 11 61r x y= + = − + =

11tan60

yx

θ −= = , so that 1 11180 tan 169.6

60θ − −⎛ ⎞= + ≈⎜ ⎟

⎝ ⎠.

(Remember to add 180 to 1 11tan60

− −⎛ ⎞⎜ ⎟⎝ ⎠

since the angle has terminal side in QII.)

Hence, ( ) ( )60 11 61 cos 169.6 sin 169.6i i⎡ ⎤− + ≈ +⎣ ⎦ .

98. In order to express 9 40 i− in polar form, observe that 9, 40x y= = − , so that the point is in QIV. Now,

( ) ( )2 22 2 9 40 41r x y= + = + − =

40tan9

yx

θ −= = , so that 1 40tan 77.3196

9θ − −⎛ ⎞= ≈ −⎜ ⎟

⎝ ⎠.

Since 0 360θ≤ < , we must use the reference angle 282.7 .

Hence, ( ) ( )9 40 41 cos 282.7 sin 282.7i i⎡ ⎤− ≈ +⎣ ⎦ .

Chapter 8

1214

99. In order to express 15 8i+ in polar form, observe that 15, 8x y= = , so that the point is in QI. Now,

( ) ( )2 22 2 15 8 17r x y= + = + =

8tan15

yx

θ = = , so that 1 8tan 28.115

θ − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

Hence, ( ) ( )15 8 17 cos 28.1 sin 28.1i i⎡ ⎤+ ≈ +⎣ ⎦ .

100. In order to express 10 24 i− − in polar form, observe that 10, 24x y= − = − , so that the point is in QIII. Now,

( ) ( )2 22 2 10 24 26r x y= + = − + − =

24tan10

yx

θ = = , so that 1 24180 tan 247.410

θ − ⎛ ⎞= + ≈⎜ ⎟⎝ ⎠

.

(Remember to add 180 to 1 24tan10

− ⎛ ⎞⎜ ⎟⎝ ⎠

since the angle has terminal side in QIII.)

Hence, ( ) ( )10 24 26 cos 247.4 sin 247.4i i⎡ ⎤− − ≈ +⎣ ⎦ .

101. ( ) ( ) 1 36 cos 300 sin 300 6 3 3 32 2

i i i⎡ ⎤

⎡ ⎤+ = − = −⎢ ⎥⎣ ⎦⎣ ⎦

102. ( ) ( ) 3 14 cos 210 sin 210 4 2 3 22 2

i i i⎡ ⎤

⎡ ⎤+ = − − = − −⎢ ⎥⎣ ⎦⎣ ⎦

103. ( ) ( ) 2 22 cos 135 sin 135 2 12 2

i i i⎡ ⎤

⎡ ⎤+ = − + = − +⎢ ⎥⎣ ⎦⎣ ⎦

104. ( ) ( ) 3 14 cos 150 sin 150 4 2 3 22 2

i i i⎡ ⎤

⎡ ⎤+ = − + = − +⎢ ⎥⎣ ⎦⎣ ⎦

105. ( ) ( )4 cos 200 sin 200 3.7588 1.3681i i⎡ ⎤+ ≈ − −⎣ ⎦

106. ( ) ( )3 cos 350 sin 350 2.9544 0.5209i i⎡ ⎤+ ≈ −⎣ ⎦

107. Using the formula ( ) ( )1 2 1 2 1 2 1 2cos sinz z r r iθ θ θ θ= + + +⎡ ⎤⎣ ⎦ yields

( )( ) ( ) ( )( ) ( ) [ ]

1 2 3 4 cos 200 70 sin 200 70

12 cos 270 sin 270 12 0 12

z z i

i i i

⎡ ⎤= + + +⎣ ⎦⎡ ⎤= + = − = −⎣ ⎦

Chapter 8 Review

1215


( )( ) ( ) ( )

( ) ( )

1 2 3 4 cos 20 220 sin 20 220

1 312 cos 240 sin 240 12 6 6 32 2

z z i

i i i

⎡ ⎤= + + +⎣ ⎦⎡ ⎤

⎡ ⎤= + = − − = − −⎢ ⎥⎣ ⎦⎣ ⎦


( )( ) ( ) ( )

( ) ( )

1 2 7 3 cos 100 140 sin 100 140

1 3 21 21 321 cos 240 sin 240 212 2 2 2

z z i

i i i

⎡ ⎤= + + +⎣ ⎦⎡ ⎤

⎡ ⎤= + = − − = − −⎢ ⎥⎣ ⎦⎣ ⎦


( )( ) ( ) ( )

( ) ( )

1 2 1 4 cos 290 40 sin 290 40

3 14 cos 330 sin 330 4 2 3 22 2

z z i

i i i

⎡ ⎤= + + +⎣ ⎦⎡ ⎤

⎡ ⎤= + = − = −⎢ ⎥⎣ ⎦⎣ ⎦

111. Using the formula ( ) ( )1 11 2 1 2

2 2

cos sinz r iz r

θ θ θ θ= − + −⎡ ⎤⎣ ⎦ yields

( ) ( )

( ) ( )

1

2

6 cos 200 50 sin 200 506

3 1cos 150 sin 1502 2

z iz

i i

⎡ ⎤= − + −⎣ ⎦

⎡ ⎤= + = − +⎣ ⎦


2 2

cos sinz r iz r

θ θ θ θ= − + −⎡ ⎤⎣ ⎦ yields

( ) ( )

( ) ( )

1

2

18 cos 190 100 sin 190 1002

9 cos 90 sin 90 9

z iz

i i

⎡ ⎤= − + −⎣ ⎦

⎡ ⎤= + =⎣ ⎦


2 2

cos sinz r iz r

θ θ θ θ= − + −⎡ ⎤⎣ ⎦ yields

( ) ( )

( ) ( )

1

2

24 cos 290 110 sin 290 1104

6 cos 180 sin 180 6

z iz

i

⎡ ⎤= − + −⎣ ⎦

⎡ ⎤= + = −⎣ ⎦

Chapter 8

1216


2 2

cos sinz r iz r

θ θ θ θ= − + −⎡ ⎤⎣ ⎦ yields

( ) ( )

( ) ( )

1

2

200 cos 93 48 sin 93 482

2 210 cos 45 sin 45 10 5 2 5 22 2

z iz

i i i

⎡ ⎤= − + −⎣ ⎦

⎡ ⎤⎡ ⎤= + = + = +⎢ ⎥⎣ ⎦

⎣ ⎦

115. In order to express ( )43 3i+ in rectangular form, follow these steps: Step 1: Write 3 3i+ in polar form. Since 3, 3x y= = , the point is in QI. So,

( ) ( )2 22 2 3 3 3 2r x y= + = + = 3tan 13

yx

θ = = = , so that ( )1tan 1 45θ −= = .

Hence, ( ) ( )3 3 3 2 cos 45 sin 45i i⎡ ⎤+ = +⎣ ⎦ .

Step 2: Apply DeMoivre’s theorem ( ) ( )cos sinn nz r n i nθ θ= +⎡ ⎤⎣ ⎦ .

( ) ( ) ( ) ( ) ( ) ( )[ ]

443 3 3 2 cos 4 45 sin 4 45 324 cos 180 sin 180

324 1 0 324

i i i

i

⎡ ⎤ ⎡ ⎤+ = ⋅ + ⋅ = +⎣ ⎦ ⎣ ⎦= − + = −

116. In order to express ( )43 3i+ in rectangular form, follow these steps:

Step 1: Write 3 3i+ in polar form. Since 3, 3x y= = , the point is in QI. So,

( ) ( )222 2 3 3 2 3r x y= + = + =

3 1tan3 3

yx

θ = = = , so that 1 1tan 303

θ − ⎛ ⎞= =⎜ ⎟⎝ ⎠

.

Hence, ( ) ( )3 3 2 3 cos 30 sin 30i i⎡ ⎤+ = +⎣ ⎦ .


( ) ( ) ( ) ( ) ( ) ( )4 43 3 2 3 cos 4 30 sin 4 30 144 cos 120 sin 120

1 3144 72 72 32 2

i i i

i i

⎡ ⎤ ⎡ ⎤+ = ⋅ + ⋅ = +⎣ ⎦ ⎣ ⎦⎡ ⎤

= − + = − +⎢ ⎥⎣ ⎦

Chapter 8 Review

1217

117. In order to express ( )51 3i+ in rectangular form, follow these steps:

Step 1: Write 1 3i+ in polar form. Since 1, 3x y= = , the point is in QI. So,

( ) ( )222 2 1 3 2r x y= + = + =

3tan1

yx

θ = = , so that ( )1tan 3 60θ −= = .

Hence, ( ) ( )1 3 2 cos 60 sin 60i i⎡ ⎤+ = +⎣ ⎦ .


( ) ( ) ( ) ( ) ( ) ( )5 51 3 2 cos 5 60 sin 5 60 32 cos 300 sin 300

1 332 16 16 32 2

i i i

i i

⎡ ⎤ ⎡ ⎤+ = ⋅ + ⋅ = +⎣ ⎦ ⎣ ⎦⎡ ⎤

= − = −⎢ ⎥⎣ ⎦

118. In order to express ( )72 2i− − in rectangular form, follow these steps: Step 1: Write 2 2i− − in polar form. Since 2, 2x y= − = − , the point is in QIII. So,

( ) ( )2 22 2 2 2 2 2r x y= + = − + − = 2tan 12

yx

θ −= = =

−, so that ( )1180 tan 1 225θ −= + = .

(Remember to add ( )1180 to tan 1− since the angle has terminal side in QII.)

Hence, ( ) ( )2 2 2 2 cos 225 sin 225i i⎡ ⎤− − = +⎣ ⎦ .


( ) ( ) ( ) ( )77 2 22 2 2 2 cos 7 225 sin 7 225 1024 22 2

1024 1024

i i i

i

⎡ ⎤⎡ ⎤− − = ⋅ + ⋅ = − +⎢ ⎥⎣ ⎦

⎣ ⎦= − +

Chapter 8

1218

119. Given 2 2 3z i= + , in order to compute 12z , follow these steps:

Step 1: Write 2 2 3i+ in polar form. Since 2, 2 3x y= = , the point is in QI. So,

( ) ( )222 2 2 2 3 4r x y= + = + =

2 3tan 32

yx

θ = = = , so that ( )1tan 3 60θ −= = .

Hence, ( ) ( )2 2 3 4 cos 60 sin 60z i i⎡ ⎤= + = +⎣ ⎦ .

Step 2: Now, apply 1 1 360 360cos sin , 0, 1, ... , 1n n k kz r i k nn n n nθ θ⎡ ⎤⎛ ⎞ ⎛ ⎞

= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

:

( ) ( ) ( ) ( )

12 60 360 60 3604 cos sin , 0, 1

2 2 2 2

2 cos 30 sin 30 , 2 cos 210 sin 210

k kz i k

i i

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤ ⎡ ⎤= + +⎣ ⎦ ⎣ ⎦

Step 3: Plot the roots in the complex plane:

Chapter 8 Review

1219

120. Given 8 8 3z i= − + , in order to compute 14z , follow these steps:

Step 1: Write 8 8 3z i= − + in polar form. Since 8, 8 3x y= − = , the point is in QII. So,

( ) ( )222 2 8 8 3 16r x y= + = − + =

8 3tan 38

yx

θ = = = −−

, so that ( )1180 tan 3 120θ −= + − = .

(Remember to add 180 to ( )1tan 3− − since the angle has terminal side in QII.)

Hence, ( ) ( )8 8 3 16 cos 120 sin 120z i i⎡ ⎤= − + = +⎣ ⎦ .


= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

:

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

14 4 120 360 120 36016 cos sin , 0, 1, 2, 3

4 4 4 4

2 cos 30 sin 30 , 2 cos 120 sin 120 ,

2 cos 210 sin 210 , 2 cos 300 sin 300

k kz i k

i i

i i

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤ ⎡ ⎤+ +⎣ ⎦ ⎣ ⎦=⎡ ⎤ ⎡ ⎤+ +⎣ ⎦ ⎣ ⎦


Chapter 8

1220

121. Given 256 0z i= − + , in order to compute 14z , follow these steps:

Step 1: Write 256 0z i= − + in polar form. Since 256, 0x y= − = , the point is on the negative x-axis. So,

256, 180r θ= =

Hence, ( ) ( )256 0 256 cos 180 sin 180z i i⎡ ⎤= − + = +⎣ ⎦ .


= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

:

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

14 4 180 360 180 360256 cos sin , 0, 1, 2, 3

4 4 4 4

4 cos 45 sin 45 , 4 cos 135 sin 135 ,

4 cos 225 sin 225 , 4 cos 315 sin 315

k kz i k

i i

i i

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤ ⎡ ⎤+ +⎣ ⎦ ⎣ ⎦=⎡ ⎤ ⎡ ⎤+ +⎣ ⎦ ⎣ ⎦


Chapter 8 Review

1221

122. Given 0 18z i= − , in order to compute 12z , follow these steps:

Step 1: Write 0 18z i= − in polar form. Since 0, 18x y= = − , the point is on the negative y-axis. So,

18, 270r θ= =

Hence, ( ) ( )0 18 18 cos 270 sin 270z i i⎡ ⎤= − = +⎣ ⎦ .


= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

:

( ) ( ) ( ) ( )

12 270 360 270 36018 cos sin , 0, 1

2 2 2 2

3 2 cos 135 sin 135 , 3 2 cos 315 sin 315

k kz i k

i i

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤ ⎡ ⎤= + +⎣ ⎦ ⎣ ⎦


Chapter 8

1222

123. We seek all complex numbers x such that 3 216 0x + = , or equivalently 3 216x = − . While it is clear that x = 6− is a solution by inspection, the remaining two complex solutions aren’t immediately discernible. As such, we shall apply the approach for computing complex roots involving the formula

1 1 2 2cos sin , 0, 1, ... , 1n n k kz r i k nn n n nθ π θ π⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

.

To this end, we follow these steps. Step 1: Write 216z = − in polar form. Since 216, 0x y= − = , the point is on the negative x-axis. So, 216,r θ π= = . Hence, ( ) ( )216 216 cos siniπ π− = +⎡ ⎤⎣ ⎦ .

Step 2: Now, apply 1 1 2 2cos sin , 0, 1, ... , 1n n k kz r i k nn n n nθ π θ π⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

:

( )

( ) ( )

13 3 2 2216 216 cos sin , 0, 1, 2

3 3 3 35 56 cos sin , 6 cos sin , 6 cos sin

3 3 3 33 3 3 , 6, 3 3 3

k ki k

i i i

i i

π π π π

π π π ππ π

⎡ ⎤⎛ ⎞ ⎛ ⎞− = + + + =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + +⎡ ⎤⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

= + − −

Step 3: Hence, the complex solutions of 3 216 0x + = are 3 3 3 , 6, 3 3 3i i+ − − .

124. We seek all complex numbers x such that 4 1 0x − = . Observe that this equation can be written equivalently as:

( )( )( )( )( )

2 2

2

1 1 0

1 1 1 0

x x

x x x

− + =

− + + =

The solutions are therefore 1,x i= ± ± . (Note: We could have alternatively used the formula for complex roots, namely

1 1 360 360cos sin , 0, 1, ... , 1n n k kz r i k nn n n nθ θ⎡ ⎤⎛ ⎞ ⎛ ⎞

= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

.

However, the left-side readily factored in a way that is easy to see what all of the solutions were. As such, factoring was the more expedient route for this problem.)

Chapter 8 Review

1223

125. We seek all complex numbers x such that 4 1 0x + = , or equivalently 4 1x = − . We shall apply the approach for computing complex roots involving the formula


.

To this end, we follow these steps. Step 1: Write 1z = − in polar form. Since 1, 0x y= − = , the point is on the negative x-axis. So, 1,r θ π= = . Hence, ( ) ( )1 1 cos siniπ π− = +⎡ ⎤⎣ ⎦ .


:

( )1

4 4 2 21 1 cos sin , 0, 1, 2, 34 4 4 4

3 3cos sin , cos sin ,4 4 4 45 5 7 7cos sin , cos sin4 4 4 4

2 2 2 2,2 2 2

k ki k

i i

i i

i

π π π π

π π π π

π π π π

⎡ ⎤⎛ ⎞ ⎛ ⎞− = + + + =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

= + − +2 2 2 2, ,

2 2 2 2 2i i i− − −

Step 3: Hence, the complex solutions of 4 1 0x + = are 2 2 2 2 2 2 2 2, , ,

2 2 2 2 2 2 2 2i i i i+ − + − − − .

Chapter 8

1224

126. We seek all complex numbers x such that 3 125 0x − = , or equivalently 3 125x = . While it is clear that x = 5 is a solution by inspection, the remaining two complex solutions aren’t immediately discernible. As such, we shall apply the approach for computing complex roots involving the formula


.

To this end, we follow these steps. Step 1: Write 125z = in polar form. Since 125, 0x y= = , the point is on the positive x-axis. So, 125, 0r θ= = . Hence, ( ) ( )125 125 cos 0 sin 0i= +⎡ ⎤⎣ ⎦ .


:

( ) ( )

13 3 0 2 0 2125 125 cos sin , 0, 1, 2

3 3 3 32 2 4 4 5 5 3 5 5 35 cos 0 sin 0 , 5 cos sin , 5 cos sin 5, ,3 3 3 3 2 2 2 2

k ki k

i i i i i

π π

π π π π

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + = − + − −⎡ ⎤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

Step 3: Hence, the complex solutions of 3 125 0x − = are 5 5 3 5 5 35, ,2 2 2 2

i i− + − − .

127. In order to convert the point ( )2, 2− (expressed in rectangular coordinates) to polar coordinates, first observe that 2, 2x y= − = , so that the point is in QII. Hence,

( ) ( )2 22 2 2 2r = − + = 2tan 1

2yx

θ −= = = − so that ( )1 3tan 1

4πθ π −= + − = .

(Remember to add π to ( )1tan 1− − since the point is in QII.)

So, the point can be expressed in polar coordinates as 32 2,4π⎛ ⎞

⎜ ⎟⎝ ⎠

, plotted below:

Chapter 8 Review

1225

128. In order to convert the point ( )4, 4 3− (expressed in rectangular coordinates) to

polar coordinates, first observe that 4, 4 3x y= = − , so that the point is in QIV. Hence,

( ) ( )224 4 3 8r = + − =

4 3tan 34

yx

θ −= = = − so that ( )1tan 3

3πθ −= − = − .


So, the point can be expressed in polar coordinates as 58,3π⎛ ⎞

⎜ ⎟⎝ ⎠

, plotted below:

Chapter 8

1226

129. In order to convert the point ( )5 3, 5− − (expressed in rectangular coordinates) to

polar coordinates, first observe that 5 3, 5x y= − = − , so that the point is in QIII. Hence,

( ) ( )2 25 3 5 10r = − + − =

5 1tan5 3 3

yx

θ −= = =

− so that 1 1 7tan

63πθ π − ⎛ ⎞= + =⎜ ⎟

⎝ ⎠.

(Remember to add π to 1 1tan3

− ⎛ ⎞⎜ ⎟⎝ ⎠

since the point is in QIII.)

So, the point can be expressed in polar coordinates as 710,6π⎛ ⎞

⎜ ⎟⎝ ⎠

, which is plotted below:

Chapter 8 Review

1227

130. In order to convert the point ( )3, 3 (expressed in rectangular coordinates) to

polar coordinates, first observe that 3, 3x y= = , so that the point is in QI. Hence,

( ) ( )2 23 3 6r = + =

3tan 13

yx

θ = = = so that ( )1tan 14πθ −= = .


⎜ ⎟⎝ ⎠

, which is plotted

below:

Chapter 8

1228

131. In order to convert the point ( )0, 2− (expressed in rectangular coordinates) to polar coordinates, first observe that 0, 2x y= = − , so that the point is on the negative y-axis. Hence,

( ) ( )2 2 30 2 2,2

r πθ= + − = =


⎜ ⎟⎝ ⎠

, which is plotted

below:

132. In order to convert the point ( )11,0 (expressed in rectangular coordinates) to polar coordinates, first observe that 11, 0x y= = , so that the point is on the positive x-axis. Hence,

( ) ( )2 211 0 11, 0r θ= + = =

So, the given point can be expressed in polar coordinates as ( )11,0 , which is plotted

below:

Chapter 8 Review

1229


⎝ ⎠ (expressed in polar coordinates) to

rectangular coordinates, first observe that 53,3


5 1 3cos 3cos 33 2 2

5 3 3 3sin 3sin 33 2 2

x r

y r

πθ

πθ

⎛ ⎞ ⎛ ⎞= = − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞= = − = − − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as 3 3 3,2 2

⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠

.


⎜ ⎟⎝ ⎠



r πθ= = . Hence,

5 2cos 4cos 4 2 24 2

5 2sin 4sin 4 2 24 2

x r

y r

πθ

πθ

⎛ ⎞⎛ ⎞= = = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞= = = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as ( )2 2, 2 2− − .


⎜ ⎟⎝ ⎠



r πθ= = . Hence,

1cos 2cos 2 13 2

3sin 2sin 2 33 2

x r

y r

πθ

πθ

⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞= = = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as ( )1, 3 .

Chapter 8

1230


⎜ ⎟⎝ ⎠



r πθ= = . Hence,

7 3cos 6cos 6 3 36 2

7 1sin 6sin 6 36 2

x r

y r

πθ

πθ

⎛ ⎞⎛ ⎞= = = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞= = = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as ( )3 3, 3− − .


⎜ ⎟⎝ ⎠



r πθ= = . Hence,

4 1cos 1cos3 2

4 3sin 1sin3 2

x r

y r

πθ

πθ

⎛ ⎞= = = −⎜ ⎟⎝ ⎠⎛ ⎞= = = −⎜ ⎟⎝ ⎠


⎛ ⎞− −⎜ ⎟⎜ ⎟⎝ ⎠

.


⎝ ⎠ (expressed in polar coordinates) to

rectangular coordinates, first observe that 73,4


7 2 3 2cos 3cos 34 2 2

7 2 3 2sin 3sin 34 2 2

x r

y r

πθ

πθ

⎛ ⎞⎛ ⎞= = − = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞= = − = − − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So, the given point can be expressed in rectangular coordinates as 3 2 3 2,2 2

⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠

.

Chapter 8 Review

1231

139. In order to graph 4cos 2r θ= , consider the following table of points:

θ 4cos 2r θ= ( ),r θ 0 4 ( )4,0

4π 0 ( )0, 4

π

2π -4 ( )4, 2

π−

34

π 0 ( )30, 4π

π 4 ( )4,π 5

4π 0 ( )50, 4

π

32

π -4 ( )34, 2π−

74

π 0 ( )70, 4π

2π 4 ( )4, 2π


Chapter 8

1232

140. In order to graph sin 3r θ= , consider the following table of points:

θ 3sin 3r θ= ( ),r θ 0 0 ( )0,0

6π 1 ( )1, 6

π

26

π 0 ( )20, 6π

2π -1 ( )1, 2

π−

46

π 0 ( )40, 6π

56

π 1 ( )51, 6π

π 0 ( )0,π 7

6π -1 ( )71, 6

π−

86

π 0 ( )80, 6π

32

π 1 ( )31, 2π

106

π 0 ( )100, 6π

116

π -1 ( )111, 6π−

2π 0 ( )0, 2π


141. In order to graph r θ= − , consider the following table of points:

θ 2r θ= − ( ),r θ 0 0 ( )0,0

2π - 2

π ( ),2 2π π− −

π -π ( ),π π− − 3

2π - 3

2π ( )3 3,2 2

π π− −

2π - 2π ( )2 , 2π π− −


Chapter 8 Review

1233

142. In order to graph 4 3sinr θ= − , consider the following table of points:

θ 4 3sinr θ= − ( ),r θ 0 4 ( )4,0

2π 1 ( )1, 2

π

π 4 ( )4,π 3

2π 7 ( )37, 2

π

2π 4 ( )4, 2π


143. The following steps constitute the program in this case: Program: ACX :Input “SIDE A =”, A :Input “SIDE C =”, C :Input “ANGLE X =”, X :sin-1(Csin(X)/A)→Y :180-Y-X →Z :Asin(Z)/sin(X)→B :Disp “ANGLE Y = “, Y :Disp “ANGLE Z = “, Z :Disp “SIDE B =”, B Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 31.6 SIDE C = 23.9 ANGLE X = 42 Now, the program will display the following information that solves the triangle: ANGLE Y = 30.40327 ANGLE Z = 107.59673 SIDE B = 45.01568

144. The following steps constitute the program in this case: Program: ABY :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Y =”, Y :sin-1(Bsin(Y)/A)→X :180-Y-X →Z :Asin(Z)/sin(Y)→C :Disp “ANGLE X = “, X :Disp “ANGLE Z = “, Z :Disp “SIDE C =”, C Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 137.2 SIDE B = 125.1 ANGLE Y = 54 Now, the program will display the following information that solves the triangle: ANGLE X = 47.53313 ANGLE Z = 78.46687 SIDE C = 166.16441

Chapter 8

1234

145. The following steps constitute the program: Program: ABZ :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Z =”, Z : (A2+B2-2ABcos(Z))→C :sin-1(Asin(Z)/C)→Y :180-Y-Z →X :Disp “SIDE C = “, C :Disp “ANGLE Y = “, Y :Disp “ANGLE X =”, X Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 33 SIDE B = 29 ANGLE Z = 41.6 Now, the program will display the following information that solves the triangle: SIDE C = ANGLE Y = ANGLE X =

146. The following steps constitute the program: Program: ABC :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “SIDE C =”, C :cos-1((B2+C2-A2)/(2BC)) →X : cos-1((A2+C2-B2)/(2AC)) →Y :180-Y-X →Z :Disp “ANGLE X = “, X :Disp “ANGLE Y = “, Y :Disp “ANGLE Z =”, Z Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 3412 SIDE B = 2178 SIDE C = 1576 Now, the program will display the following information that solves the triangle: ANGLE X = ANGLE Y = ANGLE Z =

Chapter 8 Review

1235

147. The following steps constitute the program in this case: Program: ABY :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Y =”, Y :1/2*ABsin(Y)→W :Disp “AREA W = “, W Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 312 SIDE B = 267 ANGLE Y = 189 Now, the program will display the following information that gives the area of the triangle: AREA W =

148. The following steps constitute the program in this case: Program: ABZ :Input “SIDE A =”, A :Input “SIDE B =”, B :Input “ANGLE Z =”, Z :1/2*ABsin(Z)→W :Disp “AREA W = “, W Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 12.7 SIDE B = 29.9 ANGLE Z = 104.8 Now, the program will display the following information that gives the area of the triangle: AREA W =

149. Magnitude = Sum ({25, 60}^ 2)− Direction angle = 1tan ( 60) (25)− − ÷ The output will be: Magnitude = 65 Direction Angle = -67.38014

150. Magnitude = Sum ({ 70,10 15}^ 2)− ∗

Direction angle = 1tan (10 15) ( 70)− ∗ ÷ − The output will be: Magnitude = 80 Direction Angle = -28.95502

151. ( )( ) ( )

( )1

Sum {14,37}*{9, 26} /

(Sum {14,37}^ 2 *Sum {9, 26}^ 20.76807cos 39.81911Ans−

−

− =

=

So, the angle is approximately 40 .

152. ( )( ) ( )

( )1

Sum { 23, 8}*{18, 32} /

(Sum { 23, 8}^ 2 *Sum {18, 32}^ 20.17672

cos 100.17877Ans−

− − −

− − − =

−

=


Chapter 8

1236

153. From the calculator, we obtain: 23 11 12i− − =

( )23 11 66angle i− − ≈ , so we use 246

So, ( )23 11 12 cos 246 sin 246i i− − ≈ + .

154. From the calculator, we obtain: 11 23 12i+ =

( )11 23 78angle i+ ≈

So, ( )11 23 12 cos 78 sin 78i i+ ≈ + .

155. Given 8 8 3z i= − + , in order to compute 14z , follow these steps:

Step 1: Write 8 8 3z i= − + in polar form.

Since 8, 8 3x y= − = , the point is in QII. So, using the calculator yields 16, 120r θ= =

Hence, ( ) ( )8 8 3 16 cos 120 sin 120z i i⎡ ⎤= − + = +⎣ ⎦ .


= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

:

14 4 120 360 120 36016 cos sin , 0, 1, 2, 3

4 4 4 4k kz i k

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

Step 3: Plot the roots in the complex plane and connect consecutive roots to form a square:

Chapter 8 Review

1237

156. Given 8 3 8z i= + , in order to compute 14z , follow these steps:

Step 1: Write 8 3 8z i= + in polar form.

Since 8 3, 8x y= = , the point is in QI. So, using the calculator yields 16, 30r θ= =

Hence, ( ) ( )8 8 3 16 cos 30 sin 30z i i⎡ ⎤= − + = +⎣ ⎦ .


= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

:

14 4 30 360 30 36016 cos sin , 0, 1, 2, 3

4 4 4 4k kz i k

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦


Chapter 8

1238

157. Consider the following graph of 1 2sin 3r θ= − :

Note that the curve self-intersects at the origin, multiple times. This occurs when r = 0. To find these angles for which this is true, we solve the following equation:

12

5 13 17 25 296 6 6 6 6 6

5 13 17 25 2918 18 18 18 18 18

0 1 2sin 3sin 3

3 , , , , ,

, , , , ,

π π π π π π

π π π π π π

θθθ

θ

= −=

=

=

158. Consider the following graph of 1 2cos3r θ= + :


12

5 7 11 13 173 3 3 3 3 3

5 7 11 13 179 9 9 9 9 9

0 1 2cos3cos3

3 , , , , ,

, , , , ,

π π π π π π

π π π π π π

θθθ

θ

= +=

=

=

Chapter 8 Practice Test Solutions ------------------------------------------------------------------- 1. This is AAS, so we use Law of Sines. Observe that:

( )180 30 40 110γ = − + =

( )( )10 sin 30sin sin sin 30 sin 40 7.810 sin 40

aa b aα β= ⇒ = ⇒ = ≈

( )( )10 sin110sin sin sin 40 sin110 14.610 sin 40

cb c cβ γ= ⇒ = ⇒ = ≈

2. Since only three angles are given, and none of the side lengths is prescribed, there are infinitely many such triangles by similarity.

Chapter 8 Practice Test

1239

3. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is γ .

( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 12 7 9 2 7 9 cos144 130 126cos

c a b ab γ γγ

= + − ⇒ = + −⇒ = −

Thus, 114 14cos so that cos 96.4126 126

γ γ − ⎛ ⎞= − = − ≈⎜ ⎟⎝ ⎠

.

Step 2: Find either of the remaining two angles using the Law of Sines. 1sin sin sin 96.4 sin 9sin 96.4 9sin 96.4sin sin 48.2

12 9 12 12c bγ β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find the third angle - ( )180 96.4 48.2 35.4α ≈ − + = 4. This is SSA, so we use Law of Sines: Step 1: Determineβ .

1sin sin sin 45 sin 10sin 45 10sin 45sin sin 628 10 8 8a b

α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠



( )1 180 62 45 73γ ≈ − + = 11

1 1

sinsin sin 45 sin 73 8sin 73 10.88 sin 45

ca c c

γα= ⇒ = ⇒ = ≈


( )2 180 118 45 17γ ≈ − + = 22

2 2

sinsin sin 45 sin17 8sin17 3.38 sin 45

ca c c

γα= ⇒ = ⇒ = ≈

5. There is no triangle in this case since the triangle inequality is violated (since a b+ >c ). 6. This is SAS, so begin by using Law of Cosines: Step 1: Find the side opposite the given angleβ .

( ) ( ) ( )( )2 22 2 2 2 23 5 23 57 7 7 72 cos 2 cos61.2

3.01b a c ac b

bβ= + − ⇒ = + −

⇒ ≈

Step 2: Find either of the remaining two angles using the Law of Sines. 23 23

17 723

7

sin 61.2 sin 61.2sin sin sin sin 61.2 sin sin 73.213.01 3.01 3.01a b

α β α α α − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find the third angle - ( )180 73.21 61.2 45.59γ ≈ − + =

Chapter 8

1240

7. This is AAS, so we use Law of Sines. Observe that: ( )180 110 20 50γ = − + =

( )( )5 sin 20sin sin sin110 sin 20 1.825 sin110

ba b bα β= ⇒ = ⇒ = ≈

( )( )5 sin 50sin sin sin110 sin 50 4.085 sin110

ca c cα γ= ⇒ = ⇒ = ≈

8. This is SAS, so begin by using Law of Cosines: Step 1: Find the side opposite the given angleα .

( ) ( ) ( )( )222 2 2 2 5 52 22 cos 3 5 2 3 5 cos 45

5.97

a b c bc a

a

α= + − ⇒ = + −

⇒ ≈

Step 2: Find either of the remaining two angles using the Law of Sines. 5 5

12 25

2

sin sin sin 45 sin sin 45 sin 45sin sin 7.615.97 5.97 5.97a b

α β β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find the third angle - ( )180 45 7.61 127.39γ ≈ − + =

9. ( )( )1 10 12 sin 72 572

A = ≈



( )( )( ) ( )( )( )( )15 8 5 3 34.64A s s a s b s c= − − − = ≈

11. Given that 5,12u = − , we have

( ) ( )2 25 12 13u = − + = 12tan

5θ =

− so that since the head is in QII,

1 12180 tan 112.65

θ − ⎛ ⎞= + ≈⎜ ⎟−⎝ ⎠

12.

( ) ( )2 2

3, 4 3, 453 4

3 4,5 5

vv

− − − −= =

− + −

− −=

13. a. 2 1,4 3 4,1 2,8 12,3 2 12,8 3 14,5− − = − − = − − − = −

b. ( )( ) ( )( )7, 1 2,2 7 2 1 2 16− − ⋅ = − + − = −


1241

14. We need the component form for each of the three vectors involved. Indeed, they are: 0,3 , 12,0 , 18cos330 , 18sin 330 15.59, 9A B C= = = ≈ −

As such, 27.59, 6A B C+ + = − . So, ( )2227.59 6 28.23 yardsA B C+ + = + − ≈ .

Also, the direction angle is given by 1 6tan 12.2727.59

− −⎛ ⎞ ≈ −⎜ ⎟⎝ ⎠

.

15. Given that 16 cos120 sin120z i⎡ ⎤= +⎣ ⎦ , we apply DeMoivre’s theorem

( ) ( )cos sinn nz r n i nθ θ= +⎡ ⎤⎣ ⎦ to compute 4z . Indeed, observe that

( ) ( ) ( )44 1 316 cos 4 120 sin 4 120 65,536 32,768 1 32 2

z i i i⎡ ⎤

⎡ ⎤⎡ ⎤= ⋅ + ⋅ = − + = − +⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

16. Given that 16 cos120 sin120z i⎡ ⎤= +⎣ ⎦ , we apply

1 1 360 360cos sin , 0, 1, ... , 1n n k kz r i k nn n n nθ θ⎡ ⎤⎛ ⎞ ⎛ ⎞

= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

,

to compute 14z . Indeed, observe that

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

14 4 120 360 120 36016 cos sin , 0, 1, 2, 3

4 4 4 4

2 cos 30 sin 30 , 2 cos 120 sin 120 ,

2 cos 210 sin 210 , 2 cos 300 sin 300 ,

3 , 1 3 , 3 , 1 3

k kz i k

i i

i i

i i i i

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤ ⎡ ⎤= + +⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤+ +⎣ ⎦ ⎣ ⎦

= + − + − − −



( )

( )

3 3 3cos 3cos 210 32 2

1 3sin 3sin 210 32 2

x r

y r

θ

θ

⎛ ⎞= = = − = −⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞= = = − = −⎜ ⎟⎝ ⎠

So, the given point can be expressed in rectangular coordinates as 3 3 32 2

i− − .

Chapter 8

1242


θ 6sin 2r θ= ( ),r θ 0 0 ( )0,0

4π 6 ( )6, 4

π

2π 0 ( )0, 2

π

34

π -6 ( )36, 4π−

π 0 ( )0,π 5

4π 6 ( )56, 4

π

32

π 0 ( )30, 2π

74

π -6 ( )76, 4π−

2π 0 ( )0, 2π


19. In order to graph 2 9cos 2r θ= , consider the following table of points:

θ 2 9cos 2r θ= r ( ),r θ 0 9 3± ( )3,0±

4π 0 0 ( )0, 4

π

2π -9 No

points

34

π 0 0 ( )30, 4π

π 9 3± ( )3,π± 5

4π 0 0 ( )50, 4

π

32

π -9 No points

74

π 0 0 ( )70, 4π

2π 9 3± ( )3,2π±


20. Observe that 43,1 3, 4 3 4 0x x x⋅ − = − = ⇒ =


1243

21. Observe that ( ) ( )

( )( ) ( )

( ) ( )

1 2 1 2 1 2

1 2 1 1 2 2

1 1 1 2 2 2

1 1 1 1 2 2 2 2

1 1 2 2 1 1 2 2

1 2 1 2 1 2 1 2

, , ,

, ,

, , , ,

u v w u u v v w w

u u v w v w

u v w u v wu v u w u v u wu v u v u w u w

u u v v u u w wu v u w

⋅ − = ⋅ −

= ⋅ − −

= − + −

= − + −

= + − +

= ⋅ − ⋅

= ⋅ − ⋅

22. The desired vector is

1 1 3 5 3 34 5 343,5 3,5 , ,34 343,5 34 34 34

⎛ ⎞− = − = − − = − −⎜ ⎟⎜ ⎟⎝ ⎠

23. ( ) ( )2 13 24(10)cos 40 20u v π⋅ = = − = −

24. Since ( )( ) ( )( )sin cos cos sin 0u v θ θ θ θ⋅ = − + = , we conclude that u and v are perpendicular.

25. 2 2( 1) ( 1) 2v = − + − = 26. Consider the following diagram:

We first determineβ using the Law of Cosines:

2 2 2

516

5 4 2 2(2)(4)coscos

18.209

ββ

β

= + −− =

≈ −

So, we use the reference angle 161.791β = .

Now, by the Law of Sines, we obtain

1

sin sin161.791 4sin161.791sin4 5 5

4sin161.791sin5

14.48

θ θ

θ −

= ⇒ =

⎛ ⎞⇒ = ⎜ ⎟

⎝ ⎠

≈

β

Chapter 8

1244


Begin by finding either a and b, in either order.

sin 20 sin110 2sin110 5.492 sin 20

sin 20 sin 50 2sin 50 4.482 sin 20

aa

ab

= ⇒ = ≈

= ⇒ = ≈

Next, consider the triangle with legs of lengths b, c, and d. The other two interior angles of this triangle are 70 (the angle supplementary to 110 ) and 15 (since the sum of all three angles must add to 180 ). Hence, by the Law of Sines, we have

sin 95 sin15 4.48sin15 1.164.48 sin 95

sin 95 sin 70 4.48sin 70 4.234.48 sin 95

dd

cc

= ⇒ = ≈

= ⇒ = ≈

28. True. Observe that ( ) ( ) 2 2 2 2

0

0u v u v u v u v v u u v u v=

+ ⋅ − = − + ⋅ − ⋅ = ⇒ = ⇒ = .


1245

29. Solving the equation 4 256 0z i+ = is equivalent to computing 14z , where 0 256z i= − ,

Follow these steps: Step 1: Write 0 256z i= − in polar form. Since 0, 256x y= = − , the point is on the negative y-axis. So,

256, 270r θ= =

Hence, ( ) ( )0 256 256 cos 270 sin 270z i i⎡ ⎤= − = +⎣ ⎦ .


= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

:

( ) ( )

14 4 270 360 270 360256 cos sin , 0, 1, 2, 3

4 4 4 4

4 cos 67.5 90 sin 67.5 90 , 0, 1, 2, 3

k kz i k

k i k k

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤= + + + =⎣ ⎦

30. Observe that 2

2

2 2

tansin sincos cos

rrrryx yx

θθ θθ θ

=

= =

+ =

31. ( )( ) ( )

( )1

Sum { 8, 11}*{ 16,26} /

(Sum { 8, 11}^ 2 *Sum { 16,26}^ 20.38051

cos 112.36528Ans−

− − −

− − − =

−

=


Chapter 8

1246

32. Given 8 3 8z i= − − , in order to compute 14z , follow these steps:

Step 1: Write 8 3 8z i= − − in polar form.

Since 8 3, 8x y= − = − , the point is in QIII. So, using the calculator yields 16, 210r θ= =

Hence, ( ) ( )8 8 3 16 cos 210 sin 210z i i⎡ ⎤= − − = +⎣ ⎦ .


= + + + = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

:

14 4 210 360 210 36016 cos sin , 0, 1, 2, 3

4 4 4 4k kz i k

⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦


Chapter 8 Cumulative Review ------------------------------------------------------------------------ 1.

( )22 16

2 16 4

2 4

x

x i

x i

− = −

− = ± − = ±

= ±

2. 5 3 12

5 153

xxx

− − ≤− ≤

≥ −

So, the solution is [ )3,− ∞ .

3. ( )( )

3 14 4 6 3

4 21 1 43 6 6

1m− −

= = = =− −

Chapter 8 Cumulative Review

1247

4. Completing the square on the x-terms and y-terms separately yields

( ) ( )( ) ( )

2 2

2 2

2 2

2 2

3.2 4.4 0.44 03.2 4.4 0.44

3.2 2.56 4.4 4.84 0.44 2.56 4.84

1.6 2.2 7.84

x y x yx x y y

x x y y

x y

+ − + − =

− + + =

− + + + + = + +

− + + =

Thus, the center is (1.6, -2.2) and the radius is 7.84 2.8= . 5.

( ) ( )

( )

2 2

2 2 2

5 ( ) 5( ) ( )

5 2 5

22

x h xf x h f xh h

x hx h xh

h x hx h

h

− + − −+ −=

− − − − +=

− += = − −

6. a. (0) 1f = − b. (4) 7f = c. (5) 0f = d. ( 4) 16f − = e. The domain and range are both all reals. f. increasing on (0,5) and decreasing on ( ) ( ),0 5,−∞ ∪ ∞ . Never constant.

7. Since ( 1) 1g − = − , ( )( 1) ( 1)f g f− = − , which is not defined since -1 is not in the domain of f. 8. We need to determine a, h, and k in the form 2( )y a x h k= − + . Since the vertex is (1.5, 2.5), we know that 2( 1.5) 2.5y a x= − + . In order to find a, substitute in the point (-0.5, -0.5) to obtain

20.5 ( 0.5 1.5) 2.53 4

0.75

aa

a

− = − − +− =

= −

Hence, the equation is 20.75( 1.5) 2.5y x= − − + . 9. The real zeros are 0 (multiplicity 2) and -2 (multiplicity 3). 10. Factors of -2: 1, 2± ± Factors of 6: 1, 2, 3, 6± ± ± ± Possible rational zeros:

1 1 2 12 3 3 61, 2, , , ,± ± ± ± ± ±

Observe that synthetic division yields

12

1 6 1 5 26 7 2

6 7 2 0

3 2

6 4 0

− −

−

− −

So, ( )( )( )1

2( ) 1 6 4( 1)(2 1)(3 2)

P x x x xx x x

= − + +

= − + +

Chapter 8

1248

11. Vertical asymptote: x = 2 Horizontal asymptote: none Slant asymptote:

2

2

22 0 3

( 2 )

2 3(2 4)

7

xx x x

x x

xx

+− + +

− −

+− −

So, the slant asymptote is 2y x= + .


12. y-intercept: 1 15(0) 5f −= = , so ( ( )1

50, Domain: all reals Range: (0, )∞ Horizontal asymptote: y = 0


13. Use r tA Pe= . Here,

85,000, 0.055, 15A r t= = = . Solving for P yields

0.055 (15)

0.055 (15)

85,00085,000 37, 250

Pe

pe

=

= ≈

So, about $37,250.

14. Reflect the graph of logy x= over the x-axis, then shift it up 2 units. The graph is:


1249

15. 33log10 log1010 10 3

−− = = − 16.

( )

12

12

12

1 12 2

12

2

2

3 2

2

ln 6 3 ln( 2) lnln(6 3 ) ln( 2) ln

6 3ln ln2

6 3ln ln2

6 32

6 32

6 3 ( 2)2 3 6 0

( 1) 3 6 0

1

x x xx x x

x xx

x xx

x xx

x xx

x x xx x x

x x x

x

− − + =

− − + =

−⎛ ⎞ =⎜ ⎟+⎝ ⎠

−⎛ ⎞ =⎜ ⎟+⎝ ⎠

−⎛ ⎞ =⎜ ⎟+⎝ ⎠−

=+− = +

+ + − =

− + + =

= 17. Since the leg has length x = 15 ft. and it is a 45 45 90− − triangle, the hypotenuse

is 15 2 ft.

18. Since tan 1.4285θ = , it follows that 1tan (1.4285) 0.96005θ −= ≈ . Since the

terminal side of θ is in QIII, we conclude that 0.96005θ π≈ + .

19. The period of 3cos 2y x= is π . The graph is as follows:

Chapter 8

1250


Observe that

5711

1 1 11 57seccos 57

θθ

= = = .

21. Consider the following two triangles:

Since 15cosα = and the terminal side of α is in QIV, we have 2 6sin

5α = − .

Since 35cosβ = − and the terminal side of β is in QII, we have 4sin

5β = .

Hence, using the addition formula, we obtain:

( )( ) ( )( )2 631 45 5 5 5

3 8 6cos( ) cos cos sin sin25

α β α β α β − −− = + = − + − = .

22. 2 4 8 10 2 4 51

2 3 3 3 3 3 3 3 3cos 2 2 , , , , , ,π π π π π π π πθ θ θ= − ⇒ = ⇒ =

57−

θ

β α2 6−


1251

23. This is SAS, so begin by using Law of Cosines: Step 1: Find the side opposite the given angleγ .

2 2 2 2 2 22 cos 13 8 2(13)(8)cos 216.23

c a b ab cc

γ= + − ⇒ = + −⇒ ≈

Step 2: Find either of the remaining two angles using the Law of Sines. 1sin sin sin sin 21 8sin 21 8sin 21sin sin 27.4

8 6.23 6.23 6.23b cβ γ β β β − ⎛ ⎞= ⇒ = ⇒ = ⇒ = ≈⎜ ⎟

⎝ ⎠

Step 3: Find the third angle - ( )180 21 27.4 131.6α ≈ − + = 24.

( )4 4 2 5,6 ( 3)

4 7,9 28,36

v u− = − − − −

= − = −

25. ( )3

42 2 ieπ θ

26. Magnitude = ({40, 96}^ 2) 104Sum − =

Direction angle = 1tan ( 96 / 40) 67.38014 67− − = − ≈ − 27. Consider the following graph of 2 5sin 2r θ= − :


Chapter 8

1252

( ) ( ) ( ) ( )( )( ) ( )( ) ( ) ( )( )

25

1 1 1 12 2 2 25 5 5 5

1 1 1 11 2 1 2 1 2 1 22 5 2 5 2 5 2 5

0 2 5sin 2sin 2

2 sin , 180 sin , 360 sin , 360 180 sin

sin , 180 sin , 180 sin , 180 180 sin

11.8 , 78.2 , 191.8 , 258.2

θθ

θ

θ

− − − −

− − − −

= −=

= − + + −

= − + + −

≈

cat 2e ism chapter 8 part 3 - oak park and river forest ...faculty.oprfhs.org/cavalos/book chapters...

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