Chapter 4: Circular Functionsachsprecalc.weebly.com/uploads/1/9/8/2/19823981/cpm_pre... · 2019. 8....

CPM Educational Program © 2012 Chapter 4: Page 1 Pre-Calculus with Trigonometry

Chapter 4: Circular Functions Lesson 4.1.1 4-1. a. 2! b. ! c. i.

30! ! " radians

180!= "6 radians

ii. 45! ! " radians

180!= "4 radians

iii. 60! ! " radians

180!= "3 radians

4-2. a. and b. 4-3. Possible patterns that can be seen: In Quadrant I: all x- and y-values are positive. In Quadrant II: all x-values are negative, all y-values are positive. In Quadrant III: all x- and y-values are negative In Quadrant IV: all x-values are positive, all y-values are negative. Notice there is symmetry across the x-axis and y-axis with respect to the coordinates

(without regarding the signs). 4-4. a. The coordinates are ± 22 , ±

22( ) b. i. ! 32 , 12( ) ii. 12 , ! 32( )

4-5. a. ! = 2"3 has coordinates !

12 ,

32( ) b. ! = 7"4 has coordinates 22 , ! 22( )

c. ! = 7"6 has coordinates !32 , !

12( ) d. ! = 3"4 has coordinates ! 22 , 22( )

4-6. a. ! = " #4 has coordinates

22 , !

22( ) b. ! = 7"3 has coordinates 12 , 32( )

!6 !!

32 ,

12( )

!4 !!

22 ,

22( )

!3 !!

12 ,

32( )2!3 !! " 12 , 32( )

3!4 !! "

22 ,

22( )

5!6 !! "

32 ,

12( )

7!6 !! "

32 , "

12( )

5!4 !! "

22 , "

22( )

4!3 !! "

12 , "

32( )

11!6 !!

32 , "

12( )

7!4 !!

22 , "

22( )

5!3 !!

12 , "

32( )

!2 !! 0,1( )

3!2 !! 0, "1( )

! !! "1, 0( ) 0!! 0,1( )


c. ! = " 5#6 has coordinates !32 , !

12( ) d. ! = 11"4 has coordinates ! 22 , 22( )

Review and Preview 4.1.1 4-7. a. b. c. 4-8. a. c2 + s2 = s2 + c2 = 1 b. s2 + c2 = 1! s2 = 1" c2 c. s2 + c2 = 1! c2 = 1" s2 d. Cannot be simplified.

e. s2 + c2 = 1! c2 = 1" s2 so 4(1! s2 ) = 4c2 f. 4 ! 4s2 = 4(1! s 2 ) = 4c2

4-9. 500 revolutionsminute !

2" radians1 revolution !

1 minute60 seconds =

50"3 radians per second

4-10. a. log3(9) = 2 because 32 = 9 . b. log3 13( ) = !1 because 3!1 = 13 . c. log5(625) = 4 because 54 = 625 . d. log3(0) is impossible because there is no x-value that makes 3x = 0 .

e. log5(?) = 2 means ? = 52 = 25 . f. log5(?) = 12 means ? = 51 2 = 5 .

4-11.

a. 1 23 2

= 1223= 1

3= 33 b.

1x ! x( ) 1x+1 !1( )

= 1(x+1)x !1x !

xx+1 + x

= 1!(x+1)!x2 +x2 (x+1)

(x+1)x

= !x+x3(x+1)x

= (x!1)(x+1)x(x+1)x= x !1

4-12. a. 3hours ! 45 mileshour = 135miles , 2hours !55

mileshour = 110miles , 110 +135 = 245 miles

b. See graph at right. c. Area under curve = 45 ! 3+ 55 !2 = 245

5!3 !!

12 , "

32( )

! 5"4 !! !22 ,

22( ) 8!3 !! " 12 , 32( )

spee

d (m

ph)

45

55

time (hours)


d. They are the same because the area under the curve is the distance traveled.


4-13. a.

48! ! " radians

180!= 4"15 b.

!500! " # radians180!

= ! 25#9

c. 13!12 "

180!! radians = 195

! d. !20"

9 #180!

" radians = !400!

4-14. a. y = 16(2)2x!2 = 16(22x "2!2 ) = 16(22 )x (2!2 ) = 16(22 )x 14( ) = 4(4)x b. y = (4)1 2 x+2 = (4)1 2 x (4)2 = 16(4)1 2 x = 16(41 2 )x = 16( 4 )x = 16(2)x

c. y = 60 12( )3x+1 = 60 12( )3x 12( ) = 30 12( )3!" #$x= 30 18( )x

d. y = 81 49( )12 x+2 = 81 49( )

12 x 4

9( )2 = 81 49( )12 x 16

81( ) = 16 49( )1 2!" #$x= 16 23( )x

Lesson 4.1.2 4-15. a. Because P(x, y) is a point on the unit circle, the length from the origin to the circle is 1.

Thus, the hypotenuse of the triangle has length 1. b. sin! = opphyp =

y1 = y

c. cos! = adjhyp =x1 = x

d. A circle with center (h, k) and radius r is described by: (x ! h)2 + (y ! k)2 = r2 . The unit circle has center (0, 0) and radius r = 1 so the equation of the circle is: x2 + y2 = 1.

e. Since x = cos! and y = sin! , the equation of the circle can be written: cos2 ! + sin2 ! = 1 . f. Given the triangle, the Pythagorean theorem says x2 + y2 = 1. This triangle in the unit

circle generated the equation cos2 ! + sin2 ! = 1 , thus they named it the Pythagorean Identity.

4-16. Domain: all angles; Range: [–1, 1] 4-17. a. tan! = yx b. tan! =

sin!cos!

4-18. a. The y-value sine is positive in Quadrant I and II. b. The x-value cosine is positive in Quadrants I and IV. c. Tangent is positive when sine and cosine have the same sign, i.e., in quadrants I and III. d. The mnemonic “All Students Take Control” is used to say that All three trigonometric

functions are positive in Quadrant I, the Sine function the only function that is positive in Quadrant II, the Tangent function is the only function that is positive in Quadrant III, and the Cosine function is the only function that is positive in Quadrant IV.


4-19. The coordinates are the same except for the signs. The reference angle for 4!3 is

!3 .

4-20. a. The coordinates for !3 on the unit circle are

12 ,

32( ) , so sin !3( ) = 32 .

b. 5!4 has a reference angle of !4 which has coordinates

22 ,

22( ) . Since 5!4 is in Quadrant

III, the cosine function is negative. Thus, cos(5!4 ) = "22 .

c. 7!6 has a reference angle of !6 which has coordinates

32 ,

12( ) . Since 7!6 is in Quadrant

III, the sine function is negative. Thus, sin 7!6( ) = " 12 . d. ! "3 has a reference angle of

!3 which has coordinates

12 ,

32( ) . Since ! "3 is in Quadrant

IV, the cosine function is positive. Thus, cos(! "3 ) =12 .

4-21. To find sin2 ! , find the sin of ! , then square your result.

a. sin !3( ) = 32 so sin2 !3( ) = 32( )2 = 34 . b. cos !3( ) = 12 so cos2 !3( ) = 12( )2 = 14 c. tan !3( ) = sin(! 3)cos(! 3) = 3 21 2 = 32 " 21 = 3 so tan2 !3( ) = 3( )2 = 3 . d. sin2 !3( ) + cos2 !3( ) = 34 + 14 = 44 = 1. e. Remember: sin2 ! + cos2 ! = 1 . Here, ! = "3 so sin

2 !3 + cos

2 !3 = 1 .

4-22. a. sin! is the y-coordinate of the point on the unit circle, so sin! = 0.8 in this case. b. Using sin2 ! + cos2 ! = 1 and sin! = 0.8 we get: (0.8)2 + cos2 ! = 1

cos2 ! = 1" (0.8)2

cos! = ± 1" (0.8)2 = ± 1" 0.64

= ± 0.36 = ±0.6

Since the point P is in Quadrant II, we know that the cosine function is negative, so cos! = "0.6 .


4-23. a. Label the missing side ‘A’. Using the Pythagorean theorem: b. sin! = opphyp =

35

c. Quadrant IV has values of ! : 3"2 < ! < 2" .

d. If ! is in Quadrant IV, the sine function is negative, thus sin! = " 35 . Review and Preview 4.1.2 4-24. a. Using a 30! ! 60! ! 90! triangle, we know that sin 30

! = 32 . Since 30! = !3 is the

reference angle for 4!3 which is in Quadrant III, we know the sine function is negative and

sin 4!3 = "32 .

b. Using a 30! ! 60! ! 90! triangle, we know that cos 30! = 12 . Since 30

! = !3 is the reference

angle for 11!3 which is in Quadrant IV, we know the cosine function is positive and

cos 11!3 =12 .

c. Using a 30! ! 60! ! 90! triangle, we know that sin 60! = 12 . Since 60

! = !6 is the reference

angle for ! 5"6 which is in Quadrant III, we know the sine function is negative and

sin ! 5"6( ) = ! 12 d. Using a 30! ! 60! ! 90! triangle, we know cos 30

! = 12 . Since 30! = !3 is the reference

angle for ! 2"3 which is in Quadrant III, we know the cosine function is negative and

cos ! 2"3( ) = ! 12 . 4-25. a. and b. See digrams at right.

c. sin2 ! + cos2 ! = 1 . If cos! = 35 , cos2 ! = 35( )2 = 925 .

Then, sin2 ! = 1" cos2 != 1" 925 =

1625

sin! = ± 1625 = ±45

d. tan! = sin!cos! = ±4 53 5 = ±

45 "

53 = ±

43

4-26. a. log M + 2 log N = log M + log N 2 = log(MN 2 )

A2 + 42 = 52

A2 = 25 !16 = 9

A = 9 = 3

5

3 !


b. 12 (log P ! logQ) =12 log

PQ( ) = log PQ

4-27. log 12( ) = log 2!1( ) = ! log(2) 4-28. a. The amplitude is 12 the height from peak to trough, i.e.

12 ! 4 = 2 .

b. The period is the length from peak to peak, or 4. c. Domain: all real numbers. Range: !2 " y " 2 . d. Roots, or x-intercepts, are points where the graph crosses the x-axis. This happens at all

odd integers: r = 2n +1 , n any integer. 4-29. ! is the reference angle for !, " ,# . a. sin ! is in Quadrant II where sine is positive, so sin ! = 13 . sin! is in Quadrant III where

sine is negative, so sin ! = " 13 . sin! is in Quadrant IV where sine is negative, so

sin! = " 13 .

b. sin2 ! + cos2 ! = 1 . If sin! = 13 , sin2 ! = 13( )2 = 19 . Then, cos2 ! = 1" sin2 !

= 1" 19 =89

cos! = 89 =2 23

Here cos! is positive because it is in Quadrant I. c. cos! is in quadrant II where cosine is negative, so cos ! = "

2 23 . cos! is in Quadrant III

where cosine is negative, so cos ! = " 2 23 . cos! is in Quadrant IV where cosine is

positive, so cos! = 2 23 . 4-30. a. 1y ! y

= 1y !y2y

= 1!y2

y

= x2y

b. xy +yx

= x2xy +y2xy

= 1xy


c. x ! 1xyx

= x2 !1x "xy

= !y2y

= !y

d. 1x ! x( ) y ! 1y( )= yx !

1xy ! xy +

xy

= 1xy !1xy ! xy

= !xy

4-31. a. b. c. d. Lesson 4.1.3 4-35. a. The graphs have the same shape, and are shifted versions of each other. b. Each completes one full cycle in 2! c. The maximum value is 1, minimum is –1. 4-36. a. Domain of y = sin x is !" # x # " . The range is !1 " y " 1 . b. Domain of y = cos x is !" # x # " . The range is !1 " y " 1 . 4-37. a. x-min: –6.152; x-max: 6.152; y-min: –4; y-max: 4. There are approximately 2 cycles on

the screen. 2! = 6.283 is close to 6.152. b. x-min: –352.5; x-max: 352.5; y-min: –4; y-max: 4. There are approximately 2 cycles on the

screen. The screen range is close to –360 to 360.


Review and Preview 4.1.2 4-38. a. The period for both is 2! . b. The amplitude is 1. c. The y-intercept for y = sin x is (0, 0) . The x-intercepts for y = sin x are (n! , 0) , n any

integer. d. The y-intercept for y = cos x is (0,1) . The x-intercepts for y = cos x are !2 + n! , 0( ) , n

any integer. 4-39. a. (sin!)2 = (0.6)2 = 0.36 b. sin2 ! = (sin!)2 = (0.6)2 = 0.36 c. Impossible, to calculate sin!2 , we must know the value of ! first. 4-40. Since !2 "# " ! , the cosine function is negative, so

cos! = " 1213 . tan! =sin!cos! =

5 13"12 13 = "

513 #

1312 = "

512 .

4-41. a. 3!4 has a reference angle of

!4 which has cos

!4 =

22 . Since

3!4 is in Quadrant II, the

cosine function is negative, thus cos 3!4 = "22 .

b. 11!6 has a reference angle of !6 which has sin

!6 =

12 . Since

11!6 is in Quadrant IV, the sine

function is negative, thus sin 11!6 = "12 .

c. ! 2"3 has a reference angle of !3 which has sin

!3 =

32 . Since !

2"3 is in Quadrant III, the

sine function is negative, thus sin! 2"3 = !32 .

d. 5!4 has a reference angle of !4 which has tan

!4 =

sin! 4cos! 4 =

2 22 2

= 1. Since 5!4 is in Quadrant III, the tangent function is positive, thus tan 5!4 = 1 .

4-42. a. 5 2 + 3 7 ! 2 2 + 4 7

= 5 2 ! 2 2 + 3 7 + 4 7

= (5 ! 2) 2 + (3+ 4) 7

= 3 2 + 7 7

b. 4 23( ) + 5 2 ! 7 23( ) + 2= 4 23( ) ! 7 23( ) + 5 2 + 2= (4 ! 7) 23( ) + (5 +1) 2= !3 23( ) + 6 2

cos2 ! + sin2 ! = 1cos2 ! = 1" sin2 !

cos2 ! = 1" 513( )2= 144169

cos! = ± 1213


c. 20 + 45

= 2 !2 !5 + 3 ! 3 !5

= 2 5 + 3 5 = (2 + 3) 5 = 5 5

d. 3 12 ! 32 + 27 + 2 8

= 3 2 "2 " 3 ! 2 "2 "2 "2 "2

+ 3 " 3 " 3 + 2 2 "2 "2

= 3 "2 3 ! 2 "2 2 + 3 3 + 2 "2 2

= 6 3 + 3 3 = 9 3


4-43. a. 4!7 is in Quadrant II, so the sine function (y-value) is positive and the cosine function (x-

value) is negative. The signs of the coordinates are (!,+) . b. ! 3"5 is in Quadrant III, so the sine and cosine functions are negative. The signs of the

coordinates are (!,!) . c. 37!9 is in Quadrant I, so the sine and cosine functions are positive. The signs of the

coordinates are (+,+) . d. !21" on the unit circle is the point (!1,0) , so the sign of the coordinates are (!,0) . 4-44. See graph at right. No, f (x) is not continuous

because you have to lift your pencil at x = 0 . 4-45.

a. y2x + x

= y2 +x2x

= 1x

b. y2

1 x2 !1= y

2

1!x2

x2

= y2 " x21!x2

= y2 " x2y2

= x2

c. 11!y +11+y

= (1+y)+(1!y)1!y2

= 2x2

d. 11 x!y x +1

1 x+y x

= x1!y +x1+y

= (1!y+1+y)x1!y2

= 2xx2

= 2x

4-46. ax + y = 7

3x ! y = b Add the equations together: (3+ a)x = 7 + b Solve for x: x = 7+b3+a

Substitute this value for x into any of the original two equations: 3x ! y = b becomes 3( 7+b3+a ) ! y = b . Solve this equation for y: y = 3

7+b3+a( ) ! b = 21+3b3+a ! b = 21+3b3+a ! b(3+a)3+a

= 21+3b!b(3+a)3+a =21!ab3+a

x

y


Lesson 4.1.3 4-47. a. See graph at right. b. To shift y = sin x !2 units to the right

and 3 units up, the function becomes y = sin x ! "2( ) + 3 .

c. See graph at right. d. To shift y = sin x h units to the right

and k units up, we write: y = sin(x ! h) + k .

4-48. a. The amplitude and period of sin x and cos x are the same. The shape is the same, but one

is a shifted version of the other. b. To get sin x from cos x , shift cos x to the left or right by !2 , for example.

c. y = sin x can be written as y = cos x ! "2( ) . d. y = cos x can be written as y = sin x + !2( ) . e. sin x = cos x ! "2( ) and cos x = sin x + !2( ) . 4-49. a. To give y = sin x an amplitude of 12 , write: y = 0.5 sin x . b. To reflect y = sin x across the x-axis, write: y = ! sin x . c. A sine function with an amplitude a and shifted up by k units is: y = a sin x + k . 4-50. From problem 4-47, we have y = sin(x ! h) + k . To add the vertical stretch factor, write:

y = a sin(x ! h) + k . 4-51. a. y = sin x starts in the middle and goes up. b. y = ! cos x starts at the bottom and goes up. c. y = ! sin x starts in the middle and goes down. 4-52. a. The graph of y = 3sin(x) ! 2 starts in

the middle and goes up, has an amplitude of 3, and is shifted down 2 units.


Solution continues on next page. → 4-52. Solution continued from previous page. b. The graph of y = !2cos(x) !1 starts at

the bottom and goes up, has an amplitude of 2, and is shifted down 1 unit.

c. The graph of y = ! sin x + "4( ) +1

starts in the middle and goes down, is shifted !4 units to the left, and shifted up 1 unit.

d. The graph of y = 3 cos x + !2( ) " 2

starts at the top and goes down, has an amplitude of 3, is shifted to the left by !2 units, and is shifted down 2 units.

Review and Preview 4.1.3 4-53. a. From y = sin x , the graph has an amplitude of 3 and is shifted down 2 units:

y = 3sin x ! 2 . From y = cos x , the graph has an amplitude of 3, is shifted down 2 units, and shifted to the

right !4 units: y = 3 cos x !"4( ) ! 2 .

b. From y = sin x , the graph has an amplitude of 2, is shifted up 1 unit, and is shifted to the right !4 units: y = 2 sin x !

"4( ) +1 .

From y = cos x , the graph has an amplitude of 2, is shifted up 1 unit, and is reflected across the x-axis: y = !2 cos x +1.

4-54. a. A radian is an angle measure found by wrapping a radius around a circle.

b. 240! ! " radians

180!= 4"3

c. 5!6 "

180!! radians = 150

!


4-55. a. We know the point W is in Quadrant III, so the sine function is negative. Furthermore, we

know sin! is the y-value in the coordinates for W . Since we know W has coordinates (?,!0.62) , we know sin! = "0.62 .

b. Let the origin be labeled with O. Because this is a unit circle, we know WO = 1. We are given VW = 0.62 . Then, VO2 +VW 2 =WO2

VO2 + (0.62)2 = 1VO2 + 0.384 = 1VO2 = 1! 0.384 = 0.616VO = 0.785

Now, cos! = VOWO = 0.785 . But, since we are in Quadrant III, the cosine function is negative so cos! = "0.785 .

c. The sine function is also negative in Quadrant IV, so ! is in Quadrant IV. d. The cosine function is also negative in Quadrant II, so ! is in Quadrant II. 4-56. a. 2

2= 2 22 = 2 b. !

23= ! 2 33

c. The reciprocal of 0 is an integer divided by 0, which is undefined. 4-57. a. See graph at right above. The graphs are the same. b. See graph at right below. y = cos(x + ! ) is equivalent to y = ! cos x . c. It reverses the sign because it

moves the angle 2 quadrants ahead. 4-58.

Area = !10 sin!10 k( )

k=0

9

" 4-59. (2x + 3)2 ! 3(2x + 3) !10 = 0 . Let a = 2x + 3 then we get: a2 ! 3a !10 = 0 which has

solutions: a = 3± 9!4(1)(!10)2(1) =3± 492 =

3±72 = !2, 5

Using a = !2 in a = 2x + 3 , we get: !2 = 2x + 3" x = ! 52 . Using a = 5 in a = 2x + 3 we get: 5 = 2x + 3! x = 1.


4-60. a. Ryan traveled 60 miles both ways, so 120 miles. b. 60miles ! 1 hour60 miles + 60miles !

1 hour30 miles = 1 hours + 2 hours = 3 hours

c. Total distance: 120 miles. Total time: 3 hours. Average speed: 120 miles3 hours = 40 miles per hour 4-61. a. 50x2 3 = 100

x2 3 = 2(x2 3)3 2 = (2)3 2

x = 2.828

b.

log3(x +1) + log3(x) = log312log3(x ! (x +1)) = log312x ! (x +1) = 12x2 + x "12 = 0

x = "1± 1"4(1)("12)2

= "1± 492

= "1±72

x = 3,"4log3(3) is defined, log3("4)is undefined.Answer: x = 3 only

c. 50 23( )x = 10023( )x = 2log 23( )x = log 2x log 23( ) = log 2x = log 2

log 23( )= !1.710

d. log3(x +1) ! log3(x) = log3 12

log3 x+1x( ) = log3 12x+1x = 12

x +1 = 12x1 = 11x

x = 111

Lesson 4.1.4 4-62. a. csc A = 1sin A =

1a c =

ca b. cot B =

1tan B =

1b a =

ab

c. sec A = 1cos A =1b c =

cb

4-63. b. c. i. tan B = oppositeadjacent =

32

ii. csc A = 1sin A =hypotenuseopposite =

132

iii. sin B = oppositehypotenuse =313

32 + x2 = 13( )2x = 13! 9 = 2


4-64. a. tan! = sin!cos! =

yx b. sec! =

1cos! =

1x

c. csc! = 1sin! =1y d. cot! =

cos!sin! =

xy

4-65. a. sec 2!3 =

1cos 2!3( )

= 1" 12

= "2 b. csc 3!4 =1

sin 3!4( )= 1

22

= 22= 2

c. cot 5!6 =1

tan 5!6( )=cos 5!6( )sin 5!6( )

=" 3212

= " 3 d. tan ! "3( ) =sin ! "3( )cos ! "3( )

=! 3212

= ! 3

4-66. a. csc(3) = 1sin(3) =

10.14112 = 7.086

b. cot(0.285733) = 1tan(0.285733) =1

0.293768 = 3.404

c sec(2.51334) = 1cos(2.51334) =1

!0.80906 = !1.236 4-67. Let !ABC contain !" = !B . Then, because sin! = 0.6 =

opphyp , we know that

AC = 0.6 and BC = 1. Let x = AB then x2 + (0.6)2 = 12

x2 + 0.36 = 1x2 = 1! 0.36 = 0.64

x = ± 0.64 = ±0.8

Because !2 "# " ! , we are in Quadrant II where the cosine function is negative, so we

want to let x = !0.8 . Then, cos! = ABBC =x1 = "0.8 .

4-68. a. Since (x, y) is on the unit circle, we know x2 + y2 = 1. b. We substitute x = cos! and y = sin! to get x2 + y2 = cos2 ! + sin2 ! = 1 . c. i. From sin2 ! + cos2 ! = 1 we get cos2 ! = 1" sin2 ! . ii. From sin2 ! + cos2 ! = 1 we get sin2 ! = 1" cos2 ! . iii. From sin2 ! + cos2 ! = 1 we get sin2 ! + cos2 ! "1 = 0

cos2 ! "1 = " sin2 !.


4-69. a. Let !ABC be a right triangle with !B = 90! and !A = " . Then, b. Since ! < " < 3!2 , we are in Quadrant III so the sine and

cosine functions are negative, and the tangent function is positive.

sin! = " BCAC = "441

cos! = " ABAC = "541

sec! = 1cos! =1

" 541

= " 415 csc! =1sin! =

1" 4

41

= " 414

cot! = 1tan! =145= 54

Review and Preview 4.1.4 4-70. a. sec! = sec "3( ) = 1cos "3( )

= 112= 2 b. csc! = csc "3( ) = 1sin "3( )

= 132

= 23= 2 33

c. tan! = tan "3( ) =sin "3( )cos "3( )

=3212

= 3 d. cot! = cot "3( ) = 1tan "3( )= 1

3= 33

4-71. a. sin! = 12 . The sine function is positive in Quadrants I and II. In Quadrant I this angle

corresponds to ! = "6 . In Quadrant II this angle corresponds to ! =5"6 .

b. cos! = " 12 . The cosine function is negative in Quadrants II and III. The reference angle for is ! = "3 . In Quadrant II this reference angle corresponds to ! =

2"3 . In Quadrant III

this reference angle corresponds to ! = 4"3 . c. sin! = " 32 . The sine function is negative in Quadrants III and IV. The reference angle

for which sin! = 32 is ! ="3 . In Quadrant III this corresponds to ! =

4"3 and in Quadrant

IV this corresponds to ! = 5"3 . d. cos! = 0 . The cosine function, which corresponds to the values of x, is zero on the y-axis.

These angles correspond to ! = "2 and ! =3"2 .

4-72. a. cos2 ! + sin2 ! = 1 b. cos2 ! + sin2 ! = 1

sin2 ! = 1" cos2 !

c. cos2 5 + sin2 5 = 1sin2 5 = 1! cos2 5

d. cos2 ! + sin2 ! = 1cos2 ! = 1" sin2 !

AC2 = AB2 + BC2

AC2 = 52 + 42 = 25 +16AC2 = 41

AC = 41


e. cos2(3! ) + sin2(3! ) = 1cos2(3! ) = 1" sin2(3! )

f. 4 ![cos2(3" ) + sin2(3" ) = 1]4 cos2(3" ) + 4 sin2(3" ) = 44 cos2(3" ) = 4 # 4 sin2(3" )


4-73.

a. xy !yx =

x2xy !

y2xy =

x2 !y2xy

b. 1x ! x( ) 1x + x( ) = 1x2 +1!1! x2 = 1x2 ! x4

x2= 1!x4

x2

c. 1y2

!1

1y2

= 1y2

!1"#$%&' y

2 = 1! y2

d. 1x!y +1x+y =

x+yx!y( ) x+y( ) +

x!yx!y( ) x+y( ) =

2xx2 !y2

4-74. a. c2 ! s2 = c2 ! (1! c2 ) = 2c2 !1 b. c1+s +

c1!s =

c 1!s( )1!s2

+ c 1+s( )1!s2

= 2cc2

= 2c

4-75. a. b. c. 4-76. a. A 60! sector is

60!

360!= 16 of the circle which would weigh 120g !

16 = 20g .

b. A = ! r2 . The area of the circle with radius 10cm is: A = 100! . This area weighs 120g. The area of the circle with radius 15cm is: A = (15)2! = 225! . This area weighs: x

225! =120100! " x =

120100! #225! = 270 g.

4-77. a. y = 7x

log y = log 7x

log y = x log 7

x = log ylog 7 = log7 y

b. y = x7

(y)1 7 = (x7 )1 7

(y)1 7 = x

c. y = log5 x ! 5y = x

4-78.

a. 13 +13.5 +

14 +

14.5 + ...+

19.5 =

13+0.5k

k=0

13

! b. 5 + 8 + ...+ 302 = 3k + 5k=0

99

!

c. cos(2) + 2 cos(4) + 3 cos(6) + ...+15 cos(30) = k cos(2k)k=1

15

!

f (2x)

f (12 x)

f (!x)


Lesson 4.2.2 4-79. tan! = oppositeadjacent =

opposite hypotenuseadjacent hypotenuse =

sin!cos!

cot! = 1tan! =cos!sin!

4-80. a. tan! " cos! = sin!cos! " cos! = sin!

b. cot! " sec! = cos!sin! "1

cos! =1sin! = csc!

c. sec!csc! =1 cos!1 sin! =

sin!cos! = tan!

d. sin!tan! " cot! " csc! = cos! "cos!sin! "

1sin! =

cos2 !sin2 !

= cot2 !

4-81. a. tan A ! sec A = sin Acos A !

1cos A =

sin A1"sin2 A

b. (tan2 A) 1sec2A( ) + 1sin A = sin2 Acos2 A ! cos2 A + 1sin A = sin2 A + 1sin A = sin3 A+1sin A c. 1+cos Asin A +

sin A1+cos A =

1!cos2 Asin A 1!cos A( ) +

sin A 1!cos A( )1!cos2 A

= sin A1!cos A +1!cos Asin A

= sin2 A+1!2 cos A+cos2 A1!cos A( ) sin A =2 1!cos A( )1!cos A( ) sin A =

2sin A

d. 11!cos A +1

1+cos A =1+cos A+1!cos A

1!cos2 A= 2sin2 A

4-82. a. The sine function is positive in Quadrants I and II. In Quadrant I, ! = "6 .

In Quadrant II, ! = 5"6 .

b. csc! = 1sin! = 2 =1

1 2

"! = #6 or 5#6

c. sec! = 1cos! = 2 =1

1 2

cos! = 12"! = #4 or

7#4

4-83. a. csc! = 1sin! = "2 =

1"1 2

sin! = " 12 #! =7$6 or

11$6

b. sec! = 1cos! =23

cos! = 32 "! =#6 or

11#6

c. csc! = 1sin! = "2 3

3

sin! = " 32 #! =4$3 or

5$3

d. sec! is undefined when cos! = 0

cos! = 0 "! = #2 or 3#2


4-84. a. csc! = 1sin!

sin! = 25= 2 55

b. tan! = 1cot ! = 0.8

cot! = 10.8 = 1.25

c. cos! = 35 " sin! =52 #325 =

45 d. csc! =

1sin! =

52

sin! = 25

cos! = 5"45

= 15= 55

4-85. a. Per = x measure from peak to peak = ! . Amp = y measure from peak to center = 2 b. Per = x measure from peak to peak = 6. Amp = y measure from peak to center = 5 4-86. a. log3 x + 2( ) ! log3 x = 2

log3 x+2x( ) = 2x+2x = 98x = 2

x = 14

b. 50 1.25( )x!3 = 6251.25( )x!3 = 272log1.25 272 = x ! 3

x = log1.25 272 + 3 = 14.319

4-87. a. cot ! = 73

cos ! = " 732 +7

= " 74

b. csc ! = hypotenuseopposite = "43

4-88. Sine: shifted up 1, Amp = 3! y = "3sin x( ) +1 Cosine: shifted up 1, shifted to left !2 , Amp = 3! y = 3 cos x +

"2( ) +1

4-89. a. 5 radians !

360!2" radians = 286.5

! b. c.

1200! ! 2" meters

360!= 20.944 meters

d. No, 120º is just short of the shaded region. 4-90. a. a2b5 = ab5/2 = ab2 b b. 243 = 23 ! 33 = 2 33

c. x5y63 = xy2 x23 d. x5y65 = xy y5

P


4-91. a2x + b2x + 7 = 3c

x a2 + b2( ) = 3c ! 7x = 3c!7

a2 +b2

Lesson 4.2.3 4-92. a. See graph at right, top. Period: 2!2 = !

b. See graph at right, middle. Period: 2!3 c. See graph at right, bottom.

Period: 2!0.5 =2!1/2 = 2! "2 = 4!

d. The period is 2!b . e. In part (a), 2 periods occur in 2! .

In part (b), 3 periods occur in 2! . In part (c), 0.5 periods occur in 2! .

f. p = 2!b or pb = 2! . 4-93. a. g(x) is horizontally compressed b. The amplitude and shift are the same c. b stretches the graph of y = sin x horizontally. The period is 2!b . 4-94. a. y = !2 sin(4x) + 5 has period 2!4 =

!2 . b. y = 5 cos

x3( ) !10 has period 2!1 3 = 6! .

c. y = 3sin !2 x( ) has period 2!! 2 = 4 . d. y = 4 cos 2!7 x( ) "1 has period 2!2! 7 = 7 . 4-95. Angular frequency is 2!30 =

!15 .

4-96. a. A cosine curve with period 2!3 is graph 2. b. A sine curve with amplitude 2 and shifted down by 1 unit is graph 4. c. A cosine curve with period 2!2 = ! and shifted up by 1 unit is graph 5. d. A sine curve with period 2!1 2 = 4! is graph 1. e. A sine curve with period 2! and shifted to the right by !4 is graph 8. f. A cosine curve with period 2! and shifted down by 2 units is graph 3. g. A sine curve with period 2!2 = ! , with amplitude 2, reflected across the x-axis is graph 6.


h. A cosine curve with period 2! , with amplitude 2, and shifted to the left by ! is graph 7.


Review and Preview 4.2.3 4-98. a. cos xsec x +

sin xcsc x =

cos x1

cos x+ sin x1sin x

=

cos2 x + sin2 x = 1

b. tan x+cot xsec x csc x =sin xcos x +

cos xsin x

sec x csc x =

sin2 x+cos2 xcos x sin x

1cos x sin x

=

1cos x sin x

1cos x sin x

= 1

4-99.

1+ 11! 1

1+11

= 1+ 11! 12

= 1+ 112= 1+ 2 = 3

4-100. cos! + sin! tan! = sec!

cos! + sin! " sin!cos! =

cos! " cos!cos! + sin! "sin!cos! =

1cos! cos

2 ! + sin2 !( ) =1

cos! (1) = sec!

4-101. The radius of the circular cross section, r, is: 104 =

xr !!!!!r =

4x10 =

2x5

The area of the circular cross section is: A = 2x5( )2 ! = 4x225 ! 4-102. See graph at right. By counting squares, or using the area

program we can estimate the area as ≈ 21. 4-103. a. Any angle in Quadrant 4. b. Any angle in Quadrant 3.

x

y


4-104. a. cos 7!6( ) has a reference angle of !6 . cos !6( ) = 32 . Since 7!6 is in Quadrant III, the

cosine function is negative so cos 7!6( ) = " 32 . b. Since 3!2 corresponds to the point (0, !1) , sin

3!2 = "1.

c. sec 4!3( ) = 1cos(4! 3) = 1"1 2 = "2 d. tan 2!3( ) = sin(2! 3)cos(2! 3) = 3 2"1 2 = 32 # " 21( ) = " 3 4-105. a. See graph at right above. Reflect the

function over the x-axis, halve the period, double the amplitude, shift the function up three units.

b. See graph at right below. Shift the function to the right by ! units, triple the amplitude, and shift the function down one unit.

4-106. First, log3 13( ) = !1 because 3!1 = 13 and log7 49 = 2 because 72 = 49 .

Then, log2 x ! log2 y = !12 log4 x + log4 y = 2

log2 xy = !1

log4 x2 + log4 y = 2" log4 x2y = 2

Now, log2 xy = !1" 2!1 = xy " y = 2x

!!!!!!!!!!!!!!!andlog4 x2y = 2" 42 = x2y

Substituting in y = 2x we get: 16 = x2y = x2 !2x = 2x3

8 = x3 " x = 2x = 2" y = 2 !2 = 4


Lesson 4.2.4 4-107. a. sin2 ! + cos2 ! = 1

sin2 !cos2 !

+ cos2 !cos2 !

= 1cos2 !

tan2 ! +1 = sec2 !

b. sin2 ! + cos2 ! = 1sin2 !sin2 !

+ cos2 !sin2 !

= 1sin2 !

1+ cot2 ! = csc2 !

4-108. a. tan2 ! +1 = sec2 !

tan2 ! = sec2 ! "1 b. sin2 ! + cos2 ! = 1

" sin2 ! = 1# cos2 !!!!!sin2 ! = (1# cos!)(1+ cos!)

c. sin2 ! + cos2 ! = 1" cos2 ! = 1# sin2 !tan2 ! +1 = sec2 !.Then, (1# sin2 x)(1+ tan2 x)

= cos2 x $ sec2 x

= cos2 x $ 1cos2 x

= 1

d. 1+ cot2 x = csc2 xcot2 x ! csc2 x = !1

4-109. a. (sec2 x)(1! cos2 x) = tan2 x

( 1cos2 x

)(1! cos2 x) = sin2 xcos2 x

1cos2 x

!1 = sin2 xcos2 x( ) cos2 x

1! cos2 x = sin2 x

1 = cos2 x + sin2 x

b. (cos A)(sec A ! cos A) = sin2 A

(cos A) 1cos A ! cos A( ) = sin2 A1! cos2 A = sin2 A1 = cos2 A + sin2 A

4-110. a. tan x + cot x = csc x sec x

sin xcos x +

cos xsin x =

1sin x !

1cos x

sin2 x+cos2 xcos x sin x =

1cos x sin x

1cos x sin x =

1cos x sin x

b. csc x + cot x = 1csc x!cot x1sin x +

cos xsin x =

11sin x !

cos xsin x

1+cos xsin x =

11!cos xsin x

1+cos xsin x =

sin x1!cos x

1+cos xsin x =

sin x 1+cos x( )1!cos2 x

1+cos xsin x =

1+cos xsin x


Review and Preview 4.2.3 4-111. a. tan A ! sec A = sin Acos A !

1cos A =

sin Acos2 A

= sin A1"sin2 A

b. (tan2 A) 1sec2 A( ) + 1sin A = sin2 Acos2 A( ) (cos2 A) + 1sin A = sin2 A + 1sin A = sin3 Asin A + 1sin A = sin3 A+1sin A

c. (1+ cot2 A) csc2 A = (csc2 A)(csc2 A) = csc4 A = 1sin4 A

d. 1!cos4 x1+cos2 x

= (1+cos2 x)(1!cos2 x)(1+cos2 x)

= 1! cos2 x = sin2 x

4-112.

a. cot2 x(sec2 x !1) = cot2 x tan2 x( ) = cos2 xsin2 x

sin2 xcos2 x( ) = 1

b. sin2 ! " cos2 ! = 1" cos2 ! " cos2 ! = 1" 2 cos2 !

c. tan2 ! + 6 = (tan2 ! +1) + 5 = sec2 ! + 5

d. cos y + sin y tan y = cos y + sin y sin ycos y = cos y +sin2 ycos y =

cos2 ycos y +

sin2 ycos y

= cos2 y+sin2 ycos y =

1cos y = sec y

4-113. a. csc! = 2 33 "

1sin! =

2 33 " sin! =

32 3

= 32

The sine function is positive in Quadrants I and II. Therefore, ! = "3 ,2"3 .

b. tan! = " 33 #sin!cos! = "

33 = "

13!!or !! sin!cos! = "

1/23/2

.

From the unit circle we know sin 5!6( ) = 12 and cos 5!6( ) = " 32 . Also, sin 11!6( ) = " 12 and cos 11!6( ) = 32 . Then, sin(5! 6)cos(5! 6) = 1 2" 3 2 = " 12 # 23 = " 13 = " 33 .

Also, sin(11! 6)cos(11! 6) ="1 23 2

= " 12 #23= " 1

3= " 33 . Thus, ! =

5"6 ,

11"6 .

c. sec! = 1cos! is undefined when cos! = 0 . cos! = 0 when ! ="2 ,

3"2 .

d. cot! = cos!sin! = "1 when cos! = " sin! . This is in Quadrants II and IV when ! =3"4 ,

7"4 .


4-114. a. Sinusoidal curves have either a sine or cosine base function. The amplitude is 3, and the

curve is shifted down 1 unit, and the period is ! . Using cosine: y = 3 cos(2x) !1 .

If we use sine, we need to shift the curve left by !4 units: y = 3sin 2 x +!4( )( ) "1 .

b. Since the function starts at the origin, we can use sine. The amplitude is 2, the period is 4! , and the function is reflected over the x-axis: y = !2 sin x2( ) .

4-115. a. 12 + 34 ! 27 = 4 " 3 +

14 " 3 ! 9 " 3 = 4 " 3 +

14 " 3 ! 9 " 3

= 2 " 3 + 12 " 3 ! 3 " 3 = 2 +12 ! 3( ) 3 = ! 12 " 3 = ! 32

b. 2 2 ! 3( ) 2 + 2 3( ) = 2 2( )2 + 4 3 " 2 ! 3 " 2 ! 2 3 " 3 = 2 "2 + 3 6 ! 2 " 3= 4 + 3 6 ! 6 = !2 + 3 6

c. 2 ! 3( )2 = 2( )2 ! 2 2 " 3 + 3( )2 = 2 ! 2 6 + 3 = 5 ! 2 6 4-116. Using the two inner triangles as similar right triangles: 3x =

x15 !!!!!x

2 = 45 = 3 5 4-117. a. See graph at right. b. The area under the curve can be broken up into a

rectangle with coordinates (0, 0), (3, 0), (0,18), (3,18) and a triangle with coordinates (0,18), (3,18), (3, 78) . The rectangle has an area of 18 ! 3 = 54 . The height of the triangle is: 78 !18 = 60 . The triangle has an area: 12 ! 3 !60 = 90 . Adding these together we get a total area of: 90 + 54 = 144 units.

c. The Gladiator traveled 144 miles in 3 hours. 4-118. An exponential function with a horizontal asymptote at y = 10 has the form:

y = A(Bx ) +10 because as x decreases without bound, y approaches 10, the horizontal asymptote. Since the function is increasing we know the exponent, x , will be positive. Using the points (1, 70) and (3,145) we get two equations: 70 = A(B1) +10 = AB +10145 = A(B3) +10

Solving the first equation for A we get: A = 60B .

Substitute this in for to the second equation: 135 = 60B (B)3 = 60B2 .

Solve for B: B2 = 13560B = 1.5

Solve for A: A = 60B =601.5 = 40 .

time (hours)

20

40

60

1 2 3

spee

d (m

ph)


The equation is: y = 40(1.5x ) +10 . Lesson 4.3.1 4-119. a. Let x represent the number of days since birth. b. Let y represent how positive or negative the day is. c. The domain will be the lifespan: 0 days until death. d. The range will be twice the amplitude centered at zero: (–100,100). The window settings should be x[0, 33] and y[!100,100] . 4-120. a. The team member with the highest physical cycle value P(x) . b. The team member with the lowest emotional cycle value E(x) . c. The team member with the highest intellectual cycle value I(x) . 4-121. Step 1: (2007 !1991) " 365 = 5840 Step 2: 5840 + 4 = 5844 Step 3: 5844 + (2 + 31+ 31+ 30 + 31+13) = 5982 Michelle has been alive for 5982 days. Her intellectual value is

I(5982) = 100 sin 2!33 "5982( ) = 98.98 , which is near a peak value. 4-122. No, because P(x) = –100 when x = 23 3+2n4( ) , E(x) = –100 when x = 28 3+2n4( ) , and

I(x) = –100 when x = 33 3+2n4( ) . 23, 28, and 33 do not share a common factor and thus these three solutions will never coincide.

Review and Preview 4.3.1 4-123. sin! = 23 and sin

2 ! + cos2 ! = 1!!"!!cos2 ! = 1# sin2 ! " ! cos! = # 1# sin2 !

cos! = " 1" 23( )2 = " 1" 49 = " 59 = " 53 sec! = 1cos! = 1" 5 3 = " 35 = " 3 55 tan! = sin!cos! =

23

" 5 3= " 23 #

35= " 2

5= " 2 55 cot! =

1tan! =

1"2 5 5

= " 52 5

= " 52

csc! = 1sin! =12 3 =

32

4-124. a. sin! = " 12 . The sine function is negative in Quadrants III and IV. sin! =

12 "! =

#6 is

the reference angle. sin! = " 12 #! =7$6 ,

11$6 .

b. cos! = 22 "! =#4 is the reference angle. The cosine function is positive in Quadrants I

and IV, so ! = "4 ,7"4 .

P(x) = 100 sin 2!23 x( )E(x) = 100 sin 2!28 x( )I(x) = 100 sin 2!33 x( )


c. tan! = 3"! = #3 is the reference angle. The tangent function is positive in Quadrants I and III, so ! = "3 ,

4"3 .

d. csc! = 2 " 1sin! = 2 " sin! =12= 22 "! =

#4 is the reference angle. The sine

function is positive in Quadrants I and II, so ! = "4 ,3"4 .


4-125. a. cos A + tan A sin A = cos A + sin Acos A sin A =

cos2 Acos A +

sin2 Acos A =

cos2 A+sin2 Acos A =

1cos A = sec A

b. sin2 ! + tan2 ! " cot2 ! + cos2 ! " sec2 ! + csc2 !

= sin2 ! + sin2 !cos2 !

" cos2 !sin2 !

+ cos2 ! " 1cos2 !

+ 1sin2 !

= sin2 ! + sin2 !"1cos2 !

+ cos2 ! + 1"cos2 !sin2 !

= 1+ " cos2 !cos2 !

+ sin2 !sin2 !

= 1"1+1 = 1

4-126. sec2 x + csc2 x = 1

cos2 x+ 1sin2 x

= sin2 xcos2 x sin2 x

+ cos2 xcos2 x sin2 x

= sin2 x+cos2 xcos2 x sin2 x

= 1cos2 x sin2 x

= 1cos2 x( ) 1sin2 x( ) = (sec2 x)(csc2 x)

4-127. a. Shift the sine function to the left !4 units, the amplitude is 2, the function is reflected over

the x-axis, and shifted up 1 unit: y = !2 sin x + "4( ) +1 b. Shift the cosine function to the right !4 units, the amplitude is 2, the function is reflected

over the x-axis, and shifted up 1 unit: y = !2 sin x ! "4( ) +1 4-128. a. The sine function is negative in Quadrant IV. The reference angle for sin 5!3 is ! =

"3 .

sin !3 =32 Therefore, sin

5!3 = "

32 .

b. The sine function is negative in Quadrant IV. The reference angle for cos 33!4 is ! ="4 .

cos !4 =22 Therefore, cos

33!4 =

22 .

c. cot! = cos!sin! Cosine is positive and sine is negative in Quadrant IV. Therefore cotangent will be negative. The reference angle is !6 . Therefore cot

!6( ) = cos(! /6)sin(! /6) = 3/21/2 = 3 , but

we are in Quadrant IV so cot ! "6( ) = ! 3 . d. sec! = 1cos! ,

3!2 corresponds to the point (0, –1), therefore cos

3!2 = 0 and sec

3!2 =

10 ,

which is undefined. 4-129. Let the distance from the tree to Rocky be d and the distance from Rocky to the tip of his

shadow be x , then tan! = 20+td+x and tan! =6x . Solve the latter equation for x: x =

6tan! .

Substitute this into the first equation: tan! = 20+td+ 6tan!

" d + 6tan! =20+ttan!

d = 20+ttan! #6tan! =

14+ttan!


4-130. a. If AT = BT = r, then !ATB is a right triangle and AB! = 90º .

The portion of the string that goes around the circle is 90360 !2" r = 1.5r . Therefore the total length of the string is 1.5! r + 2r " 6.712r .

b. Recall that tangent lines to circles are perpendicular to the radius. See diagram at right. The portion of the string that goes around the circle is 240360 !2" r =

4"3 r . Therefore the total length

of the string is 4! r3 + 2r 3 " 7.653r c. m!ATB =

12 "120

! = 60! Lesson 4.3.2 4-131. a. and b. See graph at right. c. The parent graph can either be y = sin x or y = cos x . d. Yes, the period is 2! . e. We know g(x) = a cos(x ! b) because

g(x) = sin x + cos x is a shift of the cosine function with a larger amplitude. The smallest positive value of b is !4 .

f. a ! 1.4 4-132. Yes, the bold graph confirms the graph drawn without a calculator. 4-133. It is also sinusoidal and it has the same period, but it has a greater amplitude. 4-134. It is also periodic, but it does not look sinusoidal. It has two small crests between each

trough. 4-135. The graph looks like a diagonal sine curve. 4-136. a. The amplitude is greater for large absolute values of x and decreases as x gets closer to

zero. b. y = x sin x is bounded by y = x and y = !x . The graph is also symmetric about the y-axis.

A

T

B 60º 60º

r 3

r 3

r r


Review and Preview 4.3.2 4-137. a. cos !5"6( ) = cos 5"6( ) = ! 32 b. tan 31!4( ) = tan 7!4( ) = sin(7! 4)cos(7! 4) = " 2 22 2 = "1 4-138. a. cos! = 1 when ! = 0, 2" . b. tan! = "1 when sin!cos! = "1# sin! = " cos! #! =

3$4 ,

7$4

c. csc! = 2 when 1sin! = 2"12 = sin! "! =

#6 ,

5#6

d. sec! = 2 when 1cos! = 2 or cos! =12= 22 . The cosine function is positive in

Quadrants I and IV: ! = "4 ,7"4 .

4-139. a. csc x ! sin x = cot x cos x

1sin x ! sin x =

cos xsin x " cos x

1!sin2 xsin x =

cos2 xsin x

cos2 xsin x =

cos2 xsin x

b. sec x + tan x = 1sec x! tan x1

cos x +sin xcos x =

11

cos x !sin xcos x

1+sin xcos x =

cos x1!sin x

1+sin xcos x =

cos x 1+sin x( )1!sin2 x

1+sin xcos x =

1+sin xcos x

.

4-140. 1

1+ tan2 x+ 11+cot2 x

= 1+cot2 x(1+ tan2 x)(1+cot2 x)

+ 1+ tan2 x(1+ tan2 x)(1+cot2 x)

= 1+cot2 x+1+ tan2 x(1+ tan2 x)(1+cot2 x)

= 2+cot2 x+ tan2 x(1+ tan2 x)(1+cot2 x)

= 2+csc2 x!1+sec2 x!1(sec2 x)(csc2 x)

= csc2 x+sec2 x(sec2 x)(csc2 x)

= csc2 x(sec2 x)(csc2 x)

+ sec2 x(sec2 x)(csc2 x)

= 1sec2 x

+ 1csc2 x

= cos2 x + sin2 x = 1

4-141. hx =

10x+5 ! h =

10xx+5

4-142. a. Starting from the cosine function, the amplitude is 2, the period is 4! , and the function is

shifted down 1 unit: y = 2 cos x2( ) !1 . b. Starting from the sine function, the amplitude is 2, the period is 2! , the function is

reflected over the x-axis, and shifted up 1 unit: y = !2 sin x +1 .


4-143. The parachute will be deployed after 10 seconds. The height will be represented by the

area under the curve v = 30t between t = 0 and t = 10 seconds. The curve maps out a triangle with base 10 and height 30(10) = 300 . The area is: 12 !10 ! 300 = 1500 ft .

Closure Merge Problems 4-144. 1. a. sin! = BD1 = BD

b. cos! = PD1 = PD 2. !A and ! are both complementary to !APB . 3. !ABP ! !PDB because of angle-angle similarity. 4. a. sin A = 1PA

b. csc A = 1sin A = PA

c. tan A = sin Acos A =1PAABPA

= 1AB

d. cot A = 1tan A = AB 5. Both !PBC and !PDC have angles !, 90

!, and ! " 90! , therefore they are similar. 6. a. tan! = BC1 = BC

b. cos! = 1PC

c. sec! = 1cos! =1

1/PC = PC

sin! = BDcos! = PDtan! = BCcsc! = PAsec! = PCcot! = AB

4-145. a. From !PDB we get: PD2 + BD2 = 1

sin2 ! + cos2 ! = 1

b. From !PBC we get: 12 + BC2 = PC2

1+ tan2 ! = sec2 !


c. From !ABP we get: 12 + AB2 = AP2

1+ cot2 ! = csc2 !


4-146.

(cot! + tan!)2 = cot2 ! + 2 cot! tan! + tan2 ! = csc2 ! "1+ 2 cot! tan! + sec2 ! "1

= csc2 ! "1+ 2 1tan! tan! + sec2 ! "1 = csc2 ! "1+ 2 + sec2 ! "1 = csc2 ! + sec2 !

This is the Pythagorean theorem applied to !PAC . 4-147. a. Juan travels the distance of the circumference of his tire: 2! (13) = 26! inches. This is the

period. b. Because the reflector is farthest from the ground at time t = 0 , we use the cosine function.

The amplitude is 11 and the function is shifted up 13 units (the radius of the tire). The period is 26! . h(x) = 11cos x13( ) +13 .

c. After riding 20ft ! 12 in.1 ft. = 240 in. h(240) = 11cos24013( ) +13 = 23.182 in.

d. Juan travels 29.33 feet per second or 29.33 ftsec !12 in1 ft " 352

insec , so h(t) = 11cos

35213 t( ) +13 .

e. After 5 minutes, or 300 seconds, h(300) = 11 cos 35213 ! 300( ) +13 = 18.168 in. Closure Problems CL 4-148. a. sin 5!3( ) has a reference angle of sin !3( ) = 32 . 5!3 is in Quadrant IV where the sine

function is negative, so sin 5!3( ) = " 32 b. cos ! "4( ) has a reference angle of cos !4( ) = 22 . ! "4 is in Quadrant IV where the cosine

function is positive, so cos ! "4( ) = 22 . c. csc 5!6( ) has a reference angle of csc !6( ) = 1sin(! 6) = 11 2 = 2 . 5!6 is in Quadrant II where

the sine and cosecant functions are positive, so csc 5!6( ) = 2 . d. tan !2( ) = sin(! 2)cos(! 2) . sin !2( ) = 1 and cos !2( ) = 0 so tan !2( ) = sin(! 2)cos(! 2) = 10 is undefined. e. cot 5!3( ) = cos(5! 3)sin(5! 3) = 1 2" 3 2 = " 12 # 23 = " 13 = " 33 f. sec 7!6( ) has a reference angle of sec !6( ) = 1cos(! 6) = 13 2 = 23 = 2 33 . 7!6 is in Quadrant

III where the cosine and secant functions are negative, so sec 7!6( ) = " 2 33 . CL 4-149. a. Starting with sin! , the amplitude is 2, the function is shifted up 2 units, and the function is

shifted right !2 units: y = 2 sin ! "#2( ) + 2 .


b. Starting with cos! , the amplitude is 2, the function is shifted up 2 units, and the function is reflected over the x-axis: y = !2 cos" + 2 .


CL 4-150. sec x!cos xsin x =

1 cos x!cos xsin x =

1sin x "

1cos x ! cos x( ) = 1sin x cos x ! cos xsin x = 1sin x cos x ! cos2 xsin x cos x

= 1!cos2 xsin x cos x =sin2 x

sin x cos x =sin xcos x = tan x

CL 4-151. a. b. c. CL 4-152. a. tan A(csc A ! sin A) = tan A csc A ! tan A sin A = sin Acos A "

1sin A !

sin Acos A " sin A =

1cos A !

sin2 Acos A

= 1!sin2 Acos A =cos2 Acos A = cos A

b. (csc x + cot x)(1! cos x) = csc x ! cos x csc x + cot x ! cot x cos x

= 1sin x ! cos x "1sin x +

1tan x !

1tan x " cos x =

1sin x !

cos xsin x +

cos xsin x !

cos2 xsin x

= 1sin x !cos2 xsin x =

1!cos2 xsin x =

sin2 xsin x = sin x

CL 4-153.

sec! = 54 " cos! =45

csc! < 0" sin! < 0 sin2 ! + cos2 ! = 1" sin2 ! + 45( )2 = 1

sin2 ! = 1# 45( )2 = 925sin! = # 35

cot! = cos!sin! =4 53 5 =

45 "

53 = #

43


CL 4-154. sin2(2x + 3) + cos2(2x + 3) = 2x + 3 Let u = 2x + 3 then we have: sin2(u) + cos2(u) = u

1 = u Substitute back in: 1 = 2x + 3

!2 = 2x!1 = x

CL 4-155. a. cos!1"sin! +

cos!1+sin! =

cos! (1+sin! )(1"sin! )(1+sin! ) +

cos! (1"sin! )(1+sin! )(1"sin! ) =

cos! (1+sin! )+cos! (1"sin! )(1"sin! )(1+sin! )

= cos! ((1+sin! )+(1"sin! ))(1"sin! )(1+sin! ) =2 cos!1"sin2 !

= 2 cos!cos2 !

= 2cos! = 2 sec!

b. 1tan x+cot x =1

sin xcos x +

cos xsin x

= sin x cos xsin2 x+cos2 x

= sin x cos x

CL 4-156. The period is 4 seconds, the amplitude is 12 !10 = 5 and since the peak is at 8 feet, the

function is shifted up 3 units: h = 5 cos !2 t( ) + 3 CL 4-157. CL 4-158. a. b.

c. h(x) =(x ! 3)3 + 4, 2 " x " 45, x > 4

#$%&

y = !x + cos x


CL 4-159.

12 !2 +x2( )3 +1"# $%"#&

$%'

x=0

9

( = 12.813 using left-endpoint rectangles.

12 !2 +x2( )3 +1"# $%"#&

$%'

x=1

10

( = 30.313 using right-endpoint rectangles.

Midpoint rectangles = 12.813+30.3132 = 21.094 CL 4-160. logb M

2NP = logb M

2 + logb N ! logb P= 2 logb M + logb N ! logb P= 2(3.2) !1.5 ! 2.4 = 2.5

Chapter 4: Circular Functionsachsprecalc.weebly.com/uploads/1/9/8/2/19823981/cpm_pre... · 2019. 8....

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Transcript of Chapter 4: Circular Functionsachsprecalc.weebly.com/uploads/1/9/8/2/19823981/cpm_pre... · 2019. 8....