Section 4.1 Solutions
1.
3. Whether the angle is positive or negative determines the direction. A positive angle is drawn in the counterclockwise direction, and a negative angle is drawn in the clockwise direction.
5. Linear speed is a measurement found by calculating distance of an arc compared to time. Angular speed is a measurement found by calculating the angle of an arc compared to time.
7.
9.
11.
13.
15.
17. 240°
19. 4π3
21. 2π3
23. 7π2≈11.00 in2
25. 9π5≈5.65 cm2
27. 20°
29. 60°
31. −75°
33. π2 radians
35. −3π radians
37. π radians
39. 5π6 radians
41. 154.795∘
43. 30.23∘
45. 2∘55′21″
47. 36∘52′12″
49. 5.02π3≈5.26 miles
51. 25π9≈8.73 centimeters
53. 21π10≈6.60 meters
55. 104.7198 cm2
57. 0.7697 in2
59. 250°
61. 320°
63. 4π3
65. 8π9
67. 1320 rad 210.085 RPM
69. 7 in./s, 4.77 RPM, 28.65 deg/s
71. 1,809,557.37 mm/min=30.16 m/s
73. 5.76 miles
75. 120∘
77. 794 miles per hour
79. 2,234 miles per hour
81. 11.5 inches
Section 4.2 Solutions
1. The unit circle is a circle of radius 1 centered at the origin.
3. Yes, when the reference angle is π4 and the terminal side of the angle is in quadrants I and III. Thus, at x=π4,5π4, the sine and cosine values are equal.
5. Substitute the sine of the angle in for y in the Pythagorean Theorem x2+y2=1. Solve for x and take the negative solution.
7. I
9. IV
11. √32 , 2√33
13. 12 , 2
15. √22 , √3
17. 0,√2
19. −1 , 0
21. 1 , 0
23. √779
25. −√154
27. sint=12,csct=2,cost=−√32,sect=−2√33,tant=−√33,cott=−√3
29. sint=−√22,csct=−√2,cost=−√22,sect=−√2,tant=1,cott=1
31. sint=√32,csct=2√33,cost=−12,sect=−2,tant=√3,cott=√33
33. sint=−√22,csct=−√2,cost=√22,sect=√2,tant=−1,cott=1
35. sint=0,csct=∅,cost=−1,sect=−1,tant=0,cott=∅
37. sint=−0.596,csct=−1.679,cost=0.803,sect=1.245,tant=−0.742,cott=−1.347
39. −0.1736
41. 0.9511
43. −0.7071
45. −0.1392
47. −0.7660
49. –0.228
51. –2.414
53. 1.556
55. √24
57. √24
59. 0
61. cos(6t)−sin(9t)
63. even
65. even
67. 13.77 hours, period: 1000π
69. 7.73 inches
Section 4.3 Solutions
1.
3. The tangent of an angle is the ratio of the opposite side to the adjacent side.
5. For example, the sine of an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement.
7. √2−44
9. 5
11. π6
13. π4
15. b=20√33,c=40√33
17. a=10,000,c=10,000.5
19. b=5√33,c=10√33
21. 5√2929
23. 52
25. √292
27. 5√4141
29. 54
31. √414
33. c=14,b=7√3
35. a=15,b=15
37. b=9.9970,c=12.2041
39. a=2.0838,b=11.8177
41. a=55.9808,c=57.9555
43. a=46.6790,b=17.9184
45. a=16.4662,c=16.8341
47. 188.3159
49. 200.6737
51. 498.3471 ft
53. 1060.09 ft
55. 27.372 ft
57. 22.6506 ft
59. 368.7633 ft
61. S29.05∘W
63. East: 13.49 inches, North: 33.38 inches
65. 18.3∘
Section 4.4 Solutions
1. Coterminal angles are angles that share the same terminal side. A reference angle is the size of the smallest acute angle, t, formed by the terminal side of the angle t and the horizontal axis.
3. The sine values are equal.
5. 60∘
7. 80∘
9. 45∘
11. π3
13. π3
15. π8
17. 60∘, Quadrant IV, sin(300∘)=−√32,cos(300∘)=12
19. 45∘, Quadrant II, sin(135∘)=√22, cos(135∘)=−√22
21. 60∘, Quadrant II, sin(480∘)=√32, cos(480∘)=−12
23. 30∘, Quadrant II, sin(−210∘)=12, cos(−210∘)=−√32
25. π6, Quadrant III, sin(7π6)=−12, cos(7π6)=−√32
27. π4, Quadrant II, sin(3π4)=√22, cos(4π3)=−√22
29. π3, Quadrant II, sin(2π3)=√32, cos(2π3)=−12
31. π4, Quadrant IV, sin(−9π4)=−√22, cos(−9π4)=√22
33. π6, Quadrant III, sec(7π6)=−2√33
35. π6, Quadrant I, cot(13π6)=√3
37. π4, Quadrant II, sec(3π4)=−√2
39. π4, Quadrant IV, cot(11π4)=−1
41. π3, Quadrant III, sec(−2π3)=−2
43. π3, Quadrant IV, cot(−7π3)=−√33
45. 60∘, Quadrant IV, sec(300∘)=2
47. 60∘, Quadrant III, cot(600∘)=√33
49. 30∘, Quadrant II, sec(−210∘)=−2√33
51. 45∘, Quadrant IV, cot(−405∘)=−1
53. If sint=−2√23,sect=−3,csct=−3√24,tant=2√2,cott=√24
55. sect=2,csct=2√33,tant=√3,cott=√33
57. √24
59. −√64
61. √24
63. √24
65. 0
Section 4.5 Solutions
1. The sine and cosine functions have the property that f(x+P)=f(x) for a certain P. This means that the function values repeat for every P units on the x-axis.
3. The absolute value of the constant A (amplitude) increases the total range and the constant D (vertical shift) shifts the graph vertically.
5. At the point where the terminal side of t intersects the unit circle, you can determine that the sin t equals the y-coordinate of the point.
7. amplitude: 23; period: 2π; midline: y=0; maximum: y=23 occurs at x=0; minimum: y=−23 occurs at x=π; for one period, the graph starts at 0 and ends at 2π
9. amplitude: 4; period: 2π; midline: y=0; maximum y=4 occurs at x=π2; minimum: y=−4 occurs at x=3π2; one full period occurs from x=0 to x=2π
11. amplitude: 1; period: π; midline: y=0; maximum: y=1 occurs at x=π; minimum: y=−1 occurs at x=π2; one full period is graphed from x=0 to x=π
13. amplitude: 4; period: 2; midline: y=0; maximum: y=4 occurs at x=0; minimum: y=−4 occurs at x=1
15. amplitude: 3; period: π4; midline: y=5; maximum: y=8 occurs at x=0.12; minimum: y=2 occurs at x=0.516; horizontal shift: −4; vertical translation 5; one period occurs from x=0 to x=π4
17. amplitude: 5; period: 2π5;midline:[latex]y=−2; maximum: y=3 occurs at x=0.08; minimum: y=−7 occurs at x=0.71; phase shift:−4; vertical translation:−2; one full period can be graphed on x=0 to x=2π5
19. amplitude: 1; period: 2π; midline: y=1; maximum:y=2 occurs at x=2.09; maximum:y=2 occurs att=2.09; minimum:y=0 occurs at t=5.24; phase shift: −π3; vertical translation: 1; one full period is from t=0 to t=2π
21. amplitude: 1; period: 4π; midline: y=0; maximum: y=1 occurs at t=11.52; minimum: y=−1 occurs at t=5.24; phase shift: −10π3; vertical shift: 0
23. amplitude: 2; midline: y=−3; period: 4; equation: f(x)=2sin(π2x)−3
25. amplitude: 2; period: 5; midline: y=3; equation: f(x)=−2cos(2π5x)+3
27. amplitude: 4; period: 2; midline: y=0; equation: f(x)=−4cos(π(x−π2))
29. amplitude: 2; period: 2; midline y=1; equation: f(x)=2cos(πx)+1
31. π6,5π6
33. π4,3π4
35. 3π2
37. π2,3π2
39. π2,3π2
41. π6,11π6
43. The graph appears linear. The linear functions dominate the shape of the graph for large values of x.
45. The graph is symmetric with respect to the y-axis and there is no amplitude because the function is not periodic.
47.
a. Amplitude: 12.5; period: 10; midline: y=13.5;
b. h(t)=12.5sin(π5(t−2.5))+13.5;
c. 26 ft
Section 4.6 Solutions
1. Since y=cscx is the reciprocal function of y=sinx, you can plot the reciprocal of the coordinates on the graph of y=sinx to obtain the y-coordinates of y=cscx. The x-intercepts of the graph y=sinx are the vertical asymptotes for the graph of y=cscx.
3. Answers will vary. Using the unit circle, one can show that tan(x+π)=tanx.
5. The period is the same: 2π.
7. IV
9. III
11. period: 8; horizontal shift: 1 unit to left
13. 1.5
15. 5
17. −cotxcosx−sinx
19. stretching factor: 2; period: π4; asymptotes: x=14(π2+πk)+8, where k is an integer
21. stretching factor: 6; period: 6; asymptotes: x=3k, where k is an integer
23. stretching factor: 1; period: π; asymptotes: x=πk, where k is an integer
25. Stretching factor: 1; period: π; asymptotes: x=π4+πk, where k is an integer
27. stretching factor: 2; period: 2π; asymptotes: x=πk, where k is an integer
29. stretching factor: 4; period: 2π3; asymptotes: x=π6k, where k is an odd integer
31. stretching factor: 7; period: 2π5; asymptotes: x=π10k, where k is an odd integer
33. stretching factor: 2; period: 2π; asymptotes: x=−π4+πk, where k is an integer
35. stretching factor: 75; period: 2π; asymptotes: x=π4+πk, where k is an integer
37. y=tan(3(x−π4))+2
39. f(x)=csc(2x)
41. f(x)=csc(4x)
43. f(x)=2cscx
45. f(x)=12tan(100πx)
For the following exercises, use a graphing calculator to graph two periods of the given function. Note: most graphing calculators do not have a cosecant button; therefore, you will need to input cscx as 1sinx.
46. f(x)=|csc(x)|
47. f(x)=|cot(x)|
48. f(x)=2csc(x)
49. f(x)=csc(x)sec(x)
51.
53.
55. a. (−π2,π2);
b.
c. x=−π2 and x=π2; the distance grows without bound as |x| approaches π2—i.e., at right angles to the line representing due north, the boat would be so far away, the fisherman could not see it;
d. 3; when x=−π3, the boat is 3 km away;
e. 1.73; when x=π6, the boat is about 1.73 km away;
f. 1.5 km; when x=0.
57. a. h(x)=2tan(π120x);
b.
c. h(0)=0: after 0 seconds, the rocket is 0 mi above the ground; h(30)=2: after 30 seconds, the rockets is 2 mi high;
d. As x approaches 60 seconds, the values of h(x) grow increasingly large. The distance to the rocket is growing so large that the camera can no longer track it.
Section 4.7 Solutions
1. The function y=sinx is one-to-one on [−π2, π2]; thus, this interval is the range of the inverse function of y=sinx, f(x)=sin−1x. The function y=cosx is one-to-one on [0,π]; thus, this interval is the range of the inverse function of y=cosx, f(x)=cos−1x.
3. π6 is the radian measure of an angle between −π2 and π2 whose sine is 0.5.
5. In order for any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval [−π2, π2] so that it is one-to-one and possesses an inverse.
7. True. The angle, θ1 that equals arccos(−x), x>0, will be a second quadrant angle with reference angle, θ2, where θ2 equals arccosx, x>0. Since θ2 is the reference angle for θ1, θ2=π(−x)=π−arccosx
9. −π6
11. 3π4
13. −π3
15. π3
17. 1.98
19. 0.93
21. 1.41
23. 0.56 radians
25. 0
27. 0.71
29. −0.71
31. −π4
33. 0.8
35. 513
37. x√1−x2
39. √x2−1x
41. 2x√4x2+1
43. √2x+1x+1
45. √2x+1x
47. x√2x+1
49. domain [−1,1]; range [0,π]
51. approximately x=0.00
53. 0.395 radians
55. 1.11 radians
57. 1.25 radians
59. 0.405 radians
61. No. The angle the ladder makes with the horizontal is 60 degrees.