9.

Section 3.4 – Inverse Trigonometric Functions

1. arccos(0.8776)≈0.5

2.

ⓐ [latex]−\frac{π}{2}[/latex];

ⓑ [latex]−\frac{π}{4}[/latex];

ⓒ π;

ⓓ [latex]\frac{π}{3}[/latex]

3. 1.9823 or 113.578°

4. [latex]sin^{−1}(0.6)=36.87​∘=0.6435[/latex] radians

5. [latex]\frac{π}{8};\frac{2π}{9}[/latex]

6. [latex]\frac{3π}{4}[/latex]

7. [latex]\frac{12}{13}[/latex]

8. [latex]\frac{4\sqrt{2}}{9}[/latex]

9. [latex]\frac{4x}{\sqrt{16x^2+1}}[/latex]

Section Exercises

Section 3.1 – Graphs of the Sine and Cosine Functions

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: [latex]\frac{2}{3}[/latex]; period: 2π; midline: y=0; maximum: [latex]y=\frac{2}{3}[/latex] occurs at x=0; minimum: [latex]y=-\frac{2}{3}[/latex] 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 [latex]x=\frac{π}{2}[/latex]; minimum: y=−4 occurs at [latex]x=\frac{3π}{2}[/latex]; 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 [latex]x=\frac{π}{2}[/latex]; 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 [latex]\frac{π}{4}[/latex]; 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 [latex]x=\frac{π}{4}[/latex]

17.

amplitude: 5; period [latex]\frac{2π}{5}[/latex]; midline: 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=[latex]\frac{2π}{5}[/latex]

19.

amplitude: 1 ; period 2π; midline: y=1; maximum: y=2 occurs at x=2.09; maximum: y=2 occurs at t=2.09; minimum: y=0 occurs at t=5.24; phase shift: [latex]−\frac{π}{3}[/latex]; 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: [latex]-\frac{10π}{3}[/latex]; vertical shift: 0

23. amplitude: 2; midline: y=−3; period: 4; equation: [latex]f(x)=2sin(\frac{π}{2}x)−3[/latex]

25. amplitude: 2; period: 5; midline: y=3; equation: [latex]f(x)=−2cos(\frac{2π}{5}x)+3[/latex]

27. amplitude: 4; period: 2; midline: y=0; equation: [latex]f(x)=−4cos(π(x−\frac{π}{2})[/latex]

29. amplitude: 2; period: 2; midline: y=1; equation: [latex]f(x)=2cos(πx)+1[/latex]

31. 0,π

33. [latex]sin(\frac{π}{2})=1[/latex]

35. [latex]\frac{π}{2}[/latex]

37. f(x)=sin(x) is symmetric

39. [latex]\frac{π}{3}[/latex],[latex]\frac{5π}{3}[/latex]

41. Maximum: 1 at x=0 ; minimum: −1 at x=π

43. A linear function is added to a periodic sine function. The graph does not have an amplitude because as the linear function increases without bound the combined function h(x)=x+sin(x) will increase without bound as well. The graph is bounded between the graphs of y=x+1 and y=x−1 because sine oscillates between −1 and 1.

45. There is no amplitude because the function is not bounded.

47. The graph is symmetric with respect to the y-axis and there is no amplitude because the function’s bounds decrease as |x| grows. There appears to be a horizontal asymptote at y=0 .

Section 3.2 – Graphs of the Other Trigonometric Functions

1. Since y=csc x is the reciprocal function of y=sin x, you can plot the reciprocal of the coordinates on the graph of y=sin x to obtain the y-coordinates of y=csc x. The x-intercepts of the graph y=sin x are the vertical asymptotes for the graph of y=csc x.

3. Answers will vary. Using the unit circle, one can show that tan(x+π)=tan x.

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. −cot(x)cos(x)−sin(x)

19.

stretching factor: 2; period [latex]\frac{π}{4}[/latex]; asymptotes: [latex]x=\frac{1}{4}(\frac{π}{2}+πk)+8[/latex], 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: [latex]x=\frac{π}{4}+πk[/latex], where k is an integer

27.

stretching factor: 2; period 2π; asymptotes: x=πk, where k is an integer

29.

stretching factor: 4; period [latex]\frac{2π}{3}[/latex]; asymptotes: [latex]x=\frac{π}{6}k[/latex], where k is an odd integer

31.

stretching factor: 7; period [latex]\frac{2π}{5}[/latex]; asymptotes: [latex]x=\frac{π}{10}k[/latex], where k is an odd integer

33.

stretching factor: 2; period 2π; asymptotes: x=[latex]−\frac{π}{4}+πk[/latex], where k is an integer

35.

stretching factor [latex]\frac{7}{5}[/latex]; period: 2π; asymptotes: [latex]x=\frac{π}{4}+πk[/latex], where k is an integer

37. [latex]y=tan(3(x−\frac{π}{4}))+2[/latex]

39. f(x)=css(2x)

41. f(x)=csc(4x)

43. f(x)=2csc(x)

45. [latex]f(x)=\frac{1}{2}tan(100πx)[/latex]

47.

49.

51.

53.

55.

ⓐ([latex]−\frac{π}{2}[/latex], [latex]\frac{π}{2}[/latex]);

ⓑ

ⓒ [latex]x=−\frac{π}{2}[/latex] and [latex]x=\frac{π}{2}[/latex]; the distance grows without bound as |x| approaches [latex]\frac{π}{2}[/latex] —i.e., at right angles to the line representing due north, the boat would be so far away, the fisherman could not see it;

ⓓ3; when [latex]x=−\frac{π}{3}[/latex], the boat is 3 km away;

ⓔ 1.73; when [latex]x=\frac{π}{6}[/latex], the boat is about 1.73 km away;

ⓕ 1.5 km; when x=0

57.

ⓐ [latex]h(x)=2tan\frac{π}{120}x[/latex];

ⓑ

ⓒ 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;

ⓓ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 3.3 – Inverse Functions

1. Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that y -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no y -values repeat and the function is one-to-one.

3. Yes. For example, [latex]f(x)=\frac{1}{x}[/latex] is its own inverse.

5. Given a function y=f(x), solve for x in terms of y. Interchange the x and y. Solve the new equation for y. The expression for y is the inverse, [latex]y=f^{−1}(x)[/latex].

7. [latex]f^{−1}(x)=x−3[/latex]

9. [latex]f^{−1}(x)=2−x[/latex]

11. [latex]f^{−1}(x)=\frac{−2x}{x−1}[/latex]

13. domain of [latex]f(x): [−7,∞); f^{−1}(x)=\sqrt{x}−7[/latex]

15. domain of [latex]f(x): [0,∞); f^{−1}(x)=\sqrt{x+5}[/latex]

16.

• ⓐ f(g(x))=x and g(f(x))=x.
• ⓑ This tells us that f and g are inverse functions

17. f(g(x))=x, g(f(x))=x

19. one-to-one

21. one-to-one

23. not one-to-one

25. 3

27. 2

29.

31. [2,10]

33. 6

35. −4

37. 0

39. 1

41.

43. [latex]f^{−1}(x)=(1+x)^{\frac{1}{3}}[/latex]

45. [latex]f^{−1}(x)=\frac{5}{9}(x−32)[/latex]. Given the Fahrenheit temperature, x, this formula allows you to calculate the Celsius temperature.

47. [latex]t(d)=\frac{d}{50}, t(180)=\frac{180}{50}[/latex]. The time for the car to travel 180 miles is 3.6 hours.

Section 3.4 – Inverse Trigonometric Functions

1. The function y=sin(x) is one-to-one on [[latex]−\frac{π}{2}[/latex][latex],\frac{π}{2}[/latex]]; thus, this interval is the range of the inverse function of [latex]y=sin(x), f(x)=sin^{−1}x[/latex]. The function y=cos(x) is one-to-one on [0,π]; thus, this interval is the range of the inverse function of [latex]y=cos(x), f(x)=cos^{−1}x[/latex].

3. [latex]\frac{π}{6}[/latex] is the radian measure of an angle between [latex]−\frac{π}{2}[/latex] and [latex]\frac{π}{2}[/latex] 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 [[latex]−\frac{π}{2}[/latex],[latex]\frac{π}{2}[/latex]] so that it is one-to-one and possesses an inverse.

7. True . The angle, [latex]θ_1[/latex] that equals arccos(−x) , x>0 , will be a second quadrant angle with reference angle, [latex]θ_2[/latex] , where [latex]θ_2[/latex] equals arccosx , x>0 . Since [latex]θ2[/latex] is the reference angle for [latex]θ_1[/latex] , [latex]θ_2=π−θ_1[/latex] and arccos(−x) = π−arccosx

9. [latex]−\frac{π}{6}[/latex]

11. [latex]\frac{3π}{4}[/latex]

13. [latex]−\frac{π}{3}[/latex]

15. [latex]\frac{π}{3}[/latex]

17. 1.98

19. 0.93

21. 1.41

23. 0.56 radians

25. 0

27. 0.71

29. -0.71

31. [latex]−\frac{π}{4}[/latex]

33. 0.8

35. 513

37. [latex]\frac{x−1}{\sqrt{−x^2+2x}}[/latex]

39. [latex]\frac{\sqrt{x^2−1}}{x}[/latex]

41. [latex]\frac{x+0.5}{\sqrt{−x^2−x+\frac{3}{4}}}[/latex]

43. [latex]\frac{\sqrt{2x+1}}{x+1}[/latex]

45. [latex]\frac{\sqrt{2x+1}}{x}[/latex]

47. [latex]t[/latex]

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.