2.E: Periodic Functions (Exercises)
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6.1: Graphs of the Sine and Cosine Functions
In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and create graphs of sine and cosine functions
Verbal
1) Why are the sine and cosine functions called periodic functions?
- Answer
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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.
2) How does the graph of y=sinx compare with the graph of y=cosx? Explain how you could horizontally translate the graph of y=sinx to obtain y=cosx.
3) For the equation Acos(Bx+C)+D
- Answer
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The absolute value of the constant A (amplitude) increases the total range and the constant D (vertical shift) shifts the graph vertically.
4) How does the range of a translated sine function relate to the equation y=Asin(Bx+C)+D
5) How can the unit circle be used to construct the graph of f(t)=sint?
- Answer
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At the point where the terminal side of t intersects the unit circle, you can determine that the sint equals the y-coordinate of the point.
Graphical
For the following exercises, graph two full periods of each function and state the amplitude, period, and midline. State the maximum and minimum y-values and their corresponding x-values on one period for x>0. Round answers to two decimal places if necessary.
6) f(x)=2sinx
7) f(x)=23cosx
- Answer
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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.;
8) f(x)=−3sinx
9) f(x)=4sinx
- Answer
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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;
10) f(x)=2cosx
11) f(x)=cos(2x)
- Answer
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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;
12) f(x)=2sin(12x)
13) f(x)=4cos(πx)
- Answer
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amplitude: 4; period: 2; midline: y=0
maximum: y=4 occurs at x=0; minimum: y=−4 occurs at x=1;
14) f(x)=3cos(65x)
15) y=3sin(8(x+4))+5
- Answer
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amplitude: 3; period: π4; midline: y=5;
maximum: y=8 occurs at x=−4+21π16≈0.123;
minimum: y=2 occurs at x=−4+23π16≈0.516;
horizontal shift: −4; vertical translation 5;
one period occurs from x=−4+22π16≈0.320 to x=−4+26π16≈1.105
16) y=2sin(3x−21)+4
17) y=5sin(5x+20)−2
- Answer
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amplitude: 5; period:2π5; midline: y=−2;
maximum: y=3 occurs at x=−4+13π10≈0.084;
minimum: y=−7 occurs at x=−4+15π10≈0.712;
phase shift: −4; vertical translation: −2;
one full period can be graphed on x=−4+7π5≈0.398 tox=−4+9π5≈1.655
For the following exercises, graph one full period of each function, starting at x=0.
For each function, state the amplitude, period, and midline.
State the maximum and minimum y-values and their corresponding x-values on one period for x>0.
State the phase shift and vertical translation, if applicable.
Round answers to two decimal places if necessary.
18) f(t)=2sin(t−5π6)
19) f(t)=−cos(t+π3)+1
- Answer
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amplitude: 1; period: 2π; midline: y=1;
maximum: y=2 occurs at t=2π3≈2.094;
minimum: y=0 occurs at t=2π3≈5.24;
phase shift: −π3; vertical translation: 1;
one full period is from t=2π3≈2.094 to t=8π3≈8.378
20) f(t)=4cos(2(t+π4))−3
21) f(t)=−sin(12t+5π3)
- Answer
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amplitude: 1; period: 4π; midline: y=0;
maximum: y=1 occurs at t=11π3≈11.52;
minimum: y=−1 occurs at t=5π3≈5.24;
phase shift: −10π3; vertical shift: 0;
one full period is from t=2π3≈2.094 to t=14π3≈14.661
22) f(x)=4sin(π2(x−3))+7
23) Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure below.

- Answer
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23. amplitude: 2; midline: y=−3 period: 4; equation: f(x)=2sin(π2x)−3
24) Determine the amplitude, midline, period, and an equation involving the cosine function for the graph shown in Figure below.

25) Determine the amplitude, midline, period, and an equation involving the cosine function for the graph shown in Figure below.

- Answer
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25. amplitude: 2; period: 5; midline: y=3 equation: f(x)=−2cos(2π5x)+3
26) Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure below.

27) Determine the amplitude, midline, period, and an equation involving the cosine function for the graph shown in Figure below.
- Answer
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27. amplitude: 4; period: 2; midline: y=0 ; equation: f(x)=−4cos(π(x−π2))
28) Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure below.

29) Determine the amplitude, midline, period, and an equation involving the cosine function for the graph shown in Figure below.

- Answer
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29. amplitude: 2; period: 2; midline y=1 equation: f(x)=2cos(πx)+1
30) Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure below.
Algebraic
For the following exercises, let f(x)=sinx
31) On [0,2π)
32) On [0,2π), solve f(x)=12.
- Answer
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π6, 5π6
33) Evaluate f(π2)
34) On [0,2π), f(x)=√22. Find all values of x.
- Answer
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π4, 3π4
35) On [0,2π)
36) On [0,2π)
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3π2
37) Show that f(−x)=−f(x)
For the following exercises, let f(x)=cosx
38) On [0,2π)
- Answer
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π2, 3π2
39) On [0,2π)
40) On [0,2π)
- Answer
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π2, 3π2
41) On [0,2π)
42) On [0,2π)
- Answer
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π6, 11π6
Technology
43) Graph h(x)=x+sinx on [0,2π]
44) Graph h(x)=x+sinx on [−100,100]
- Answer
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The graph appears linear. The linear functions dominate the shape of the graph for large values of x.
45) Graph f(x)=xsinx on [0,2π] and verbalize how the graph varies from the graph of f(x)=xsinx.
46) Graph f(x)=xsinx on the window [−10,10] and explain what the graph shows.
- Answer
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The graph is symmetric with respect to the y-axis and there is no amplitude because the function is not periodic.
47) Graph f(x)=sinxx on the window [−5π,5π] and explain what the graph shows.
Real-World Applications
48) A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. The function h(t) gives a person’s height in meters above the ground t minutes after the wheel begins to turn
- Find the amplitude, midline, and period of h(t).
- Find a formula for the height function h(t).
- How high off the ground is a person after 5 minutes?
- Answer
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- Amplitude: 12.5; period: 10; midline: y=13.5
- h(t)=12.5sin(π5(t−2.5))+13.5
- 26 ft
6.2: Graphs of the Other Trigonometric Functions
This section addresses the graphing of the Tangent, Cosecant, Secant, and Cotangent curves.
Verbal
1) Explain how the graph of the sine function can be used to graph y=cscx.
Answer
Since y=cscx is the reciprocal function of y=sinx
2) How can the graph of y=cosx be used to construct the graph of y=secx?
3) Explain why the period of y=tanx is equal to π.
Answer
Answers will vary. Using the unit circle, one can show that y=tan(x+π)=tanx.
4) Why are there no intercepts on the graph of y=cscx?
5) How does the period of y=cscx compare with the period of y=sinx?
Answer
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The period is the same: 2π
Algebraic
For the exercises 6-9, match each trigonometric function with one of the following graphs.
6) f(x)=tanx
7) f(x)=secx
- Answer
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IV
8) f(x)=cscx
9) f(x)=cotx
- Answer
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III
For the exercises 10-16, find the period and horizontal shift of each of the functions.
10) f(x)=2tan(4x−32)
11) h(x)=2sec(π4(x+1))
- Answer
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period: 8; horizontal shift: 1 unit to left
12) m(x)=6csc(π3x+π)
13) If tanx=−1.5
- Answer
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1.5
14) If secx=2, find sec(−x).
15) If cscx=−5, find csc(−x).
- Answer
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5
16) If xsinx=2, find (−x)sin(−x).
For the exercises 17-18, rewrite each expression such that the argument x is positive.
17) cot(−x)cos(−x)+sin(−x)
- Answer
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−cotxcosx−sinx
18) cos(−x)+tan(−x)sin(−x)
Graphical
For the exercises 19-36, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes.
19) f(x)=2tan(4x−32)
- Answer
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stretching factor: 2; period: π3
asymptotes: x=14(π2+πk)+8, where k is an integer;
20) h(x)=2sec(π4(x+1))
21) m(x)=6csc(π3x+π)
- Answer
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stretching factor: 6; period: 6; asymptotes: x=k, where k is an integer
22) j(x)=tan(π2x)
23) p(x)=tan(x−π2)
- Answer
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stretching factor: 1; period: π
asymptotes: x=πk, where k is an integer;
24) f(x)=4tan(x)
25) f(x)=tan(x+π4)
- Answer
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Stretching factor: 1; period: π
asymptotes: x=π4+πk, where k is an integer;
26) f(x)=πtan(πx−π)−π
27) f(x)=2csc(x)
- Answer
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stretching factor: 2; period: 2π
asymptotes: x=πk, where k is an integer;
28) f(x)=−14csc(x)
29) f(x)=4sec(3x)
- Answer
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stretching factor: 4; period: 2π3
asymptotes: x=π6k, where k is an odd integer;
30) f(x)=−3cot(2x)
31) f(x)=7sec(5x)
- Answer
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stretching factor: 7; period: 2π5
asymptotes: x=π10k, where k is an odd integer;
32) f(x)=910csc(πx)
33) f(x)=2csc(x+π4)−1
- Answer
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Stretching factor: 2; period: 2π ; asymptotes: x=−π4+πk, where k is an integer
34) f(x)=−sec(x−π3)−2
35) f(x)=75csc(x−π4)
- Answer
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Stretching factor: 75; period: 2π ; asymptotes: x=π4+πk, where k is an integer
36) f(x)=5(cot(x+π2)−3)
37) A tangent curve, A=1
- Answer
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y=tan(3(x−π4))+2
38) A tangent curve, A=−2, period of π4; and phase shift (h,k)=(−π4,−2)
For the exercises 39-45, find an equation for the graph of each function.
39)
- Answer
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f(x)=csc(2x)
40)
41)
- Answer
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f(x)=csc(4x)
42)
43)
- Answer
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f(x)=2cscx
44)
45)
- Answer
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f(x)=12tan(100πx)
Technology
For the exercises 46-53, 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)|
- Answer
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48) f(x)=2csc(x)
49) f(x)=csc(x)sec(x)
- Answer
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50) Graph f(x)=1+sec2(x)−tan2(x)
51) f(x)=sec(0.001x)
- Answer
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52) f(x)=cot(100πx)
53) f(x)=sin2x+cos2x
- Answer
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Real-World Applications
54) The function f(x)=20tan(π10x) marks the distance in the movement of a light beam from a police car across a wall for time x
- Graph on the interval [0,5]
- Find and interpret the stretching factor, period, and asymptote.
- Evaluate f(10) and f(2.5) and discuss the function’s values at those inputs.
55) Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let x
- What is a reasonable domain for d(x)?
- Graph d(x) on this domain.
- Find and discuss the meaning of any vertical asymptotes on the graph of d(x).
- Calculate and interpret d(−π3)
Round to the second decimal place.. - Calculate and interpret d(π6)
Round to the second decimal place.. - What is the minimum distance between the fisherman and the boat? When does this occur?
- Answer
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- (−π2,π2)
- 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;; - 3; when x=−π3
the boat is 3 km away;, - 1.73; when x=π6
the boat is about 1.73 km away;, - 1.5 km; when x=0
56) A laser rangefinder is locked on a comet approaching Earth. The distance g(x)
- Graph g(x) on the interval [0,35].
- Evaluate g(5) and interpret the information.
- What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond?
- Find and discuss the meaning of any vertical asymptotes.
57) A video camera is focused on a rocket on a launching pad 2 miles from the camera. The angle of elevation from the ground to the rocket after x seconds is π120x.
- Write a function expressing the altitude h(x)
in miles, of the rocket above the ground after x seconds. Ignore the curvature of the Earth., - Graph h(x) on the interval (0,60).
- Evaluate and interpret the values h(0) and h(30).
- What happens to the values of h(x) as x approaches 60 seconds? Interpret the meaning of this in terms of the problem.
- Answer
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- h(x)=2tan(π120x)
- 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.
6.3: Inverse Trigonometric Functions
In this section, we will explore the inverse trigonometric functions. Inverse trigonometric functions “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa.
Verbal
1) Why do the functions f(x)=sin−1x and g(x)=cos−1x have different ranges?
- Answer
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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;
2) Since the functions y=cosx and y=cos−1x are inverse functions, why is cos−1(cos(−π6))
3) Explain the meaning of π6=arcsin(0.5).
- Answer
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π6 is the radian measure of an angle between −π2 and π2 whose sine is 0.5.
4) Most calculators do not have a key to evaluate sec−1(2)
5) Why must the domain of the sine function, sinx
- Answer
-
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.
6) Discuss why this statement is incorrect: arccos(cosx)=x for all x.
7) Determine whether the following statement is true or false and explain your answer: arccos(−x)=π−arccosx
- Answer
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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−θ1 and arccos(−x)=π−arccosx−
Algebraic
For the exercises 8-16, evaluate the expressions.
8) sin−1(√22)
9) sin−1(−12)
- Answer
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−π6
10) cos−1(−12)
11) cos−1(−√22)
- Answer
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3π4
12) tan−1(1)
13) tan−1(−√3)
- Answer
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−π3
14) tan−1(−1)
15) tan−1(√3)
- Answer
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π3
16) tan−1(−1√3)
For the exercises 17-21, use a calculator to evaluate each expression. Express answers to the nearest hundredth.
17) cos−1(−0.4)
- Answer
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1.98
18) arcsin(0.23)
19) arccos(35)
- Answer
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0.93
20) cos−1(−0.8)
21) tan−1(6)
- Answer
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1.41
For the exercises 22-23, find the angle θ in the given right triangle. Round answers to the nearest hundredth.
22)
23)
- Answer
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0.56 radians
For the exercises 24-36, find the exact value, if possible, without a calculator. If it is not possible, explain why.
24) sin−1(cos(π))
25) tan−1(sin(π))
- Answer
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0
26) cos−1(sin(π3))
27) tan−1(sin(π3))
- Answer
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0.71
28) sin−1(cos(−π2))
29) tan−1(sin(4π3))
- Answer
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−0.71
30) sin−1(sin(5π6))
31) tan−1(sin(−5π2))
- Answer
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−π4
32) cos(sin−1(45))
33) sin(cos−1(35))
- Answer
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0.8
34) sin(tan−1(43))
35) cos(tan−1(125))
- Answer
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513
36) cos(sin−1(12))
For the exercises 37-41, find the exact value of the expression in terms of x with the help of a reference triangle.
37) tan(sin−1(x−1))
- Answer
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x−1√−x2+2x
38) sin(sin−1(1−x))
39) cos(sin−1(1x))
- Answer
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√x2−1x
40) cos(tan−1(3x−1))
41) tan(sin−1(x+12))
- Answer
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x+0.5√−x2−x+34
Extensions
For the exercise 42, evaluate the expression without using a calculator. Give the exact value.
2) sin−1(12)−cos−1(√22)+sin−1(√32)−cos−1(1)cos−1(√32)−sin−1(√22)+cos−1(12)−sin−1(0)
For the exercises 43-47, find the function if \sin t = \dfrac{x}{x+1}
43) \cos t
- Answer
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\dfrac{\sqrt{2x+1}}{x+1}
44) \sec t
45) \cot t
- Answer
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\dfrac{\sqrt{2x+1}}{x}
46) \cos \left(\sin^{-1} \left(\dfrac{x}{x+1}\right)\right)
47) \tan^{-1} \left(\dfrac{x}{\sqrt{2x+1}}\right)
- Answer
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t
Graphical
48) Graph y=\sin^{-1} x and state the domain and range of the function.
49) Graph y=\arccos x
- Answer
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domain [-1,1]
range [0,\pi ];
50) Graph one cycle of y=\tan^{-1} x and state the domain and range of the function.
51) For what value of x does \sin x=\sin^{-1} x? Use a graphing calculator to approximate the answer.
- Answer
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approximately x=0.00
52) For what value of x does \cos x=\cos^{-1} x? Use a graphing calculator to approximate the answer.
Real-World Applications
53) Suppose a 13-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 12 feet above the ground. What angle, in radians, does the ladder make with the building?
Answer
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0.395 radians
54) Suppose you drive 0.6 miles on a road so that the vertical distance changes from 0 to 150 feet. What is the angle of elevation of the road?
55) An isosceles triangle has two congruent sides of length 9 inches. The remaining side has a length of 8 inches. Find the angle that a side of 9 inches makes with the 8-inch side.
- Answer
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1.11 radians
56) Without using a calculator, approximate the value of \arctan (10,000)
57) A truss for the roof of a house is constructed from two identical right triangles. Each has a base of 12 feet and height of 4 feet. Find the measure of the acute angle adjacent to the 4-foot side.
- Answer
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1.25 radians
58) The line y=\dfrac{3}{5}x passes through the origin in the x,y-plane. What is the measure of the angle that the line makes with the positive x-axis?
59) The line y=\dfrac{-3}{7}x passes through the origin in the x,y-plane. What is the measure of the angle that the line makes with the negative x-axis?
- Answer
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0.405 radians
60) What percentage grade should a road have if the angle of elevation of the road is 4 degrees? (The percentage grade is defined as the change in the altitude of the road over a 100-foot horizontal distance. For example a 5\% grade means that the road rises 5 feet for every 100 feet of horizontal distance.)
61) A 20-foot ladder leans up against the side of a building so that the foot of the ladder is 10 feet from the base of the building. If specifications call for the ladder's angle of elevation to be between 35 and 45 degrees, does the placement of this ladder satisfy safety specifications?
- Answer
-
No. The angle the ladder makes with the horizontal is 60 degrees.
62) Suppose a 15-foot ladder leans against the side of a house so that the angle of elevation of the ladder is 42 degrees. How far is the foot of the ladder from the side of the house?
Contributors and Attributions
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.
Contributors and Attributions
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.