Graphs of the Other Trigonometric Functions
Analyzing the Graph of y = tan x and Its Variations
We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that[latex]\begin{array} \tan(−x)=\frac{\sin(−x)}{\cos(−x)} \hfill& \text{Definition of tangent.} \\ =\frac{−\sin x}{\cos x} \hfill& \text{Sine is an odd function, cosine is even.} \\ =−\frac{\sin x}{\cos x} &\hfill \text{The quotient of an odd and an even function is odd.} \hfill \\ =−\tan x \hfill& \text{Definition of tangent.} \end{array}\\[/latex]
Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in the table below.x | [latex]−\frac{\pi}{2}\\[/latex] | [latex]−\frac{\pi}{3}\\[/latex] | [latex]−\frac{\pi}{4}\\[/latex] | [latex]−\frac{\pi}{6}\\[/latex] | 0 | [latex]\frac{\pi}{6}\\[/latex] | [latex]\frac{\pi}{4}\\[/latex] | [latex]\frac{\pi}{3}\\[/latex] | [latex]\frac{\pi}{2}\\[/latex] |
tan (x) | undefined | [latex]−\sqrt{3}\\[/latex] | –1 | [latex]−\frac{\sqrt{3}}{3}\\[/latex] | 0 | [latex]\frac{\sqrt{3}}{3}\\[/latex] | 1 | [latex]\sqrt{3}\\[/latex] | undefined |
x | 1.3 | 1.5 | 1.55 | 1.56 |
tan x | 3.6 | 14.1 | 48.1 | 92.6 |
x | −1.3 | −1.5 | −1.55 | −1.56 |
tan x | −3.6 | −14.1 | −48.1 | −92.6 |
Graphing Variations of y = tan x
As with the sine and cosine functions, the tangent function can be described by a general equation.A General Note: Features of the Graph of y = Atan(Bx)
- The stretching factor is |A| .
- The period is [latex]P=\frac{\pi}{|B|}\\[/latex].
- The domain is all real numbers x, where [latex]x\ne \frac{\pi}{2|B|} + \frac{\pi}{|B|} k\\[/latex] such that k is an integer.
- The range is (−∞, ∞).
- The asymptotes occur at [latex]x=\frac{\pi}{2|B|} + \frac{\pi}{|B|}k\\[/latex], where k is an integer.
- [latex]y = A \tan (Bx)\\[/latex] is an odd function.
Graphing One Period of a Stretched or Compressed Tangent Function
We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/or compressed tangent function of the form [latex]f(x)=A\tan(Bx)\\[/latex]. We focus on a single period of the function including the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we wish. Our limited domain is then the interval [latex](−\frac{P}{2}, \frac{P}{2})\\[/latex] and the graph has vertical asymptotes at [latex]\pm \frac{P}{2}\\[/latex] where [latex]P=\frac{\pi}{B}\\[/latex]. On [latex](−\frac{\pi}{2}, \frac{\pi}{2})\\[/latex], the graph will come up from the left asymptote at [latex]x=−\frac{\pi}{2}\\[/latex], cross through the origin, and continue to increase as it approaches the right asymptote at [latex]x=\frac{\pi}{2}\\[/latex]. To make the function approach the asymptotes at the correct rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can useHow To: Given the function [latex]f(x)=A\tan(Bx)\\[/latex], graph one period.
- Identify the stretching factor, |A|.
- Identify B and determine the period, [latex]P=\frac{\pi}{|B|}\\[/latex].
- Draw vertical asymptotes at [latex]x=−\frac{P}{2}\\[/latex] and [latex]x=\frac{P}{2}\\[/latex].
- For A > 0 , the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for A < 0 ).
- Plot reference points at [latex]\left(\frac{P}{4}\text{, }A\right)\\[/latex] (0, 0), and ([latex]−\frac{P}{4}\\[/latex],− A), and draw the graph through these points.
Example 1: Sketching a Compressed Tangent
Sketch a graph of one period of the function [latex]y=0.5\tan(\frac{\pi}{2}x)\\[/latex].Solution
First, we identify A and B.[latex]\begin{array}f(0.5)=0.5\tan(\frac{0.5\pi}{2}) \hfill& \\ =0.5\tan(\frac{\pi}{4}) \hfill& \\ =0.5 \end{array}\\[/latex]
This means the curve must pass through the points(0.5,0.5),(0,0),and(−0.5,−0.5).The only inflection point is at the origin. Figure shows the graph of one period of the function.Try It 1
Sketch a graph of [latex]f(x)=3\tan\left(\frac{\pi}{6}x\right)\\[/latex]. SolutionGraphing One Period of a Shifted Tangent Function
Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add C and D to the general form of the tangent function.A General Note: Features of the Graph of y = Atan(Bx−C)+D
- The stretching factor is |A|.
- The period is [latex]\frac{\pi}{|B|}\\[/latex].
- The domain is [latex]x\ne\frac{C}{B}+\frac{\pi}{|B|}k\\[/latex], where k is an integer.
- The range is (−∞,−|A|] ∪ [|A|, ∞).
- The vertical asymptotes occur at [latex]x=\frac{C}{B}+\frac{\pi}{2|B|}k\\[/latex], where k is an odd integer.
- There is no amplitude.
- [latex]y=A\tan(Bx)[/latex] is an odd function because it is the quotient of odd and even functions (sin and cosine perspectively).
How To: Given the function [latex]y=A\tan(Bx−C)+D\\[/latex], sketch the graph of one period.
- Express the function given in the form [latex]y=A\tan(Bx−C)+D\\[/latex].
- Identify the stretching/compressing factor, |A|.
- Identify B and determine the period, [latex]P=\frac{\pi}{|B|}\\[/latex].
- Identify C and determine the phase shift, [latex]\frac{C}{B}\\[/latex].
- Draw the graph of [latex]y=A\tan(Bx)\\[/latex] shifted to the right by [latex]\frac{C}{B}\\[/latex] and up by D.
- Sketch the vertical asymptotes, which occur at [latex]x=\frac{C}{B}+\frac{\pi}{2|B|}k\\[/latex], where k is an odd integer.
- Plot any three reference points and draw the graph through these points.
Example 2: Graphing One Period of a Shifted Tangent Function
Graph one period of the function [latex]y=−2\tan(\pi x+\pi)−1\\[/latex].Solution
Step 1. The function is already written in the form [latex]y=A\tan(Bx−C)+D\\[/latex]. Step 2. [latex]A=−2[/latex], so the stretching factor is [latex]|A|=2[/latex]. Step 3. [latex]B=\pi[/latex], so the period is [latex]P=\frac{\pi}{|B|}=\frac{\pi}{\pi}=1\\[/latex]. Step 4. [latex]C=−\pi[/latex], so the phase shift is [latex]\frac{C}{B}=\frac{−\pi}{\pi}=−1\\[/latex]. Step 5–7. The asymptotes are at [latex]x=−\frac{3}{2}\\[/latex] and [latex]x=−\frac{1}{2}\\[/latex] and the three recommended reference points are (−1.25, 1), (−1,−1), and (−0.75, −3). The graph is shown in Figure 4.Analysis of the Solution
Note that this is a decreasing function because A < 0.Try It 2
How would the graph in Example 2 look different if we made A = 2 instead of −2 ? SolutionHow To: Given the graph of a tangent function, identify horizontal and vertical stretches.
- Find the period P from the spacing between successive vertical asymptotes or x-intercepts.
- Write [latex]f(x)=A\tan(\frac{\pi}{P}x)[/latex].
- Determine a convenient point (x, f(x)) on the given graph and use it to determine A.
Example 3: Identifying the Graph of a Stretched Tangent
Find a formula for the function graphed in Figure 5.Solution
The graph has the shape of a tangent function. Step 1. One cycle extends from –4 to 4, so the period is [latex]P=8[/latex]. Since [latex]P=\frac{\pi}{|B|}\\[/latex], we have [latex]B=\frac{\pi}{P}=\frac{\pi}{8}\\[/latex]. Step 2. The equation must have the [latex]\text{form}f(x)=A\tan(\frac{\pi}{8}x)\\[/latex]. Step 3. To find the vertical stretch A, we can use the point (2,2).[latex]2=A\tan(\frac{\pi}{8}\times2)=A\tan(\frac{\pi}{4})\\[/latex]
Because [latex]\tan(\frac{\pi}{4})=1\\[/latex], A = 2. This function would have a formula [latex]f(x)=2\tan(\frac{\pi}{8}x)\\[/latex].Using the Graphs of Trigonometric Functions to Solve Real-World Problems
Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function .Example 4: Using Trigonometric Functions to Solve Real-World Scenarios
Suppose the function [latex]y=5\tan\left(\frac{\pi}{4}t\right)\\[/latex] marks the distance in the movement of a light beam from the top of a police car across a wall where t is the time in seconds and y is the distance in feet from a point on the wall directly across from the police car.- Find and interpret the stretching factor and period.
- Graph on the interval [0, 5].
- Evaluate f(1) and discuss the function’s value at that input.
Solution
- We know from the general form of [latex]y=A\tan(Bt)\\[/latex] that |A| is the stretching factor and π B is the period. We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period. The period is [latex]\frac{\pi}{\frac{\pi}{4}}=\frac{\pi}{1}\times \frac{4}{\pi}=4\\[/latex]. This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.
- To graph the function, we draw an asymptote at [latex]t=2[/latex] and use the stretching factor and period. See Figure 8.
- period: [latex]f(1)=5\tan \left(\frac{\pi}{4}\left(1\right)\right)=5\left(1\right)=5\\[/latex]; after 1 second, the beam of has moved 5 ft from the spot across from the police car.
Analyzing the Graphs of y = sec x and y = cscx and Their Variations
The secant was defined by the reciprocal identity [latex]\sec x=\frac{1}{\cos x}[/latex]. Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at [latex]\frac{\pi}{2}\text{, }\frac{3\pi}{2}\text{, etc}[/latex]. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value. We can graph [latex]y=\sec x[/latex] by observing the graph of the cosine function because these two functions are reciprocals of one another. See Figure 9. The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined. The secant graph has vertical asymptotes at each value of x where the cosine graph crosses the x-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant. Note that, because cosine is an even function, secant is also an even function. That is, [latex]\sec(−x)=\sec x[/latex]. As we did for the tangent function, we will again refer to the constant |A| as the stretching factor, not the amplitude.A General Note: Features of the Graph of y = Asec(Bx)
- The stretching factor is |A|.
- The period is [latex]\frac{2\pi}{|B|}[/latex].
- The domain is [latex]x\ne \frac{\pi}{2|B|}k[/latex], where k is an odd integer.
- The range is (−∞, −|A|] ∪ [|A|, ∞).
- The vertical asymptotes occur at [latex]x=\frac{\pi}{2|B|}k [/latex], where k is an odd integer.
- There is no amplitude.
- [latex]y=A\sec(Bx)[/latex] is an even function because cosine is an even function.
A General Note: Features of the Graph of [latex]y=A\csc(Bx)
- The stretching factor is |A|.
- The period is [latex]\frac{2\pi}{|B|}[/latex].
- The domain is [latex]x\ne\frac{\pi}{|B|}k[/latex], where k is an integer.
- The range is ( −∞, −|A|] ∪ [|A|, ∞).
- The asymptotes occur at [latex]x=\frac{\pi}{|B|}k[/latex], where k is an integer.
- [latex]y=A\csc(Bx)[/latex] is an odd function because sine is an odd function.
Graphing Variations of y = sec x and y = csc x
For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions. The equations become the following.A General Note: Features of the Graph of [latex]y=A\sec(Bx−C)+D[/latex]
- The stretching factor is |A|.
- The period is [latex]\frac{2\pi}{|B|}[/latex].
- The domain is [latex]x\ne \frac{C}{B}+\frac{\pi}{2|B|}k[/latex], where k is an odd integer.
- The range is (−∞, −|A|] ∪ [|A|, ∞).
- The vertical asymptotes occur at [latex]x=\frac{C}{B}+\frac{\pi}{2|B|}k[/latex], where k is an odd integer.
- There is no amplitude.
- [latex]y=A\sec(Bx)[/latex] is an even function because cosine is an even function.
A General Note: Features of the Graph of [latex]y=A\csc(Bx−C)+D[/latex]
- The stretching factor is |A|.
- The period is [latex]\frac{2\pi}{|B|}[/latex].
- The domain is [latex]x\ne\frac{C}{B}+\frac{\pi}{2|B|}k[/latex], where k is an integer.
- The range is (−∞, −|A|] ∪ [|A|, ∞).
- The vertical asymptotes occur at [latex]x=\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where k is an integer.
- There is no amplitude.
- [latex]y=A\csc(Bx)[/latex] is an odd function because sine is an odd function.
How To: Given a function of the form [latex]y=A\sec(Bx)[/latex], graph one period.
- Express the function given in the form [latex]y=A\sec(Bx)[/latex].
- Identify the stretching/compressing factor, |A|.
- Identify B and determine the period, [latex]P=\frac{2\pi}{|B|}[/latex].
- Sketch the graph of [latex]y=A\cos(Bx)[/latex].
- Use the reciprocal relationship between [latex]y=\cos x[/latex] and [latex]y=\sec x[/latex] to draw the graph of [latex]y=A\sec(Bx)[/latex].
- Sketch the asymptotes.
- Plot any two reference points and draw the graph through these points.
Example 6: Graphing a Variation of the Secant Function
Graph one period of [latex]f(x)=2.5\sec(0.4x)[/latex].Solution
Step 1. The given function is already written in the general form, [latex]y=A\sec(Bx)[/latex]. Step 2. [latex]A=2.5[/latex] so the stretching factor is 2.5. Step 3. [latex]B=0.4[/latex], so [latex]P=\frac{2\pi}{0.4}=5\pi[/latex]. The period is 5π units. Step 4. Sketch the graph of the function [latex]g(x)=2.5\cos(0.4x)[/latex]. Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function. Steps 6–7. Sketch two asymptotes at [latex]x=1.25\pi[/latex] and [latex]x=3.75\pi[/latex]. We can use two reference points, the local minimum at (0, 2.5) and the local maximum at (2.5π, −2.5). Figure 11 shows the graph.Try It 4
Graph one period of [latex]f(x)=−2.5\sec(0.4x)[/latex]. SolutionQ & A
Do the vertical shift and stretch/compression affect the secant’s range?
Yes. The range of f(x) = A sec(Bx − C) + D is ( −∞, −|A| + D] ∪ [|A| + D, ∞).How To: Given a function of the form [latex]f(x)=A\sec (Bx−C)+D[/latex], graph one period.
- Express the function given in the form [latex]y=A\sec(Bx−C)+D[/latex].
- Identify the stretching/compressing factor, |A|.
- Identify B and determine the period, [latex]\frac{2\pi}{|B|}[/latex].
- Identify C and determine the phase shift, [latex]\frac{C}{B}[/latex].
- Draw the graph of [latex]y=A\sec(Bx)[/latex]. but shift it to the right by [latex]\frac{C}{B}[/latex] and up by D.
- Sketch the vertical asymptotes, which occur at [latex]x=\frac{C}{B}+\frac{\pi}{2|B|}k[/latex], where k is an odd integer.
Example 7: Graphing a Variation of the Secant Function
Graph one period of [latex]y=4\sec \left(\frac{\pi}{3}x−\frac{\pi}{2}\right)+1[/latex].Solution
Step 1. Express the function given in the form [latex]y=4\sec \left(\frac{\pi}{3}x−\frac{\pi}{2}\right)+1[/latex]. Step 2. The stretching/compressing factor is |A| = 4. Step 3. The period is[latex]\begin{array} \frac{2\pi}{|B|}=\frac{2\pi}{\frac{\pi}{3}} \hfill& \\ =\frac{2\pi}{1}\times\frac{3}{\pi} \hfill& \\ =6 \end{array}[/latex]
Step 4. The phase shift is[latex]\begin{array}\frac{C}{B}=\frac{\frac{\pi}{2}}{\frac{\pi}{3}} \hfill& \\ =\frac{\pi}{2} \times \frac{3}{\pi} \hfill& \\ =1.5 \end{array}[/latex]
Step 5. Draw the graph of [latex]y=A\sec(Bx)[/latex],but shift it to the right by [latex]\frac{C}{B}=1.5[/latex] and up by D = 6. Step 6. Sketch the vertical asymptotes, which occur at x = 0, x = 3, and x = 6. There is a local minimum at (1.5, 5) and a local maximum at (4.5, −3). Figure 12 shows the graph.Try It 5
Graph one period of [latex]f(x)=−6\sec(4x+2)−8[/latex]. SolutionQ & A
The domain of csc x was given to be all x such that [latex]x\ne k\pi[/latex] for any integer k. Would the domain of [latex]y=A\csc(Bx−C)+D[/latex] be [latex]x\ne\frac{C+k\pi}{B}[/latex]?
Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input.How To: Given a function of the form [latex]y=A\csc(Bx)[/latex], graph one period.
- Express the function given in the form [latex]y=A\csc(Bx)[/latex].
- |A|.
- Identify B and determine the period, [latex]P=\frac{2\pi}{|B|}[/latex].
- Draw the graph of [latex]y=A\sin(Bx)[/latex].
- Use the reciprocal relationship between [latex]y=\sin x[/latex] and [latex]y=\csc x[/latex] to draw the graph of [latex]y=A\csc(Bx) [/latex].
- Sketch the asymptotes.
- Plot any two reference points and draw the graph through these points.
Example 8: Graphing a Variation of the Cosecant Function
Graph one period of [latex]f(x)=−3\csc(4x)[/latex].Solution
Step 1. The given function is already written in the general form, [latex]y=A\csc(Bx)[/latex]. Step 2. [latex]|A|=|−3|=3[/latex], so the stretching factor is 3. Step 3. [latex]B=4\text{, so}P=\frac{2\pi}{4}=\frac{\pi}{2}[/latex].The period is [latex]\frac{\pi}{2}[/latex] units. Step 4. Sketch the graph of the function [latex]g(x)=−3\sin(4x)[/latex]. Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function. Steps 6–7. Sketch three asymptotes at [latex]x=0\text{, }x=\frac{\pi}{4}\text{, and }x=\frac{\pi}{2}[/latex].We can use two reference points, the local maximum at [latex]\left(\frac{\pi}{8}\text{, }−3\right)[/latex] and the local minimum at [latex]\left(\frac{3\pi}{8}\text{, }3\right)[/latex]. Figure 13 shows the graph.Try It 6
Graph one period of [latex]f(x)=0.5\csc(2x)[/latex]. SolutionHow To: Given a function of the form [latex]f(x)=A\csc(Bx−C)+D[/latex], graph one period.
- Express the function given in the form [latex]y=A\csc(Bx−C)+D[/latex].
- Identify the stretching/compressing factor, |A|.
- Identify B and determine the period, [latex]\frac{2\pi}{|B|}[/latex].
- Identify C and determine the phase shift, [latex]\frac{C}{B}[/latex].
- Draw the graph of [latex]y=A\csc(Bx)[/latex] but shift it to the right by and up by D.
- Sketch the vertical asymptotes, which occur at [latex]x=\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where k is an integer.
Example 9: Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant
Sketch a graph of [latex]y=2\csc\left(\frac{\pi}{2}x\right)+1[/latex]. What are the domain and range of this function?Solution
Step 1. Express the function given in the form [latex]y=2\csc\left(\frac{\pi}{2}x\right)+1[/latex]. Step 2. Identify the stretching/compressing factor, [latex]|A|=2[/latex]. Step 3. The period is [latex]\frac{2\pi}{|B|}=\frac{2\pi}{\frac{\pi}{2}}=\frac{2\pi}{1}\times \frac{2}{\pi}=4[/latex]. Step 4. The phase shift is [latex]\frac{0}{\frac{\pi}{2}}=0[/latex]. Step 5. Draw the graph of [latex]y=A\csc(Bx)[/latex] but shift it up [latex]D=1[/latex]. Step 6. Sketch the vertical asymptotes, which occur at x = 0, x = 2, x = 4. The graph for this function is shown in Figure 14.Analysis of the Solution
The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of [latex]f(x)=2\sin\left(\frac{\pi}{2}x\right)+1[/latex], shown as the orange dashed wave.Try It 7
Given the graph of [latex]f(x)=2\cos\left(\frac{\pi}{2}x\right)+1[/latex] shown in Figure 15, sketch the graph of [latex]g(x)=2\sec\left(\frac{\pi}{2}x\right)+1[/latex] on the same axes.Analyzing the Graph of y = cot x and Its Variations
The last trigonometric function we need to explore is cotangent. The cotangent is defined by the reciprocal identity [latex]\cot x=\frac{1}{\tan x}[/latex]. Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers. We can graph [latex]y=\cot x[/latex] by observing the graph of the tangent function because these two functions are reciprocals of one another. See Figure 16. Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases. The cotangent graph has vertical asymptotes at each value of x where [latex]\tan x=0[/latex]; we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent, [latex]\cot x[/latex] has vertical asymptotes at all values of x where [latex]\tan x=0[/latex] , and [latex]\cot x=0[/latex] at all values of x where tan x has its vertical asymptotes.A General Note: Features of the Graph of y = Acot(Bx)
- The stretching factor is |A|.
- The period is [latex]P=\frac{\pi}{|B|}[/latex].
- The domain is [latex]x\ne\frac{\pi}{|B|}k[/latex], where k is an integer.
- The range is (−∞, ∞).
- The asymptotes occur at [latex]x=\frac{\pi}{|B|}k[/latex], where k is an integer.
- [latex]y=A\cot(Bx)[/latex] is an odd function.
Graphing Variations of y = cot x
We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following.A General Note: Properties of the Graph of y = Acot(Bx−C)+D
- The stretching factor is |A|.
- The period is [latex]\frac{\pi}{|B|}[/latex].
- The domain is [latex]x\ne\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where k is an integer.
- The range is (−∞, −|A|] ∪ [|A|, ∞).
- The vertical asymptotes occur at [latex]x=\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where k is an integer.
- There is no amplitude.
- [latex]y=A\cot(Bx)[/latex] is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively)
How To: Given a modified cotangent function of the form [latex]f(x)=A\cot(Bx)[/latex], graph one period.
- Express the function in the form [latex]f(x)=A\cot(Bx)[/latex].
- Identify the stretching factor, |A|.
- Identify the period, [latex]P=\frac{\pi}{|B|}[/latex].
- Draw the graph of [latex]y=A\tan(Bx)[/latex].
- Plot any two reference points.
- Use the reciprocal relationship between tangent and cotangent to draw the graph of [latex]y=A\cot(Bx)[/latex].
- Sketch the asymptotes.
Example 10: Graphing Variations of the Cotangent Function
Determine the stretching factor, period, and phase shift of [latex]y=3\cot(4x)[/latex], and then sketch a graph.Solution
Step 1. Expressing the function in the form [latex]f(x)=A\cot(Bx)[/latex] gives [latex]f(x)=3\cot(4x)[/latex]. Step 2. The stretching factor is [latex]|A|=3[/latex]. Step 3. The period is [latex]P=\frac{\pi}{4}[/latex]. Step 4. Sketch the graph of [latex]y=3\tan(4x)[/latex]. Step 5. Plot two reference points. Two such points are [latex]\left(\frac{\pi}{16}\text{, }3\right)[/latex] and [latex]\left(\frac{3\pi}{16}\text{, }−3\right)[/latex]. Step 6. Use the reciprocal relationship to draw [latex]y=3\cot(4x)[/latex]. Step 7. Sketch the asymptotes, [latex]x=0[/latex], [latex]x=\frac{\pi}{4}[/latex]. The orange graph in Figure 17 shows [latex]y=3\tan(4x)[/latex] and the blue graph shows [latex]y=3\cot(4x)[/latex].How To: Given a modified cotangent function of the form [latex]f(x)=A\cot(Bx−C)+D[/latex], graph one period.
- Express the function in the form [latex]f(x)=A\cot(Bx−C)+D[/latex].
- Identify the stretching factor, |A|.
- Identify the period, [latex]P=\frac{\pi}{|B|}[/latex].
- Identify the phase shift, [latex]\frac{C}{B}[/latex].
- Draw the graph of [latex]y=A\tan(Bx)[/latex] shifted to the right by [latex]\frac{C}{B}[/latex] and up by D.
- Sketch the asymptotes [latex]x =\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where k is an integer.
- Plot any three reference points and draw the graph through these points.
Example 11: Graphing a Modified Cotangent
Sketch a graph of one period of the function [latex]f(x)=4\cot(\frac{\pi}{8}x−\frac{\pi}{2})−2[/latex].Solution
Step 1. The function is already written in the general form [latex]f(x)=A\cot(Bx−C)+D[/latex]. Step 2. [latex]A=4[/latex], so the stretching factor is 4. Step 3. [latex]B=\frac{\pi}{8}[/latex], so the period is [latex]P=\frac{\pi}{|B|}=\frac{\pi}{\frac{\pi}{8}}=8[/latex]. Step 4. [latex]C=\frac{\pi}{2}[/latex], so the phase shift is [latex]\frac{C}{B}=\frac{\frac{\pi}{2}}{\frac{\pi}{8}}=4[/latex]. Step 5. We draw [latex]f(x)=4\tan\left(\frac{\pi}{8}x−\frac{\pi}{2}\right)−2[/latex]. Step 6-7. Three points we can use to guide the graph are (6,2), (8,−2), and (10,−6). We use the reciprocal relationship of tangent and cotangent to draw [latex]f(x)=4\cot(\frac{\pi}{8}x−\frac{\pi}{2})−2[/latex]. Step 8. The vertical asymptotes are [latex]x=4[/latex] and [latex]x=12[/latex]. The graph is shown in Figure 18.Key Equations
Shifted, compressed, and/or stretched tangent function | [latex]y=A\tan(Bx−C)+D[/latex] |
Shifted, compressed, and/or stretched secant function | [latex]y=A\sec(Bx−C)+D[/latex] |
Shifted, compressed, and/or stretched cosecant | [latex]y=A\csc(Bx−C)+D[/latex] |
Shifted, compressed, and/or stretched cotangent function | [latex]y=A\cot(Bx−C)+D[/latex] |
Key Concepts
- The tangent function has period π.
- [latex]f(x)=A\tan(Bx−C)+D[/latex] is a tangent with vertical and/or horizontal stretch/compression and shift.
- The secant and cosecant are both periodic functions with a period of2π. [latex]f(x)=A\sec(Bx−C)+D[/latex] gives a shifted, compressed, and/or stretched secant function graph.
- [latex]f(x)=A\csc(Bx−C)+D[/latex] gives a shifted, compressed, and/or stretched cosecant function graph.
- The cotangent function has period π and vertical asymptotes at 0, ±π,±2π,....
- The range of cotangent is (−∞,∞),and the function is decreasing at each point in its range.
- The cotangent is zero at [latex]\ne\frac{\pi}{2}\text{, }\ne\frac{3\pi}{2}[/latex],....
- [latex]f(x)=A\cot(Bx−C)+D[/latex] is a cotangent with vertical and/or horizontal stretch/compression and shift.
- Real-world scenarios can be solved using graphs of trigonometric functions.
Section Exercises
1. Explain how the graph of the sine function can be used to graph [latex]y=\csc x[/latex]. 2. How can the graph of [latex]y=\cos x[/latex] be used to construct the graph of [latex]y=\sec x[/latex]? 3. Explain why the period of [latex]\tan x[/latex] is equal to π. 4. Why are there no intercepts on the graph of [latex]y=\csc x[/latex]? 5. How does the period of [latex]y=\csc x[/latex] compare with the period of [latex]y=\sin x[/latex]? For the following exercises, match each trigonometric function with one of the following graphs. 6. [latex]f(x)=\tan x[/latex] 7. [latex]f(x)=\sec x[/latex] 8. [latex]f(x)=\csc x[/latex] 9. [latex]f(x)=\cot x[/latex] For the following exercises, find the period and horizontal shift of each of the functions. 10. [latex]f(x)=2\tan(4x−32)[/latex] 11. [latex]h(x)=2\sec\left(\frac{\pi}{4}(x+1)\right)[/latex] 12. [latex]m(x)=6\csc\left(\frac{\pi}{3}x+\pi\right)[/latex] 13. If tan x = −1.5, find tan(−x). 14. If sec x = 2, find sec(−x). 15. If csc x = −5, find csc(−x). 16. If [latex]x\sin x=2[/latex], find [latex](−x)\sin(−x)[/latex]. For the following exercises, rewrite each expression such that the argument x is positive. 17. [latex]\cot(−x)\cos(−x)+\sin(−x)[/latex] 18. [latex]\cos(−x)+\tan(−x)\sin(−x)[/latex] For the following exercises, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes. 19. [latex]f(x)=2\tan(4x−32)[/latex] 20. [latex]h(x)=2\sec\left(\frac{\pi}{4}\left(x+1\right)\right)[/latex] 21. [latex]m(x)=6\csc\left(\frac{\pi}{3}x+\pi\right)[/latex] 22. [latex]j(x)=\tan\left(\frac{\pi}{2}x\right)[/latex] 23. [latex]p(x)=\tan\left(x−\frac{\pi}{2}\right)[/latex] 24. [latex]f(x)=4\tan(x)[/latex] 25. [latex]f(x)=\tan\left(x+\frac{\pi}{4}\right)[/latex] 26. [latex]f(x)=\pi\tan\left(\pi x−\pi\right)−\pi[/latex] 27. [latex]f(x)=2\csc(x)[/latex] 28. [latex]f(x)=−\frac{1}{4}\csc(x)[/latex] 29. [latex]f(x)=4\sec(3x)[/latex] 30. [latex]f(x)=−3\cot(2x)[/latex] 31. [latex]f(x)=7\sec(5x)[/latex] 32. [latex]f(x)=\frac{9}{10}\csc(\pi x)[/latex] 33. [latex]f(x)=2\csc \left(x+\frac{\pi}{4}\right)−1[/latex] 34. [latex]f(x)=−\sec \left(x−\frac{\pi}{3}\right)−2[/latex] 35. [latex]f(x)=\frac{7}{5}\csc \left(x−\frac{\pi}{4}\right)[/latex] 36. [latex]f(x)=5\left(\cot\left(x+\frac{\pi}{2}\right)−3\right)[/latex] For the following exercises, find and graph two periods of the periodic function with the given stretching factor, |A|, period, and phase shift. 37. A tangent curve, [latex]A=1[/latex], period of [latex]\frac{\pi}{3}[/latex]; and phase shift [latex](h\text{,}k)=\left(\frac{\pi}{4}\text{,}2\right)[/latex] 38. A tangent curve, [latex]A=−2[/latex], period of [latex]\frac{\pi}{4}[/latex], and phase shift [latex](h\text{,}k)=\left(−\frac{\pi}{4}\text{,}−2\right)[/latex] For the following exercises, find an equation for the graph of each function. 39. 40. 41. 42. 43. 44. 45. 47. 49. 50. Graph [latex]f(x)=1+\sec^{2}(x)−\tan^{2}(x)[/latex]. What is the function shown in the graph? 51. [latex]f(x)=\sec(0.001x)[/latex] 52. [latex]f(x)=\cot(100\pi x)[/latex] 53. [latex]f(x)=\sin^{2}x+\cos^{2}x[/latex] 54. The function [latex]f(x)=20\tan\left(\frac{\pi}{10}x\right)[/latex] marks the distance in the movement of a light beam from a police car across a wall for time x, in seconds, and distance [latex]f(x)[/latex], in feet.Licenses & Attributions
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