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# 2.7: Graphing Tangent, Cotangent, Secant, and Cosecant

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Apply transformations to the remaining four trigonometric functions.
• Identify the equation, given a basic graph.

## Tangent and Cotangent

From Section 2.3, the graph of tangent looks like the picture below, where the period is π\begin{align*}\pi\end{align*} and vertical asymptotes are at 2πn±π2\begin{align*}2\pi n \pm \frac{\pi}{2}\end{align*} and 2πn±3π2\begin{align*}2\pi n \pm \frac{3\pi}{2}\end{align*}, where n\begin{align*}n\end{align*} is any integer. Notice that the period is only π\begin{align*}\pi\end{align*} and the function repeats after every asymptote. The x\begin{align*}x-\end{align*}intercepts are ,π,0,π,2π,\begin{align*}\ldots, -\pi, 0, \pi, 2\pi, \ldots\end{align*} The general equation of a tangent function is just like sine and cosine, f(x)=D±Atan(B(x±C))\begin{align*}f(x)=D \pm A \tan (B(x \pm C))\end{align*}, where A\begin{align*}A\end{align*}, B\begin{align*}B\end{align*}, C\begin{align*}C\end{align*}, and D\begin{align*}D\end{align*} represent the same transformations as they did before.

Cotangent also has a period of π\begin{align*}\pi\end{align*}, but the asymptotes and x\begin{align*}x-\end{align*}intercepts are reversed. What this means is that the vertical asymptotes are now at 0 and ±nπ\begin{align*}\pm n \pi\end{align*}, and the x\begin{align*}x-\end{align*}intercepts are at 2πn±π2\begin{align*}2 \pi n \pm \frac{\pi}{2}\end{align*} and 2πn±3π2\begin{align*}2 \pi n \pm \frac{3\pi}{2}\end{align*}, where n\begin{align*}n\end{align*} is an integer. The general equation of a cotangent function is just like sine and cosine, f(x)=D±Acot(B(x±C))\begin{align*}f(x)=D \pm A \cot (B(x \pm C))\end{align*}, where A\begin{align*}A\end{align*}, B\begin{align*}B\end{align*}, C\begin{align*}C\end{align*}, and D\begin{align*}D\end{align*} represent the same transformations as they did before.

One important difference: the period of sine and cosine is defined as 2πB\begin{align*}\frac{2\pi}{B}\end{align*}. The period of tangent and cotangent is only π\begin{align*}\pi\end{align*}, so the period would be πB\begin{align*}\frac{\pi}{B}\end{align*}.

Example 1: Sketch the graph of g(x)=2+cot13x\begin{align*}g(x)=-2+ \cot \frac{1}{3}x\end{align*} over the interval [0,6π]\begin{align*}[0, 6\pi]\end{align*}.

Solution: Starting with y=cotx\begin{align*}y=\cot x\end{align*}, g(x)\begin{align*}g(x)\end{align*} would be shifted down two and frequency is 13\begin{align*}\frac{1}{3}\end{align*}, which means the period would be 3π\begin{align*}3\pi\end{align*}, instead of π\begin{align*}\pi\end{align*}. So, in our interval of [0,6π]\begin{align*}[0, 6\pi]\end{align*} there would be two complete repetitions. The red graph is y=cotx\begin{align*}y=\cot x\end{align*} .

Example 2: Sketch the graph of y=3tan(xπ4)\begin{align*}y=-3 \tan \left(x-\frac{\pi}{4}\right)\end{align*} over the interval [π,2π]\begin{align*}[-\pi, 2\pi]\end{align*}.

Solution: If you compare this graph to y=tanx\begin{align*}y=\tan x\end{align*}, it will be stretched and flipped. It will also have a phase shift of π4\begin{align*}\frac{\pi}{4}\end{align*} to the right. The red graph is y=tanx\begin{align*}y=\tan x\end{align*}.

## Secant and Cosecant

Because secant is the reciprocal of cosine, it will have the same period, 2π\begin{align*}2\pi\end{align*}. Notice that an entire period encompasses an upward \begin{align*}\bigcup\end{align*} and downward \begin{align*}\bigcap\end{align*} and the asymptote between them. There are no x\begin{align*}x-\end{align*}intercepts and only one y\begin{align*}y-\end{align*}intercept at (0,1)\begin{align*}(0, 1)\end{align*}. The vertical asymptotes are everywhere cosine is zero, so πn±π2\begin{align*}\pi n \pm \frac{\pi}{2}\end{align*} and πn±3π2\begin{align*}\pi n \pm \frac{3\pi}{2}\end{align*}, where n\begin{align*}n\end{align*} is any integer. The general equation of a secant function is just like the others, \begin{align*}f(x)=D \pm A \sec (B(x \pm C))\end{align*}, where \begin{align*}A\end{align*}, \begin{align*}B\end{align*}, \begin{align*}C\end{align*}, and \begin{align*}D\end{align*} represent the same transformations as they did before.

The cosecant is the reciprocal of sine and it has the same period, \begin{align*}2\pi\end{align*}. Notice that an entire period encompasses an upward \begin{align*}\bigcup\end{align*} and downward \begin{align*}\bigcap\end{align*} and the asymptote between them, just like secant. There are no \begin{align*}x-\end{align*}intercepts and no \begin{align*}y-\end{align*}intercepts. The vertical asymptotes are everywhere sine is zero, so \begin{align*}\pm n \pi\end{align*}, where \begin{align*}n\end{align*} is any integer. The general equation of a cosecant function is just like the others, \begin{align*}f(x)=D \pm A \csc (B(x \pm C))\end{align*}, where \begin{align*}A\end{align*}, \begin{align*}B\end{align*}, \begin{align*}C\end{align*}, and \begin{align*}D\end{align*} represent the same transformations as they did before.

Recall that the period of sine and cosine is defined as \begin{align*}\frac{2\pi}{B}\end{align*}. The period of secant and cosecant will also be defined this way.

Example 3: Sketch a graph of \begin{align*}h(x)=5+\frac{1}{2} \sec 4x\end{align*} over the interval \begin{align*}[0, 2\pi]\end{align*}.

Solution: If you compare this example to \begin{align*}f(x)=\sec x\end{align*}, it will be translated 5 units up, with an amplitude of \begin{align*}\frac{1}{2}\end{align*} and a frequency of 4. This means in our interval of 0 to \begin{align*}2\pi\end{align*}, there will be 4 secant curves.

## Graphing Calculator Note

For the two examples above, it might seem difficult to graph these on a graphing calculator. Most graphing calculators do not have sec, csc, or cot buttons. However, we do know that these three functions are reciprocals of cosine, sine, and tangent, respectively. So, you must enter them into the calculator in this way. For example, the equation \begin{align*}f(x)=2+3 \csc \left(\frac{3}{4} (x-2) \right)\end{align*} would be entered like \begin{align*}2+\frac{3}{\sin \left(\frac{3}{4}(x-2)\right)}\end{align*} in the \begin{align*}y =\end{align*} menu.

Find the Equation from a Graph

For tangent, cotangent, secant, and cosecant it can be difficult to determine the equation from a graph, so to simplify this section amplitude changes will not be included.

Example 4: Find the equation of the graph below.

Solution: From the graph, we can see this is tangent. Usually tangent intercepts the origin, but here it intercepts at \begin{align*}(0, 2)\end{align*}. Therefore, we know that there is no horizontal shift and the vertical shift is up 2. Because we have eliminated amplitude from this section, the only thing left to find is the period. Normally, the period of tangent is \begin{align*}\pi\end{align*}, but as you can see from the graph, there are three curves from \begin{align*}[0, \pi]\end{align*}. So, the frequency is 3. The equation is \begin{align*}y=2+ \tan 3x\end{align*}.

Example 5: Find the equation for the graph below.

Solution: First of all, this could be either a secant or cosecant function. Let’s say this is a secant function. Secant usually intersects the \begin{align*}y-\end{align*}axis at \begin{align*}(0, 1)\end{align*} at a minimum. Now, that corresponding minimum is \begin{align*}\left(\frac{\pi}{2}, -2\right)\end{align*}. Because there is no amplitude change, we can say that the vertical shift is the difference between the two \begin{align*}y-\end{align*}values, -3. It looks like there is a phase shift and a period change. From minimum to minimum is one period, which is \begin{align*}\frac{9\pi}{2}-\frac{\pi}{2}=\frac{8\pi}{2}=4\pi\end{align*} and \begin{align*}B=\frac{2\pi}{4\pi}=\frac{1}{2}\end{align*}. Lastly, we need to find the horizontal shift. Since secant usually intersects the \begin{align*}y-\end{align*}axis at \begin{align*}(0, 1)\end{align*} at a minimum, and now the corresponding minimum is \begin{align*}\left(\frac{\pi}{2}, -2\right)\end{align*}, we can say that the horizontal shift is the difference between the two \begin{align*}x-\end{align*}values, \begin{align*}\frac{\pi}{2}\end{align*}. Therefore, our equation is \begin{align*}f(x)=-3+ \sec \left(\frac{1}{2}(x-\frac{\pi}{2} \right))\end{align*}.

## Points to Consider

• How can you shift or change tangent to make it look like cotangent?
• Are secant and cosecant “out of phase” like sine and cosine?
• Why do tangent and cotangent have a different period than the other four trig functions?

## Review Questions

For questions 1-6, graph the following functions. Determine the amplitude, period, frequency, phase shift and vertical translation.

1. \begin{align*}y=-1+\frac{1}{3} \cot 2x\end{align*}
2. \begin{align*}g(x)=5 \csc \left( \frac{1}{4}(x+\pi \right))\end{align*}
3. \begin{align*}f(x)=4+ \tan (0.5 (x - \pi))\end{align*}
4. \begin{align*}y=-2+\frac{1}{2} \sec(4(x-1))\end{align*}
5. \begin{align*}y=-2 \tan 2x\end{align*}
6. \begin{align*}h(x)=-\cot \frac{1}{3}x +1\end{align*}
7. We know that sine and cosine (and secant and cosecant) would be the same graph with a shift of \begin{align*}\frac{\pi}{2}\end{align*}. How can we manipulate the graph of \begin{align*}y=\cot x\end{align*} to match up with \begin{align*}y=\tan x\end{align*}?

For problems 8 and 9, determine the equation of the trig functions below. All amplitudes are 1.

## Review Answers

1. \begin{align*}y=-1+\frac{1}{3} \cot 2x\end{align*}
2. \begin{align*}g(x)=5 \csc \left( \frac{1}{4}(x+\pi) \right)\end{align*}
3. \begin{align*}f(x)=4+ \tan (0.5(x - \pi))\end{align*}
4. \begin{align*}y=-2+\frac{1}{2} \sec (4(x-1))\end{align*}
5. \begin{align*}y=-2 \tan 2x\end{align*}
6. \begin{align*}h(x)=-\cot \left(\frac{1}{3} x \right) +1\end{align*}
7. To make cotangent match up with tangent, it is helpful to graph the two on the same set of axis. First, cotangent needs to be flipped, which would make the amplitude of -1. Once cotangent is flipped, it also needs a phase shift of \begin{align*}\frac{\pi}{2}\end{align*}. So, \begin{align*}\tan x=-\cot \left(x-\frac{\pi}{2}\right)\end{align*}.
8. This is a tangent graph. From the two points we are given, we can determine the phase shift, vertical shift and frequency. There is no phase shift, the vertical shift is 3 and the frequency is 6. \begin{align*}y=3+ \tan 6x\end{align*}
9. This could be either a secant or cosecant function. We will use a cosecant model. First, the vertical shift is -1. The period is the difference between the two given \begin{align*}x-\end{align*}values, \begin{align*}\frac{7\pi}{4}-\frac{3\pi}{4}=\pi\end{align*}, so the frequency is \begin{align*}\frac{2\pi}{\pi}=2\end{align*}. The horizontal shift incorporates the frequency, so in \begin{align*}y=\csc x\end{align*} the corresponding \begin{align*}x-\end{align*}value to \begin{align*}\left(\frac{3\pi}{4}, 0\right)\end{align*} is \begin{align*}\left(\frac{\pi}{2}, 1\right)\end{align*}. The difference between the \begin{align*}x-\end{align*}values is \begin{align*}\frac{3\pi}{4}-\frac{\pi}{2}=\frac{3\pi}{4}-\frac{2\pi}{4}=\frac{\pi}{4}\end{align*} and then multiply it by the frequency, \begin{align*}2 \cdot \frac{\pi}{4}=\frac{\pi}{2}\end{align*}. The equation is \begin{align*}y=-1+ \csc \left(2(x-\frac{\pi}{2}\right))\end{align*}.

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CK.MAT.ENG.SE.2.Trigonometry.2.7