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2.11: Tangent and Cotangent Graphs

Difficulty Level: At Grade Created by: CK-12
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Practice Tangent and Cotangent Graphs

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What if your instructor gave you a set of graphs like these:

and asked you to identify which were the graphs of the tangent and cotangent functions?

After completing this Concept, you'll be able to identify the graphs of tangent and cotangent.

Guidance

The name of the tangent function comes from the tangent line of a circle. This is a line that is perpendicular to the radius at a point on the circle so that the line touches the circle at exactly one point.

If we extend angle θ\begin{align*}\theta\end{align*} through the unit circle so that it intersects with the tangent line, the tangent function is defined as the length of the red segment.

The dashed segment is 1 because it is the radius of the unit circle. Recall that the tanθ=yx\begin{align*}\tan \theta = \frac{y}{x}\end{align*}, and it can be verified that this segment is the tangent by using similar triangles.

tanθtanθ=yx=t1=t=t\begin{align*}\tan \theta &=\frac{y}{x}=\frac{t}{1}=t\\ \tan \theta &=t\end{align*}

So, as we increase the angle of rotation, think about how this segment changes. When the angle is zero, the segment has no length. As we rotate through the first quadrant, it will increase very slowly at first and then quickly get very close to one, but never actually touch it.

As we get very close to the y\begin{align*}y-\end{align*}axis the segment gets infinitely large, until when the angle really hits 90\begin{align*}90^\circ\end{align*}, at which point the extension of the angle and the tangent line will actually be parallel and therefore never intersect.

This means there is no finite length of the tangent segment, or the tangent segment is infinitely large.

Let’s translate this portion of the graph onto the coordinate plane. Plot (θ,tanθ)\begin{align*}(\theta, \tan \theta)\end{align*} as (x,y)\begin{align*}(x, y)\end{align*}.

In fact as we get infinitely close to 90\begin{align*}90^\circ\end{align*}, the tangent value increases without bound, until when we actually reach 90\begin{align*}90^\circ\end{align*}, at which point the tangent is undefined. Recall there are some angles (90\begin{align*}90^\circ\end{align*} and 270\begin{align*}270^\circ\end{align*}, for example) for which the tangent is not defined. Therefore, at these points, there are going to be vertical asymptotes.

Rotating past 90\begin{align*}90^\circ\end{align*}, the intersection of the extension of the angle and the tangent line is actually below the x\begin{align*}x-\end{align*}axis. This fits nicely with what we know about the tangent for a 2nd\begin{align*}2^{nd}\end{align*} quadrant angle being negative. At first, it will have very large negative values, but as the angle rotates, the segment gets shorter, reaches 0, then crosses back into the positive numbers as the angle enters the 3rd\begin{align*}3^{rd}\end{align*} quadrant. The segment will again get infinitely large as it approaches 270\begin{align*}270^\circ\end{align*}. After being undefined at 270\begin{align*}270^\circ\end{align*}, the angle crosses into the 4th\begin{align*}4^{th}\end{align*} quadrant and once again changes from being infinitely negative, to approaching zero as we complete a full rotation.

The graph y=tanx\begin{align*}y=\tan x\end{align*} over several rotations would look like this:

Notice the x\begin{align*}x-\end{align*}axis is measured in radians. Our asymptotes occur every π\begin{align*}\pi\end{align*} radians, starting at π2\begin{align*}\frac{\pi}{2}\end{align*}. The period of the graph is therefore π\begin{align*}\pi\end{align*} radians. The domain is all reals except for the asymptotes at π2,3π2,π2,etc.\begin{align*}\frac{\pi}{2}, \frac{3\pi}{2}, -\frac{\pi}{2}, etc.\end{align*} and the range is all real numbers.

Cotangent is the reciprocal of tangent, xy\begin{align*}\frac{x}{y}\end{align*}, so it would make sense that where ever the tangent had an asymptote, now the cotangent will be zero. The opposite of this is also true. When the tangent is zero, now the cotangent will have an asymptote. The shape of the curve is generally the same, so the graph looks like this:

When you overlap the two functions, notice that the graphs consistently intersect at 1 and -1. These are the angles that have 45\begin{align*}45^\circ\end{align*} as reference angles, which always have tangents and cotangents equal to 1 or -1. It makes sense that 1 and -1 are the only values for which a function and it’s reciprocal are the same. Keep this in mind as we look at cosecant and secant compared to their reciprocals of sine and cosine.

The cotangent function has a domain of all real angles except multiples of π{2π,π,0,π,2π}\begin{align*}\pi \left \{\ldots -2\pi, -\pi, 0, \pi, 2\pi \ldots \right \}\end{align*} The range is all real numbers.

Example A

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 B

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

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

Example C

Sketch the graph of \begin{align*}h(x)=4 \tan \left(x+\frac{\pi}{2}\right) + 3 \end{align*} over the interval \begin{align*}[0, 2\pi]\end{align*}.

Solution: The constant in front of the tangent function will cause the graph to be stretched. It will also have a phase shift of \begin{align*}\frac{\pi}{2}\end{align*} to the left. Finally, the graph will be shifted up three. Here you can see both graphs, where the red graph is \begin{align*}y=\tan x\end{align*}.

Vocabulary

Circular Function: A circular function is a function that is measured by examining the angle of rotation around the coordinate plane.

Guided Practice

1. Graph \begin{align*}y=-1+\frac{1}{3} \cot 2x\end{align*}.

2. Graph \begin{align*}f(x)=4+ \tan (0.5 (x - \pi))\end{align*}.

3. Graph \begin{align*}y=-2 \tan 2x\end{align*}.

Solutions:

1.

2.

3.

Concept Problem Solution

As you can tell after completing this Concept, when presented with the graphs:

1 2 3

4 5 6

The tangent and cotangent graphs are the third and sixth graphs.

Practice

Graph each of the following functions.

1. \begin{align*}f(x)=\tan(x)\end{align*}.
2. \begin{align*}h(x)=\tan(2x)\end{align*}.
3. \begin{align*}k(x)=\tan(2x+\pi)\end{align*}.
4. \begin{align*}m(x)=-\tan(2x+\pi)\end{align*}.
5. \begin{align*}g(x)=-\tan(2x+\pi)+3\end{align*}.
6. \begin{align*}f(x)=\cot(x)\end{align*}.
7. \begin{align*}h(x)=\cot(2x)\end{align*}.
8. \begin{align*}k(x)=\cot(2x+\pi)\end{align*}.
9. \begin{align*}m(x)=3\cot(2x+\pi)\end{align*}.
10. \begin{align*}g(x)=-2+3\cot(2x+\pi)\end{align*}.
11. \begin{align*}h(x)=\tan(\frac{x}{2})\end{align*}.
12. \begin{align*}k(x)=\tan(\frac{x}{2}+\frac{\pi}{4})\end{align*}.
13. \begin{align*}m(x)=3\tan(\frac{x}{2}+\frac{\pi}{4})\end{align*}.
14. \begin{align*}g(x)=3\tan(\frac{x}{2}+\frac{\pi}{4})-1\end{align*}.
15. \begin{align*}h(x)=\cot(\frac{x}{2})\end{align*}.
16. \begin{align*}k(x)=\cot(\frac{x}{2}+\frac{3\pi}{2})\end{align*}.
17. \begin{align*}m(x)=-3\cot(\frac{x}{2}+\frac{3\pi}{2})\end{align*}.
18. \begin{align*}g(x)=2-3\cot(\frac{x}{2}+\frac{3\pi}{2})\end{align*}.

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Color Highlighted Text Notes

Vocabulary Language: English

Circular Function

A circular function is a function measured by examining the angle of rotation around the coordinate plane.

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