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?

### Tangent and Cotangent Graphs

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

Why? The dashed segment is 1 because it is the radius of the unit circle. Recall that in general,

As the value of

As we get ** very** close to the

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

In fact as we get infinitely close to

Rotating past

The graph

Notice the

Cotangent is the reciprocal of tangent,

When you overlap the two functions, notice that the graphs consistently intersect at 1 and -1. These are the angles that have

The cotangent function has a domain of all real angles except multiples of

#### Sketching Graphs

1. Sketch the graph of

Starting with

2. Sketch the graph of

If you compare this graph to

3. Sketch the graph of

The constant in front of the tangent function will cause the graph to be stretched. It will also have a phase shift of

### Examples

#### Example 1

Earlier, you were asked to identify which graphs are tangent and cotangent.

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

1 2 3

4 5 6

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

#### Example 2

Graph

#### Example 3

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

#### Example 4

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

### Review

Graph each of the following functions

- \begin{align*}f(x)=\tan(x)\end{align*}
- \begin{align*}h(x)=\tan(2x)\end{align*}
- \begin{align*}k(x)=\tan(2x+\pi)\end{align*}
- \begin{align*}m(x)=-\tan(2x+\pi)\end{align*}
- \begin{align*}g(x)=-\tan(2x+\pi)+3\end{align*}
- \begin{align*}f(x)=\cot(x)\end{align*}
- \begin{align*}h(x)=\cot(2x)\end{align*}
- \begin{align*}k(x)=\cot(2x+\pi)\end{align*}
- \begin{align*}m(x)=3\cot(2x+\pi)\end{align*}
- \begin{align*}g(x)=-2+3\cot(2x+\pi)\end{align*}
- \begin{align*}h(x)=\tan(\frac{x}{2})\end{align*}
- \begin{align*}k(x)=\tan(\frac{x}{2}+\frac{\pi}{4})\end{align*}
- \begin{align*}m(x)=3\tan(\frac{x}{2}+\frac{\pi}{4})\end{align*}
- \begin{align*}g(x)=3\tan(\frac{x}{2}+\frac{\pi}{4})-1\end{align*}
- \begin{align*}h(x)=\cot(\frac{x}{2})\end{align*}
- \begin{align*}k(x)=\cot(\frac{x}{2}+\frac{3\pi}{2})\end{align*}
- \begin{align*}m(x)=-3\cot(\frac{x}{2}+\frac{3\pi}{2})\end{align*}
- \begin{align*}g(x)=2-3\cot(\frac{x}{2}+\frac{3\pi}{2})\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.11.