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Cotangent

Cotangent is simply the reciprocal of tangent; in mathematical notation, $\cot{\theta}=\frac{1}{\tan{\theta}}$.

Since $\tan{\theta}=\frac{\text{Opposite Side}}{\text{Adjacent Side}}$, the reciprocal would be $\cot{\theta}=\frac{\text{Adjacent Side}}{{\text{Opposite Side}}}$.

Strategies involving Cotangent

As with inverse trig functions, it is often useful to convert cotangents into tangents, by replacing $\cot{\rm expression}$ into $\frac{1}{\tan{\rm expression}}$.

The Cotangent Function

The next in our series of trigonometric functions is the cotangent function! This function is part of our group of functions, the trigonometric functions. These functions help us to understand angles and triangles.

Cotangent is the reciprocal of the tangent function. In a right-angled triangle, the ratio of tangent is $\text{opposite} : \text{adjacent}$.

The reciprocal of this ratio is the cotangent: $\text{adjacent} : \text{opposite}$.

Cotangent is written as $\text{cot}(\theta)$, where $\theta$ (theta) is the angle. Cotangent is the reciprocal of tangent. Let's see what that means mathematically:

$$\text{cot}(\theta) = 1 \div \text{tan}(\theta) = \text{cos}(\theta) \div \text{sin}(\theta)$$

We use cotangent when working with right angles or in graphing curves. Cotangent goes up and down in a repeating pattern, similar to what we've seen with sine and cosine. Unlike sine and cosine, though, cotangent has vertical asymptotes, usually when sine is zero.

You can also use the unit circle to find cotangent.

(Video) Graphing the Cotangent Function

by Brian McLogan

In this video by Brian McLogan, you will learn how to graph the cotangent function.

Summary

Graph $y=\text{ cot }(\frac{x}{2})$

With cotangent, we are not as concerned with the amplitude. We are concerned with the period, though. We know that the period of the function is $\frac{\pi}{b} = \frac{\pi}{\frac{1}{2}} = 2\pi$.

In the parent function of cotangent, $\text{ cot }(x)$, the vertical asymptotes are $x=0$ and $x=\pi$. Because we are changing what is inside of the parentheses, we are also changing the vertical asymptotes. We calculated earlier that the period is $2\pi$, so we know that our vertical asymptotes are at $x=0$ and $x=2\pi$.

Now we need to tackle the $x$-intercept. We can find the first $x$-intercept by dividing the period, $2\pi$ by $2$. So, our $x$-intercept is at $\pi$.

Ultimately, our graph looks like: