Introduction

Welcome back to AP Precalc! Today we’ll be discussing topic 3.11 of AP Precalculus, which will cover the secant, cosecant, and cotangent functions. Over the past unit, we’ve been primarily discussing three main functions: $\sin (x)$, $\cos (x)$, and $\tan (x)$. In this topic, we’ll be discussing three new functions that are the reciprocal of these functions. We’re almost done with trigonometry, so hang in there while we soar to the finish. Without further adieu, let’s get started.

TL;DR

We go into a lot of depth in this article, but not all of it is necessarily required for the AP exam. If you want to get the key takeaways from the article that is needed for the exam, a summary is provided below.

$\begin{array}{cccc}\text{Function}& \text{Definition}& \text{Domain}& \text{Range}\\ \sec \left(\theta \right)& \frac{1}{\cos \left(\theta \right)}& \text{All values of }\theta \text{ where }\cos \left(\theta \right)\ne 0& \left(-\infty ,-1\right]\cup \left[1,\infty \right)\\ \csc \left(\theta \right)& \frac{1}{\sin \left(\theta \right)}& \text{All values of }\theta \text{ where }\sin \left(\theta \right)\ne 0& \left(-\infty ,-1\right]\cup \left[1,\infty \right)\\ \cot \left(\theta \right)& \begin{array}{c}\frac{\cos \left(\theta \right)}{\sin \left(\theta \right)}\text{ or }\\ \cot \left(\theta \right)=\{\begin{array}{ll}\frac{1}{\tan \left(\theta \right)},& \text{if }\tan \left(\theta \right)\ne 0,\\ 0,& \text{if }\tan \left(\theta \right)=0.\end{array}\end{array}& \text{All values of }\theta \text{ where }\sin \left(\theta \right)\ne 0& \left(-\infty ,\infty \right)\end{array}$

Secant

The secant function is equal to the reciprocal of the cosine function. In other words, $\sec (\theta )=\frac{1}{\cos (\theta )}$.

As you can see in the graph below, the secant function (red) is actually tangent to the extrema of the cosine function (blue).

Quick note: Although it may look like it, the secant functions are not parabolas. Since secant is the reciprocal of cosine, meaning $\frac{1}{\cos \theta }$, it means that whenever $\cos \theta =0$, the secant function will be… you guessed it… undefined! As you can see on the graph above, whenever the cosine function touches zero, there is an asymptote for the secant function, because the secant function will be undefined in those areas. 

Domain

Since $\sec (\theta )=\frac{1}{cos(\theta )}$, $\cos (\theta )$ is defined for all values where $\cos (\theta )$ is both defined and not equal to zero, as if $\cos (\theta )=0$, $\sec (\theta )=\frac{1}{0}$, which is undefined. $\cos (\theta )$ is defined for all real values of $\theta$, so our only concern is when $\cos (\theta )=0$. Referring to our unit circle, we can see that $\cos (\theta )=0$ when $\theta =k\pi +\frac{\pi }{2}$, where $k$ is an arbitrary integer.

Range

The range of $\sec (\theta )=(-\infty ,-1]\cup [1,\infty )$. In other words, no value of $\sec (\theta )$ lies between $-1$ and $1$ because in order for that to be the case, $\cos (\theta )$ would have to be greater than $1$ or less than $-1$, which is outside of the range of $\cos (\theta )$. If you don’t understand, think about it like this. The cosine function goes from $-1$ to $1$ and oscillates between those two numbers. If you replace $\cos (\theta )$ with numbers from $-1$ to $1$ into the secant function, which is $\frac{1}{\cos (\theta )}$, you will notice that the value is never greater than $-1$ or less than $1$. However, you may also notice that $\sec (\theta )$ can be any other real number.

Cosecant

The cosecant function is equal to the reciprocal of the sine function. In other words, $\csc (\theta )=\frac{1}{\sin (\theta )}$.

Domain

$csc(\theta )$ is defined for all values of $\theta$ where $\sin (\theta )$ is both defined and not equal to zero, as if $\sin (\theta )=0$, $\csc (\theta )=\frac{1}{0}$, which is undefined. $\sin (\theta )$ is defined for all real values of $\theta$, so our only concern is when $\sin (\theta )=0$. Referring to our unit circle, we can see that $\sin(\theta) = 0$ when $\theta =k\pi$, where $k$ is an integer.

Range

The range of $\csc (\theta )=(-\infty ,-1]\cup [1,\infty )$. Since the range of $\sin (\theta )$ is $[-1,1]$, $\csc (\theta )$ has the same range as $\sec (\theta )$ for the same reason. The sine function oscillates between $-1$ and $1$, so, if we replace $\sin (\theta )$ with numbers in between $-1$ and $1$, the cosescant function will return a value that is smaller than $-1$ or larger than $1$.

Caution: Even though cosecant feels like it should be the inverse of cosine, and secant should be the inverse of sine, it’s actually the opposite way around.

Cotangent

The cosecant function is equal to the reciprocal of the tangent function. In other words, $\cot (\theta )=\frac{1}{\tan (\theta )}$ or $\cot (\theta )=\frac{\cos (\theta )}{\sin (\theta )}$.

A special case

You might notice that $\frac{1}{\tan \left(\theta \right)}$ and $\frac{\cos \left(\theta \right)}{\sin \left(\theta \right)}$ are not both defined when $\theta =\pi \cdot k+\frac{\pi }{2}$, where $k$ is an arbitrary integer. If we say that $\cot \left(\theta \right)=\frac{1}{\tan \left(\theta \right)}$, then $\cot \left(\frac{\pi }{2}\right)$ would be undefined, since $\tan \left(\frac{\pi }{2}\right)$ is undefined. However, if we say that $\cot \left(\theta \right)=\frac{\cos \left(\theta \right)}{\sin \left(\theta \right)}$, then $\cot \left(\frac{\pi }{2}\right)=\frac{0}{1}=0$. $\cot \left(\theta \right)$ is formally defined as $\frac{\cos \left(\theta \right)}{\sin \left(\theta \right)}$, so $\cot \left(\frac{\pi }{2}\right)=0$. In other words, $\cot \left(x\right)$ will have a vertical asymptote whenever $\tan \left(x\right)=0$ at $x=k\pi$. For the rest of this lesson, we’ll be using $\cot \left(\theta \right)=\frac{\cos \left(\theta \right)}{\sin \left(\theta \right)}$, but excluding cases where $\theta =k\pi$, where $k$ is an integer. The two definitions are identical wherever both are defined.

Domain

Since $cot(\theta )=\frac{\cos (\theta )}{sin(\theta )}$, $\cot (\theta )$ is defined for all values where $\sin (\theta )$ is both defined and not equal to zero, as if $\sin (\theta )=0$, $\cot (\theta )=\frac{\pm 1}{0}$. $\sin (\theta )$ is defined for all real values of $\theta$, so our only concern is when $\sin (\theta )=0$. Referring to our unit circle, we can see that $\sin (\theta )=0$ when $\theta =k\pi$, where k is an integer.

Range

Since $\tan (x)$ can be all real numbers, $\frac{1}{\tan (\theta )}$ can also be all real numbers, with the exception of $0$. However, we’ve previously acknowledged that when $\theta =k\pi +\frac{\pi }{2}$, where k is an integer, $\cot (\theta )=0$. Therefore, the range of $\cot (x)$ is all real numbers.

Practice Problems