Graphing cot, sec and cosec curves

To understand how the graphs of the cotangent, secant and cosecant functions are sketched and manipulated, it will be helpful to refer to other chapters dealing with these functions by refreshing the key features, transformations, and domain and range.

We can determine the essential features of a graph if we know the forms of the basic graphs of $\cot$cot, $\sec$sec and $\csc$csc and how constants included in the function definitions cause vertical and horizontal dilations and shifts.

          

The graphs of the basic functions can be seen in more detail here. They should be referred to while reading the following paragraphs.

 

Cotangent

The cotangent function has asymptotes at $0$0, $\pi$π and at every positive and negative integer multiple of $\pi$π. It is periodic with period $\pi$π. It takes the value $0$0 at $\frac{\pi}{2}\pm n\pi$π2​±nπ for every integer $n$n. It is a decreasing function, meaning the graph always slopes downwards to the right.

The cotangent function is the reciprocal of the tangent function. That is, $\cot x=\frac{1}{\tan x}$cotx=1tanx​.

It is also true that $\cot x=\tan\left(\frac{\pi}{2}-x\right)=-\tan\left(x-\frac{\pi}{2}\right)$cotx=tan(π2​−x)=−tan(x−π2​). The form on the right of this identity makes it clear that the graph of the cotangent function is obtained by translating the graph of the tangent function to the right by $\frac{\pi}{2}$π2​ and then reflecting the result in the horizontal axis.

 

Secant

The secant function has asymptotes at $\frac{\pi}{2}\pm n\pi$π2​±nπ. It is periodic with period $2\pi$2π. It takes no values in the open interval between $-1$−1 and $1$1. It has local minima at $0\pm2n\pi$0±2nπ and local maxima at $\pi\pm2n\pi$π±2nπ. These correspond with the maxima and minima respectively of the cosine function since, by definition, $\sec x=\frac{1}{\cos x}$secx=1cosx​. 

The function $\sec x$secx is an even function, meaning $\sec(-x)=\sec x$sec(−x)=secx for all values of $x$x. This follows from the fact that $\cos x$cosx is an even function.

From the corresponding relations between the sine and cosine functions, we see that $\sec x=\csc\left(\frac{\pi}{2}-x\right)=-\csc\left(x-\frac{\pi}{2}\right)$secx=csc(π2​−x)=−csc(x−π2​). This identity shows that the graph of $\sec x$secx is obtained from the graph of $\csc x$cscx by shifting it to the right by $\frac{\pi}{2}$π2​ and reflecting it in the $x$x-axis.

 

Cosecant

The cosecant function has asymptotes at $0\pm n\pi$0±nπ. It has local minima at $\frac{\pi}{2}\pm2n\pi$π2​±2nπ and local maxima at $\frac{3\pi}{2}\pm2n\pi$3π2​±2nπ. 

Since $\sin(-x)=-\sin x$sin(−x)=−sinx it follows that $\csc(-x)=-\csc x$csc(−x)=−cscx and so, $\csc x$cscx is an odd function. Its symmetry is captured by the statement $\csc x=-\csc(-x)$cscx=−csc(−x) which says that the graph of $\csc x$cscx looks the same after reflection in the vertical axis and then, reflection in the horizontal axis.

Starting from the graph of $\sec x$secx, we can obtain the graph of $\csc x$cscx by a transformation. Since $\csc x=\sec\left(\frac{\pi}{2}-x\right)$cscx=sec(π2​−x), we have $\csc x=\sec\left(x-\frac{\pi}{2}\right)$cscx=sec(x−π2​) (because $\sec$sec is an even function).

This says that we can obtain the graph of $\csc$csc by shifting the graph of $\sec$sec to the right by $\frac{\pi}{2}$π2​. 

 

Example 1

Sketch the graph of $y(x)=2\csc x-1$y(x)=2cscx−1.

The form of the graph is the same as that of $\csc$csc but there is a dilation by the factor $2$2 in the vertical direction and a vertical translation by $-1$−1. The dilation by itself would mean the local minima have the value $2$2 and the local maxima the value $-2$−2. But with the translation, these will be $1$1 and $-3$−3 respectively. There is no change in the positions of the asymptotes since there is no horizontal shift or dilation.

 

Example 2

Describe the graph of $y(x)=\sec\left(3x+\frac{\pi}{4}\right)$y(x)=sec(3x+π4​).

The form of the graph is the same as that of $\sec$sec. There is no vertical shift and no vertical dilation. There is a horizontal dilation by the factor $\frac{1}{3}$13​ so that the function has period $\frac{2\pi}{3}$2π3​.

As well, there is a horizontal shift. If the function is rewritten $y(x)=\sec3\left(x+\frac{\pi}{12}\right)$y(x)=sec3(x+π12​), we see that the shift is to the left by $\frac{\pi}{12}$π12​.

The asymptotes occur at intervals of $\frac{\pi}{3}$π3​ instead of $\pi$π because of the dilation. The first asymptote in the positive region occurs at $x=\frac{1}{3}\times\frac{\pi}{2}-\frac{\pi}{12}=\frac{\pi}{12}$x=13​×π2​−π12​=π12​.

Worked Examples

Question 1

Question 2

Question 3