Trigonometric Graphs

Lesson

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 $\cot$`c``o``t`, $\sec$`s``e``c` and $\csc$`c``s``c` 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.

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±`n`π 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, $\cot x=\frac{1}{\tan x}$`c``o``t``x`=1`t``a``n``x`.

It is also true that $\cot x=\tan\left(\frac{\pi}{2}-x\right)=-\tan\left(x-\frac{\pi}{2}\right)$`c``o``t``x`=`t``a``n`(π2−`x`)=−`t``a``n`(`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.

The secant function has asymptotes at $\frac{\pi}{2}\pm n\pi$π2±`n`π. 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±2`n`π and local maxima at $\pi\pm2n\pi$π±2`n`π. These correspond with the maxima and minima respectively of the cosine function since, by definition, $\sec x=\frac{1}{\cos x}$`s``e``c``x`=1`c``o``s``x`.

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

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

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

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

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

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

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

The form of the graph is the same as that of $\csc$`c``s``c` 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.

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

The form of the graph is the same as that of $\sec$`s``e``c`. 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`)=`s``e``c`3(`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.

What is the lowest positive value $k$`k` for which $x=k$`x`=`k` is a vertical asymptote of $y=\operatorname{cosec}x$`y`=`c``o``s``e``c``x`?

What is the greatest negative value that is in the range of the function $y=\operatorname{cosec}x$`y`=`c``o``s``e``c``x`?

Consider the graph of the function $y=\operatorname{cosec}x$`y`=`c``o``s``e``c``x`.

The graph of $y=\operatorname{cosec}x$

`y`=`c``o``s``e``c``x`has been provided.Use this to graph the function $y=\operatorname{cosec}\left(x+\frac{3\pi}{4}\right)$

`y`=`c``o``s``e``c`(`x`+3π4).Loading Graph...The graph of $y=\operatorname{cosec}\left(x+\frac{3\pi}{4}\right)$

`y`=`c``o``s``e``c`(`x`+3π4) from part (a) has been drawn on a new set of axes.Use this to graph the function $y=\frac{3}{2}\operatorname{cosec}\left(x+\frac{3\pi}{4}\right)$

`y`=32`c``o``s``e``c`(`x`+3π4).Loading Graph...

Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions