Find the equation of a cot, sec and cosec curve

We have previously looked at how to sketch the graphs of secant, cosecant and cotangent curves given their equations. We are now going to look at the reverse of this - how to recover the equation of a secant, cosecant or cotangent curve given its graph.

To do so, we first want to identify which of the three functions is most appropriate for the given curve.

 

A graph of $y=\sec x$y=secx		A graph of $y=\csc x$y=cscx

Notice that the graphs of $y=\sec x$y=secx and $y=\csc x$y=cscx both have local minima and maxima, but no points of inflection. They differ by a phase shift (a horizontal translation): $y=\sec x$y=secx has a local minimum on the $y$y-axis, while $y=\csc x$y=cscx has an asymptote along the $y$y-axis.

 

On the other hand, the graph of $y=\cot x$y=cotx has points of inflection, but no local minima or maxima.

 

Once we have identified the most appropriate base function, we look at the features of the graph in order to determine the particular coefficients of the equation. These features include the location of the asymptotes, the points of inflection or local minima and maxima and the median value of the function. Let's go through an example.

 

Worked example

Consider the graph below

We can immediately see that this curve has the shape of a secant or cosecant function. In particular it has an asymptote along the $y$y-axis, which matches the graph of $y=\csc x$y=cscx. So we can write this function in the form $y=a\csc\left(b\left(x-c\right)\right)+d$y=acsc(b(x−c))+d.

To determine the values of the coefficients, recall that:

• the value of $a$a represents a vertical dilation,
• the value of $b$b represents a horizontal dilation, and is related to the period of the function,
• the value of $c$c represents a phase shift (a horizontal translation), and
• the value of $d$d represents a vertical translation.

Now, we chose to use $y=\csc x$y=cscx as our base function because the feature on the $y$y-axis matches up (an asymptote in this case). This means that there is no phase shift, and so we have that $c=0$c=0.

Also, we can see that the median value of this function is $y=0$y=0 (halfway between the local minima and maxima), which also matches the median value of $y=\csc x$y=cscx. This means that there is no vertical translation, and so we have that $d=0$d=0.

Looking at the graph of $y=\csc x$y=cscx, we can see that the local minima and maxima are $1$1 unit above or below the median value respectively. On our graph, however, these points are $3$3 units above or below the median value instead. So there is a vertical dilation by a factor of $\frac{3}{1}=3$31​=3, and so we have that $a=3$a=3.

Finally, the period of our function is $\pi$π, while the graph of $y=\csc x$y=cscx has a period of $2\pi$2π. From this we get the relation $\frac{2\pi}{b}=\text{Period}=\pi$2πb​=Period=π, and so we have that $b=2$b=2.

Putting this together, the equation of our graph is

$y=3\csc\left(2x\right)$y=3csc(2x)

 

Note that we could have also expressed this function in the form $y=3\sec\left(2\left(x-\frac{\pi}{4}\right)\right)=3\sec\left(2x-\frac{\pi}{2}\right)$y=3sec(2(x−π4​))=3sec(2x−π2​), using $y=\sec x$y=secx as the base function. Observe that this form of the equation has a phase shift, since the graph did not line up with that of $y=\sec x$y=secx on the $y$y-axis.

 

 

Practice questions

Question 1

Question 2

Question 3