Question

Plot the graph of ${\cos }^{-1}(\cos x)$ and write its domain and range.

Answer 1

Hint: In this question, we will first understand the relation of $\cos x$ and ${\cos }^{-1}x$. Then, observe the graph of $\cos x$, and how they change sign and use it to plot a graph of ${\cos }^{-1}(\cos x)$. From the graph, we will find its range and domain.


Complete step-by-step answer:

Firstly, let us understand what is meant by the inverse function of cosine.
Suppose, $y={\cos }^{-1}x$.
Then, for each value of $x$ there will exist some value of $y$. Then, the cosine inverse of this value of $y$ will be $x$.
For example, ${\displaystyle \frac{1}{2}}=\cos {\displaystyle \frac{\pi }{6}}$.
Then, ${\cos }^{-1}{\displaystyle \frac{1}{2}}={\displaystyle \frac{\pi }{6}}$.
Now, $\cos x$ is a periodic function with period $2\pi$, which means its values repeat in the same pattern after $2\pi$ increases in $x$. That is, $\cos x=\cos (2\pi +x)$.
Since, $\cos x$ is periodic with period $2\pi$. Therefore, ${\cos }^{-1}(\cos x)$ is also period with period $2\pi$.
Also, the domain here is set of those values of $x$ for which ${\cos }^{-1}(\cos x)$ is defined. And, range is the set of values where ${\cos }^{-1}(\cos x)$ lies.
Now, for all real values of $x$, $\cos x$lies between -1 and 1. And, between -1 and 1, the inverse function of cosine is defined. Therefore, ${\cos }^{-1}(\cos x)$ is defined for all real values of $x$. Hence, the domain of ${\cos }^{-1}(\cos x)$ is $(-\mathrm{\infty },\mathrm{\infty })$ .
We know, graph of $y=\cos x$ is:

We see that, in the interval $[-\pi ,\pi ]$, for two different values of $x$, we have the same value of $y$.
Also, from definition of cosine inverse, in this graph, we get,
${\cos }^{-1}y=x$
If we substitute $y=\cos x$ here, we get,
${\cos }^{-1}(\cos x)=x$
Now, in graph of ${\cos }^{-1}(\cos x)$, we have,
 $y={\cos }^{-1}(\cos x)$
$\Rightarrow y=x$
But, in interval $[-\pi ,\pi ]$, for two different values of $x$, we have the same value of $y$.
Let those two different values be represented by $y_1,y_2$.
Now, as $x$ increases from $-\pi$ to 0, $\cos x$ increases from -1 to 1, and hence, ${\cos }^{-1}(\cos x)$ decreases from $\pi$ to 0. Therefore, here we will have, $y_1=-x$.
And as $x$ increases from 0 to $\pi$, $\cos x$ decreases from 1 to -1, and hence, ${\cos }^{-1}(\cos x)$ increases from 0 to$\pi$. Therefore, here we will have, $y_2=x$.
Also, from $-\pi$ to $\pi$, length of interval is $2\pi$ and ${\cos }^{-1}(\cos x)$ periodic with period $2\pi$. Therefore, the rest of the graph will repeat the same as in the interval $[-\pi ,\pi ]$.
Hence, the graph of ${\cos }^{-1}(\cos x)$ is given by:

Here, values of ${\cos }^{-1}(\cos x)$ lies between 0 to $\pi$.
Hence for the graph of ${\cos }^{-1}(\cos x)$ plotted above, the domain is $(-\mathrm{\infty },\mathrm{\infty })$ and the range is $[0,\pi ]$.

Note: While plotting the graph, keep in mind that for two different values of $x$, ${\cos }^{-1}(\cos x)$ will have the same value in interval of length $2\pi$. So, looking at $y=x$, do not directly plot a graph of an infinite straight line.