Question

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

Answer 1

Hint: To draw the graph of ${\sin }^{-1}(\sin x)$, we must know about the following concepts,
Inverse of a function means a function which returns back the original value applied on the given function. Inverse of a particular function exists if the function is bijective.i.e., for each unique value from the domain of the function, we should have a corresponding unique value from the range of the function.

Complete step-by-step solution:
For a function $f$ , if its inverse exists, then
$f^{-1}\left(f\left(x\right)\right)=x\text{ where }x\in Range\text{ of }{f}^{-1}\left(x\right)$
Another point to be noted is
$\begin{array}{rl}& \text{If, }\sin y=\sin x\\ & \Rightarrow y=n\pi +{(-1)}^{n}x\end{array}$
Range of $y={\sin }^{-1}x$ $\mathrm{\forall }x\in (-1,1),y\in ({\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}})$
Now, let us proceed with the question,
We have, $f\left(x\right)={\sin }^{-1}(\sin x)$
Let us assume $f\left(x\right)=\theta$
$\begin{array}{rl}& \Rightarrow {\sin }^{-1}(\sin x)=\theta \\ & \Rightarrow \sin x=\sin \theta \end{array}$
Using,
$\begin{array}{rl}& \text{If, }\sin y=\sin x\\ & \Rightarrow y=n\pi +{(-1)}^{n}x\end{array}$
We get
$\Rightarrow \theta =n\pi +{(-1)}^{n}x$
For different values of $x$ , we will get different values of $\theta$ .i.e., for different values of $x$, we will get different values of $f\left(x\right)$ .
For $n=-1$
$\theta ={\sin }^{-1}(\sin (-\pi -x))$
For $\theta$ to exist,
$\begin{array}{rl}& {\displaystyle \frac{-\pi }{2}}\le -\pi -x\le {\displaystyle \frac{\pi }{2}}\\ & \Rightarrow {\displaystyle \frac{\pi }{2}}\le -x\le {\displaystyle \frac{3\pi }{2}}\\ & \Rightarrow {\displaystyle \frac{-\pi }{2}}\ge x\ge {\displaystyle \frac{-3\pi }{2}}\end{array}$
For $n=0,$
$\theta ={\sin }^{-1}(\sin x)$
For $\theta$ to exist,
${\displaystyle \frac{-\pi }{2}}\le x\le {\displaystyle \frac{\pi }{2}}$
For $n=1,$
$\theta ={\sin }^{-1}(\sin (\pi -x))$
For $\theta$ to exist,
$\begin{array}{rl}& {\displaystyle \frac{-\pi }{2}}\le \pi -x\le {\displaystyle \frac{\pi }{2}}\\ & \Rightarrow {\displaystyle \frac{-3\pi }{2}}\le -x\le {\displaystyle \frac{-\pi }{2}}\\ & \Rightarrow {\displaystyle \frac{3\pi }{2}}\ge x\ge {\displaystyle \frac{\pi }{2}}\end{array}$
So, we can redefine the function $f\left(x\right)$ as
$f\left(x\right)=\{\begin{array}{ccc}& -\pi -x& {\displaystyle \frac{-3\pi }{2}}\le x\le {\displaystyle \frac{-\pi }{2}}\\ & x& {\displaystyle \frac{-\pi }{2}}\le x\le {\displaystyle \frac{\pi }{2}}\\ & \pi -x& {\displaystyle \frac{\pi }{2}}\le x\le {\displaystyle \frac{3\pi }{2}}\end{array}$
Now, using these inequalities we will draw the graph of $f\left(x\right)={\sin }^{-1}(\sin x)$

So, for the function $f\left(x\right)={\sin }^{-1}(\sin x)$
$Domain:x\in R$
$Range:x\in ({\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}})$
The graph is periodic with a fundamental period of $2\pi$.

Note: While drawing the graph of $f\left(x\right)={\sin }^{-1}(\sin x)$, remember that the different values of $x$ should correspond to the range of ${\sin }^{-1}x$ .i.e.,$x\in ({\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}})$. Here, since $x$ is the domain of $\sin x$, hence $x\in R$. This means that $\sin x\in (-1,1)$ and hence ${\sin }^{-1}(\sin x)\in ({\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}})$. So, the given function has a range of $({\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}})$ for every $x\in R$.