Question

Determine the domain and range of ${\sin }^{-1}x$.

Answer 1

Hint: To solve this question, we will start by assuming ${\sin }^{-1}x=\theta$. Also, while solving the question, we have to remember that a function always has one to one mapping, which means that one particular value of x will give a particular value of $\theta$. Also, we should have some knowledge regarding $x=\sin \theta$.

Complete step-by-step answer:
In this question, we have been asked to find the domain and range of ${\sin }^{-1}x$. For that, we will consider, $\theta ={\sin }^{-1}x$. We know that such a type of function can also be written as $\sin \theta =x$. Now, according to the wave of $\sin \theta$, as shown in the figure below,

We can say that $-1\le \sin \theta \le 1\Rightarrow x\in [-1,1]$. So, we can say that after a period of time, value starts repeating. So, from the curve, we can see that if $\theta \in [{\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}]$, then only we are getting all the different possible values of x, otherwise values are repeating. Therefore the range and domain of $\sin \theta =x$ is $[-1,1]$ and $[{\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}]$ respectively.
Now, if we talk about $\theta ={\sin }^{-1}x$, then we can say range and domain of the function will interchange because we are talking about inverse function here.
Hence, we can say that the range of ${\sin }^{-1}x$ can be given as $[{\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}]$ and domain can be given as, $[-1,1]$.

Note: We can also solve this question from the graph of ${\sin }^{-1}x$ also, which looks like the figure below.

From the curve, we can say that the curve has one to one mapping for $x\in [-1,1]$ and values of ${\sin }^{-1}x$ goes from ${\displaystyle \frac{-\pi }{2}}$ to ${\displaystyle \frac{\pi }{2}}$, so the range is $[{\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}]$.