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 $-1\le \sin \theta \le 1\Rightarrow x\in \left[ -1,1 \right]$. So, we can say that after a period of time, value starts repeating. So, from the curve, we can see that if $\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$, 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 $\left[ -1,1 \right]$ and $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ 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 $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ and domain can be given as, $\left[ -1,1 \right]$.

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 $x\in \left[ -1,1 \right]$ and values of ${{\sin }^{-1}}x$ goes from $\dfrac{-\pi }{2}$ to $\dfrac{\pi }{2}$, so the range is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$.