Chapter 7, Problem 1RE

In problems $1-6,$ state the domain and range of each function.

$y={\sin }^{-1}x$

To determine

The domain and range of function $y={\sin }^{-1}x$.

Answer to Problem 1RE

Solution:

The domain of the function $y={\sin }^{-1}x$ is $\left[-1,1\right]$ and its range is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.

Explanation of Solution

Given information:

$y={\sin }^{-1}x$.

Explanation:

Let a function $y={\sin }^{-1}x\Rightarrow x=\sin y$.

As, the domain of the function $x=\sin y$ is $-\infty <y<\infty$ and its range is $-1\le x\le 1$.

If, the domain of $x=\sin y$ is restricted to the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, then the restricted function $x=\sin y;-\frac{\pi }{2}\le y\le \frac{\pi }{2}$ is one-one and has an inverse function.

Thus, $y={\sin }^{-1}x$ is an inverse of a function $x=\sin y$.

Now, the restricted sine function $x=\sin y$ receives as input an angle or real number in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ and output a real number in the interval $\left[-1,1\right]$.

Therefore, the inverse sine function $y={\sin }^{-1}x$ receives as input a real number $x$ in the interval $\left[-1,1\right]$, its domain and output an angle or real number in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, its range.

Hence, the domain of the function $y={\sin }^{-1}x$ is $\left[-1,1\right]$ and its range is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.