Chapter 4.3, Problem 35E

To determine

To State: The domain of ${\sin }^{-1}x$ .

Answer to Problem 35E

True, the domain of ${\sin }^{-1}x\text{is}[-1,1]$ .

Explanation of Solution

Given information: Inverse of sine function that is ${\sin }^{-1}x$ .

If we consider $f:R\to R\text{given}\text{by}f(x)=\sin x$ , it is a many-one function as it attains same value at infinitely many points and its range $[-1,1]$ is not same as its co-domain.

Now, to compute the inverse of the given function, the function has to be bijective so we restrict the domain as $\left[\frac{-\pi }{2},\frac{\pi }{2}\right]$ and replace co-domain by its range,

Thus, $f:\left[\frac{-\pi }{2},\frac{\pi }{2}\right]\to [-1,1]\text{given}\text{by}f(x)=\sin x$ is a bijective and hence invertible.

The inverse of the sine function with domain $[-1,1]$ and range $\left[\frac{-\pi }{2},\frac{\pi }{2}\right]$ .