Chapter 5, Problem 9RCC

a.

To determine

Define the function along with its range and domain.

Answer to Problem 9RCC

  $y={\sin }^{-1}x$ ; $\left[-1,1\right]$ ; $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Explanation of Solution

Calculation:

Here, sine function is not one-to-one function. To make it so we have to restrict its domain to one period.

Now, we will choose the interval $A=\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ because the sine function is one-to-one in that interval.

Now, here the range of restricted function $B=\left[-1,1\right]$ is unchanged.

Thus, from this we will define the inverse sine function as:

  ${\sin }^{-1}x=y\iff \sin y=x$

We know that the domain and range of an inverse function are reversed, so the domain of the function becomes the range of its inverse and its range becomes the domain of the inverse.

Hence, the domain $A=\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ and range $B=\left[-1,1\right]$ of the restricted sine function, conclude that the inverse function $y={\sin }^{-1}x$ has a domain $\left[-1,1\right]$ and range $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ .

b.

To determine

Find if the equation is true or not.

Answer to Problem 9RCC

It is true.

Explanation of Solution

Calculation:

Here, we know the cancellation property of an inverse function.

  $f\left(f^{-1}\left(x\right)\right)=x$ , for every $x$ in $B$

Hence, $\sin \left({\sin }^{-1}x\right)=x$ is true for every value of $x$ in $\left[-1,1\right]$ .

c.

To determine

Find if the equation is true or not.

Answer to Problem 9RCC

It is true.

Explanation of Solution

Calculation:

Here, we know the cancellation property of an inverse function.

  $f^{-1}\left(f\left(x\right)\right)=x$ , for every $x$ in $A$

Hence, ${\sin }^{-1}\left(\sin x\right)=x$ is true for every value of $x$ in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ .