Chapter 4.7, Problem 56E

Solution

(a)

To calculate: find domain and range.

Domain: $[-1,1]$ , range: $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Given information: $f(x)=\sin ({\sin }^{-1}x)$

Formula used: trigonometry

Calculation: Domain

Inner function: the arcsine function can only take in values in the domain $[-1,1]$

Outer function: the sine function can take in any values, so the domain of the function as a whole is limited solely by the inner function. Thus the domain of the function is $[-1,1]$

Range

Inner function: the sine function only outputs values in the range $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Outer function: considering the output of the arcsine function, the only values that will be output are $[-1,1]$

This is the range of the function overall.

(b)

To calculate: find domain and range.

Domain: $[-1,1]$ , range: $\frac{\pi }{2}$

Given information: $g(x)={\sin }^{-1}(x)+{\cos }^{-1}(x)$

Formula used: trigonometryCalculation: Domain

Inner function: both the sine and arcsine function can only take in values in the domain $[-1,1]$

Range

Let us consider the endpoints of the domain to give us a clue about the nature of the range of the function. At $x=-1$

  $g(x)=\arcsin (-1)+\arcsin (-1)=-\frac{\pi }{2}+\pi =\frac{\pi }{2}$

At $x=1$ ,

  $g(x)=\arcsin (1)+\arccos (1)=\frac{\pi }{2}-0=\frac{\pi }{2}$

In fact, at every value of $x$ , $g(x)=\frac{\pi }{2}$ .

Therefore, the range of $g(x)$ is simply $\frac{\pi }{2}$ .

(c)

To calculate: find domain and range.

Domain: $x\in ℝ$ , range: $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Given information: $h(x)={\sin }^{-1}(\sin x)$

Formula used: trigonometry

Calculation:

Domain:

Inner function: the sine function can take in any value of $x$ .

However, it outputs only $[-1,1]$ which will become the input for the outer function.

Outer function: the arcsine function can take in values in the domain $[-1,1]$ .

Since the inner function cannot output any value that the outer function cannot evaluate, the domain of the function is thus $x\in ℝ$

Range:

Inner function: the sine function outputs only $[-1,1]$ , which will become the input for the outer function.

Outer function: since the output of the inner function is the entire domain of the outside function, the range of $h(x)$ is the same as the range of the arcsine function, which is:

  $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

(d)

To calculate: find domain and range.

Domain: $[-1,1]$ , range: $[0,1]$

Given information: $k(x)=\sin ({\cos }^{-1}x)$

Formula used: trigonometry

Calculation:

Domain:

Inner function: the arccosine function can take in values in the domain $[-1,1]$ . This results in an output of $[0,\pi ]$ , which will be the input for the outer function.

Outer function: the arcsine function can take in values thus the domain as a whole is limited only by the inner function, making the overall domain $[-1,1]$ .

Range:

Inner function: the arccosine function only outputs values of $[0,\pi ]$ , which will be the input for the outer function.

Outer function: in the domain $[0,\pi ]$ , the sine function outputs values of $[0,1]$

(e)

To calculate: find domain and range.

Domain: $x\in ℝ$ , range: $[0,1]$

Given information: $q(x)={\cos }^{-1}(\sin x)$

Formula used: trigonometry

Calculation:

Domain:

Inner function: the arccosine function can take in values of x but will output the range $[-1,1]$ , which will be the input for the outer function.

Outer function: given the input $[-1,1]$ , the arccosine function can output values for that entire domain. Thus there is no value of x that the function cannot generate an output for,

Thus, the domain is $x\in ℝ$

Range:

Inner function: the sine function only outputs values of $[-1,1]$ , which will be the input for the outer function.

Outer function: when presented with the domain $[-1,1]$ , the arccosine function outputs values in the range $[0,\pi ]$