Answer
(a) $f(x) = sin(sin^{-1}x) = x$
The domain of $f(x)$ is $[-1,1]$.
(b) $g(x) = sin^{-1}(sin~x) = x$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$
On the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$, the value of $~sin~x = sin(\pi-x)~$. Thus, on this interval, $g(x) = g(\pi-x)$.
The function $~sin~x~$ repeats over a period of $2\pi$, so $g(x)$ also repeats over a period of $2\pi$.
Work Step by Step
(a) $f(x) = sin(sin^{-1}x)$
Since $sin$ and $sin^{-1}$ are inverse functions, then $f(x) = sin(sin^{-1}x) = x$
However, the domain of $sin^{-1}$ is $[-1,1]$. Therefore the domain of $f(x)$ is $[-1,1]$.
(b) Note that the range of $sin^{-1}~x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$
$g(x) = sin^{-1}(sin~x) = x$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$
On the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$, the value of $~sin~x = sin(\pi-x)~$. Thus, on this interval, $g(x) = g(\pi-x)$. This explains the symmetry on the interval $[-\frac{\pi}{2}, \frac{3\pi}{2}]$
The function $~sin~x~$ repeats over a period of $2\pi$. Therefore, the pattern we see on the interval $[-\frac{\pi}{2}, \frac{3\pi}{2}]$ repeats in both directions.
