Inverse of $f(x)=\sin x$ when restricted to intervals other than $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Question

So $f^{-1}(x)=\arcsin x$ is defined as the inverse of $f(x)=\sin x$ when $f$ is restricted to the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

My question is, what if we restricted $f$ to, say, $\left[\frac{3\pi}{2},\frac{5\pi}{2}\right]$? Would such an inverse resemble something like $g(x)=\arcsin x + 2\pi$? Here's a graph.

Thanks for your time.

Answer 1

Since translating by $2\pi$ does not change the sine function, you can define $f(x)=\sin(x-2\pi)$. Then, $g(x)=\arcsin(x)+2\pi$ will effectively be the inverse you are looking for, since $f(g(x))=x$ and $g(f(x))=x$.