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

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.

• Isn't it $\arcsin(x-2\pi)+2\pi?$ – saulspatz Nov 22 '18 at 23:37
• Why is that? Take $x=2\pi$. So $\sin(2\pi)=0$, but your suggestion is undefined at $x=0$. Shouldn't it be the case that if $f(x)=y$ then $f^{-1}(y)=x$? – Euler's Friend Nov 23 '18 at 0:00
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 $$g(f(x))=x$$.