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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.

graph

Thanks for your time.

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  • $\begingroup$ Isn't it $\arcsin(x-2\pi)+2\pi?$ $\endgroup$ – saulspatz Nov 22 '18 at 23:37
  • $\begingroup$ 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 $f^{-1}(y)=x$? $\endgroup$ – Euler's Friend Nov 23 '18 at 0:00
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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 $g(f(x))=x$.

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