Finding a shift of $f^{-1}(x)$, where $f(x) = \sin{x}, \; x \in [{\pi \over 2}; {3\pi \over 2}]$

Question

How to analytically find the shift of the inverse of $f(x) = \sin{x}, \; x\in [{\pi \over 2}; {3\pi \over 2}]$ ?

Given a function $f(x) = \sin{x}$ i need to find the inverse function. The inverse of $\sin{x}$ is $\arcsin(x)$ which is defined for $x \in [-1, 1]$, and its values are in the range of $[-{\pi \over 2}; {\pi \over 2}]$.

I've used some tools to draw the graph of the function and its inverse, from which I see that $\arcsin(x)$ should be shifted by $\pi$, otherwise its not the inverse of $\sin(x)$ for $x \in [{\pi \over 2}; {3\pi \over 2}]$.

In my case based on the graph i see that for:

$$ f(x) = \sin x, \; \left[ \frac{\pi}{2}+2\pi n\le x\le\frac{3\pi}{2}+2\pi n\right], \;\; n\in \mathbb Z $$

The inverse is:

$$ f^{-1}(x) = \left(2n+1\right)\pi-\arcsin x, \;\; n\in \mathbb Z $$

But how do I show that analytically without graphing the functions?

Answer 1

Solution

Denote $$y=f(x)=\sin x,x \in \left[\frac{\pi}{2},\frac{3\pi}{2}\right].$$ Then, $$\sin(x-\pi)=-\sin x=-y,x \in \left[\frac{\pi}{2},\frac{3\pi}{2}\right].$$ Notice that $x-\pi \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$ Hence $$x-\pi=2n\pi+\arcsin(-y)=2n\pi-\arcsin y,n \in \mathbb{Z},$$ namely,$$x=(2n+1)\pi-\arcsin y, n \in \mathbb{Z},$$ Exchanging $x,y$, we have $$f^{-1}(x)=y=(2n+1)\pi-\arcsin x, n \in \mathbb{Z}.$$