Simplify $\arcsin(\sin(x))$ when $\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$

Question

Simplify $\arcsin(\sin(x))$ when $\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$

I realize that $\arcsin(\theta)$ is restriced to $\frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}$ in order to be one to one and therefore invertible; however, I can't seem to connect the new domain given in the question to the provided result of $\pi - x$.

How does the shifted domain change the result to $\pi - x$ instead of just $x$? Isn't the function just the same in the new interval, i.e., it's still one to one?

I have seen a few explanations of this topic in textbooks; however, I can't seem to connect the logic. How do I go about solving these type of problems in general?

Answer 1

The formula $\arcsin(\sin(x))=x$, with the standard definition of $\arcsin$, holds only if $x$ is in the range of $\arcsin$, that is only if $-\pi/2\le x\le\pi/2$.

So saying, for example, $\arcsin(\sin (\pi))=\pi$ is wrong ( $\arcsin(\sin (\pi)=\arcsin(0)=0$).

When computing $\arcsin(\sin( x))$, you want to find an angle $\theta$ in the interval $[-\pi/2,\pi/2]$ such that $\sin\theta=\sin x$, and then $\arcsin(\sin(x))=\arcsin(\sin\theta))=\theta$.

If $\pi/2\le x\le3\pi/2$, then $\pi-x$ is in the interval $[-\pi/2,\pi/2]$ and $\sin(x)=\sin(\pi-x)$.

So, for $\pi/2\le x\le3\pi/2$, $ \arcsin(\sin(x))= \arcsin(\sin(\pi-x))=\pi-x. $

Answer 2

Because $\sin(x)$ is symmetric under the reflection $x\mapsto\frac{\pi}{2}-(x-\frac{\pi}{2})$ you get $$\arcsin(\sin(x))=\frac{\pi}{2}-(x-\frac{\pi}{2}),\ \forall x\in[\frac{\pi}{2},\frac{3\pi}{2}]$$