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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?

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You've tried plotting $\arcsin(\sin\,x)$ for starters? – J. M. Apr 25 '12 at 13:31
@J.M.: I wasn't sure how to. I guess the domain would be [1,-1] and the range would be $\frac{\pi}{2} \leq y \leq \frac{3\pi}{2}$? Maybe being able to plot such functions is the missing link :D I'll try and find some info on that now. – stariz77 Apr 25 '12 at 14:12
up vote 1 down vote accepted

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. $

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Ah, ok. So, you need to find $\theta$ $st$ the values of $x$ produce output within the standard definition's range? If the interval was $\frac{3\pi}{2} \leq x \leq \frac{5\pi}{2}$ you would need $\theta$ to be $2\pi$ - x, right? – stariz77 Apr 25 '12 at 14:19
@stariz77 Yes, If I understand you correctly. By definition $\arcsin(x)$ is the angle in the interval $[-\pi/2,\pi/2]$ whose $\sin$ is $x$, so $\arcsin(\sin(x))$ is the angle in $[-\pi/2,\pi/2]$ whose $\sin$ is $\sin (x)$. – David Mitra Apr 25 '12 at 14:22
@stariz77 For $3\pi/2\le x \le5\pi/2$, I think you want to use $x-2\pi$, not $2\pi-x$. – David Mitra Apr 25 '12 at 14:34

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

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