Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

2 Answers 2

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. $

share|improve this answer
    
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 $$\arcsin(\sin(x))=\frac{\pi}{2}-(x-\frac{\pi}{2}),\ \forall x\in[\frac{\pi}{2},\frac{3\pi}{2}]$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.