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A little something I'm trying to understand:

$\sin(\arcsin{x})$ is always $x$, but $\arcsin(\sin{x})$ is not always $x$

So my question is simple - why? Since each cancels the other, it would make sense that $\arcsin(\sin{x})$ would always result in $x$.

I'd appreciate any explanation. Thanks!

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Because there are (infinitely) many angles that give the same sine. Arcsine will only pick one of them. –  J. M. Nov 8 '11 at 10:42
    
So is there any easy method to calculate it? –  yotamoo Nov 8 '11 at 10:43
1  
Look at the plot of $\arcsin(\sin(x))$. –  Sasha Nov 8 '11 at 10:48
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You do know that $\sin(x+2\pi k)=\sin\,x$ for any integer $k$, right? –  J. M. Nov 8 '11 at 10:50

1 Answer 1

up vote 2 down vote accepted

It is a result of deriving an inverse function for non-bijective one. Let $f:X\to Y$ be some function, i.e. for each $x\in X$ we have $f(x)\in Y$. If $f$ is not a bijection then we cannot find a function $g:Y\to X$ such that $g(f(x)) = x$ for all $x\in X$ and $f(g(y)) =y$ for all $y\in Y$.

Consider your example, $f = \sin:\mathbb R\to\mathbb R$. It is neither a surjection (since $\sin (\mathbb R) = [0,1]$) nor an injection (since $\sin x = \sin(x+2\pi k)$ for all $k\in \mathbb Z$). As a result you cannot say that $\sin$ has an inverse. On the other hand, if you consider a restriction of $f^* = \sin|_{X}$ with $X = [-\pi/2,\pi/2]$ and a codomain $Y = [-1,1]$ then $f^*$ has an inverse since $$ \sin|_{[-\pi/2,\pi/2]}:[-\pi/2,\pi/2]\to[-1,1] $$ is an injection. As a result you obtain a function $\arcsin:[-1,1]\to [-\pi/2,\pi/2]$ which is the inverse for $f^* = \sin|_{[-\pi/2,\pi/2]}$.

In particular it means that $\sin (\arcsin{y})=f^*(\arcsin{y}) = y$ for all $y\in[-1,1]$. On the other hand, if you take $x = \frac{\pi}2+2\pi$ then $\sin x = 1$ and hence $\arcsin(\sin{x}) = \frac{\pi}2\neq x$. More precisely, $\arcsin$ is the partial inverse for $\sin$ with all following properties.

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