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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
Look at the plot of $\arcsin(\sin(x))$. – Sasha Nov 8 '11 at 10:48
You do know that $\sin(x+2\pi k)=\sin\,x$ for any integer $k$, right? – J. M. Nov 8 '11 at 10:50
up vote 3 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.

share|cite|improve this answer

Your Answer


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.