Why is $\arcsin(\sin(x))$ equal to $x$?

Question

Why is $\arcsin(\sin(x))$ equal to $x$?

Most videos only state this but give no explanation other than: "They cancel." As these are not reciprocals, how do they "cancel"?

Answer 1

If $f:A \to B$ is a bijective map, then $f^{-1}(f(x))=x$ for all $x \in A$.

This is meant by "cancel".

Answer 2

By definition, $\arcsin\colon[-1,1]\longrightarrow\left[-\frac\pi2,\frac\pi2\right]$ is the inverse of the restriction to $\left[-\frac\pi2,\frac\pi2\right]$ of the sine function. Therefore, for each $x\in\left[-\frac\pi2,\frac\pi2\right]$ we have $\arcsin\bigl(\sin(x)\bigr)=x$ because that's part of the definition of inverse functions. The other part of the definition says that, if $x\in[-1,1]$, then $\sin\bigl(\arcsin(x)\bigr)=x$.

Answer 3

Perhaps seeing this graphically will help.

Restricting $x \in [-\pi/2, \pi/2]$ in order that a unique value for $\arcsin(x)$ exists the black arrows on the plot below show the effect of $x\to \sin(x) = y$. The the red arrows then show the effect of $y \to \arcsin(y)=\arcsin(\sin(x)) = x$ and you end up back where you started.

You might imagine extending the graph beyond $x \in [-\pi/2, \pi/2]$ to see why a unique value of $\arcsin(x)$ would not exist.

Answer 4

It is not valid for all $x$! E.g. $\arcsin(\sin(10 \tfrac{\pi}{4})) = \arcsin(1) = \tfrac{\pi}{2}.$ Since $\sin(x)$ is periodic, you have to restrict $x$ to $[- \tfrac{\pi}{2}, \tfrac{\pi}{2}].$ For other $x$ you get a saw-tooth function for $\arcsin(\sin(x))$