Why is $\arcsin(x) = \pi$ an impossible equation?

Question

Why is $\arcsin(x) = \pi$ an impossible equation, if $\sin(\arcsin(x)) = \sin(\pi)$ for $x = \sin(\pi)$?

Answer 1

This would imply $x=\sin\pi=0$, giving the contradiction $\arcsin x=0$.

Answer 2

The function $\sin$ maps each $I_k = [k\pi - \frac{\pi}{2},k\pi + \frac{\pi}{2}] $, $k \in \mathbb{Z}$, bijectively onto $J = [-1,1]$. Therefore each of the functions $\sin_k = \sin \mid_{I_k} : I_k \to J$ has an inverse $\arcsin_k : J \to I_k$.

The functions $\arcsin_k$ are the branches of the arcsine and for $k=0$ we obtain the principal value which is simply denoted by

$$\arcsin : [-1,1] \to [- \frac{\pi}{2}, \frac{\pi}{2}] .$$

This is a universally accepted convention. We have $$\sin(\arcsin(y)) = y \text{ for all } y \in [-1,1] ,$$ $$\arcsin(\sin(x)) = x \text{ iff } x \in [- \frac{\pi}{2}, \frac{\pi}{2}] .$$ Note that the equation $\arcsin_k(y) = \pi$ has a solution for $k=1$ ($y = 0$) and no solution for $k \ne 1$.