The definition of inverse function is that the function must be one to one. However, as the sine function isn't one to one, how is $\sin^{-1}$, the inverse of the sine function?
Is it due to the periodic property of trigonometric curves? $\iff \sin(0)=0$ and $\sin(2\pi)=0$ (in radians).
If I throw the output into the inverse function I get $0$. Why not $2\pi$?
$\sin x$ takes a value $y$ at many $x$. If we define an inverse relation, that relation can take many values $x$ at its input $y$. In order for that relation to be a function by definition, it has to take only one value of $x$ at its input $y$. Hence we define $\sin^{-1}y=x$, where $y\in\left[-1,1\right]$ is what sine could map to only, such that $x$ belongs to $\left[-\frac{\pi}2,\frac{\pi}2\right]$ only. Outside these boundaries for $x$, inverse sine is not a function but a mere relation.