Because they're not inverses, not exactly. [tex]\sin x[/tex] is not invertible on its entire domain because it's not one-to-one. There are infinitely many values of [tex]x[/tex] such that [tex]\sin x=0[/tex], for example.
The standard function [tex]\sin^{-1}x[/tex] has a domain of [tex]-1\le x\le1[/tex] and outputs values between [tex]-\dfrac\pi2[/tex] and [tex]\dfrac\pi2[/tex]. This means that its inverse, [tex]\sin x[/tex], is indeed its inverse as long as [tex]-\dfrac\pi2\le x\le\dfrac\pi2[/tex].
[tex]\dfrac{3\pi}4[/tex] is larger than [tex]\dfrac\pi2[/tex] and thus does not fall in the "invertible part" of the domain of [tex]\sin x[/tex]. We have
[tex]\sin\dfrac{3\pi}4=\dfrac1{\sqrt2}[/tex]
which is a value between -1 and 1, so that
[tex]\sin^{-1}\left(\sin\dfrac{3\pi}4\right)=\sin^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4[/tex]
If we wanted to recover [tex]\dfrac{3\pi}4[/tex], we'd have to redefine [tex]\sin^{-1}x[/tex] or define a new inverse function that works on a different branch of the domain of [tex]\sin x[/tex].
