Question

Explain why sin^-1(sin(3pi/4))does not equal 3pi/4 when y=sin x and y=sin^-1 x are inverses.

Answer 1

Because they're not inverses, not exactly. $\sin x$ is not invertible on its entire domain because it's not one-to-one. There are infinitely many values of $x$ such that $\sin x=0$, for example.

The standard function $\sin^{-1}x$ has a domain of $-1\le x\le1$ and outputs values between $-\dfrac\pi2$ and $\dfrac\pi2$. This means that its inverse, $\sin x$, is indeed its inverse as long as $-\dfrac\pi2\le x\le\dfrac\pi2$.

$\dfrac{3\pi}4$ is larger than $\dfrac\pi2$ and thus does not fall in the "invertible part" of the domain of $\sin x$. We have

$\sin\dfrac{3\pi}4=\dfrac1{\sqrt2}$

which is a value between -1 and 1, so that

$\sin^{-1}\left(\sin\dfrac{3\pi}4\right)=\sin^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4$

If we wanted to recover $\dfrac{3\pi}4$, we'd have to redefine $\sin^{-1}x$ or define a new inverse function that works on a different branch of the domain of $\sin x$.