Problem related to question similar to $\sin^{-1}\sin(x)$ where $x$ is an integer?

Question

So I wantes to ask that why can't we write $\sin^{-1}\sin(x)=x$, as in my text book, we have to first convert $x$ to $\pi -x$? For example, in my textbook we have

$$\sin^{-1}\sin(x)=\sin^{-1}\sin(\pi-x)=\pi-x$$

why do we have to do it like that?

Thanks, Akash

Answer 1

The equation $y=\sin^{-1}a$ means $$\sin y=a\quad\hbox{and}\quad -\frac{\pi}{2}\le y\le\frac{\pi}{2}\ .$$ In particular, $y=\sin^{-1}(\sin x)$ means $$\sin y=\sin x\quad\hbox{and}\quad -\frac{\pi}{2}\le y\le\frac{\pi}{2}\ .$$ So for example $\sin^{-1}(\sin 2)$ is not $2$ because $2$ is not between $-\pi/2$ and $\pi/2$. However from the basic formula $$\sin\theta=\sin(\pi-\theta)$$ we have $$\sin2=\sin(\pi-2)\ .$$ Therefore $\sin^{-1}(\sin 2)=\sin^{-1}(\sin(\pi-2))$: we are looking for $$\sin y=\sin(\pi-2)\quad\hbox{and}\quad -\frac{\pi}{2}\le y\le\frac{\pi}{2}\ ,$$ and since $\pi-2$ is between $-\pi/2$ and $\pi/2$, it is the correct answer.

Answer 2

Because $\arcsin({\sin{x}})=x$ if and only if $x$ is in the principal domain of $\sin{x}$, $[-\pi/2,\pi/2]$,