Inverse trigonometric function of trigonometric function

Question

This is a simple question.

We know that

$\sin( arcsin(2\pi) ) = 2\pi$

Because $arcsin$ is the inverse function of $sin$

So logically, the following should be true as well, since $sin$ is the inverse of $arcsin$

$\arcsin( sin(2\pi) ) = 2\pi$

but it actually equals $0$. Why so?

Answer 1

No, $\arcsin$ is not the inverse of the sine function. Rather, it is the inverse of the restriction of that function to $\left[-\frac\pi2,\frac\pi2\right]$. And the domain of $\arcsin$ is $[-1,1]\left(=\sin\left(\left[-\frac\pi2,\frac\pi2\right]\right)\right)$. So, $\arcsin(2\pi)$ isn't even defined.

Answer 2

On the unit circle $2\pi=0$ degrees, but the range of $\arcsin{x}$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$. Therefore, the answer is $0$ because $2\pi$ is outside of the range of $\arcsin{x}$.

Answer 3

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

is a triangular wave function whose range is $\pm \dfrac{\pi}{2}$ and whose roots $x_k =k \pi$ occur periodically.

When $x= 2 \pi, y=0 $ is the only correct value of ordinate in its graph.

Can be plotted on Mathematica/ Wolfram-Alpha etc.