$\arcsin(\sin x)$ explanation?

Since $\sin(x)$ is always in the interval $[-1,1]$ where $\arcsin$ is defined, $\arcsin(\sin(x))$ is defined for all real $x$.

$\arcsin(y)$ is defined as the number $x$ within the interval $[-\pi/2, \pi/2]$ where $\sin(x) = y$. So for $x$ in the interval $[-\pi/2, \pi/2]$, $\arcsin(\sin(x)) = x$ satisfies that definition. For $x$ in the interval $[\pi/2, 3 \pi/2]$, $\arcsin(\sin(x))$ can't be $x$, but it can be $\pi - x$ which is in the interval $[-\pi/2, \pi/2]$ (notice that $\sin(\pi - x) = \sin(x)$). So that takes care of this part of the graph:

Now notice that $\sin(x)$ is periodic with period $2\pi$, so the graph looks the same if you shift it left or right by $2\pi$. That takes care of the rest of it.

So let $f(x)=\sin(x)$ and let $g(x)=\arcsin(x)$. Let $\operatorname{Dom}(f)$ be the domain of $f$ and let $\operatorname{Rng}(f)$ be the range of $f$. Thus we have $$ \begin{align} \operatorname{Dom}(f)&=\mathbb{R}\\ \operatorname{Rng}(f)&=[-1,1]\\ \operatorname{Dom}(g)&=[-1,1]\\ \operatorname{Rng}(g)&=\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \end{align} $$

If we restrict the $\operatorname{Dom}(f)$ to $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, then $f$ and $g$ are inverses of one another and

$$ (f\circ g)(x)=(g\circ f)(x) = x $$

on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Now let $\operatorname{Dom}(f)=\mathbb{R}$ and consider $(f\circ g)(x)$. This graph should remain unchanged since for any $x\not\in[-1,1]$, $(f\circ g)(x)$ is not defined and so no picture can be drawn on $\mathbb{R}\setminus[-1,1]$.

On the other hand, when we let $\operatorname{Dom}(f)=\mathbb{R}$ and consider $(g\circ f)(x)$, this composition is defined over all of $\mathbb{R}$, but it behaves in two different ways:

1.) On the intervals $\left[\frac{\pi(2k-1)}{2},\frac{\pi\cdot(2k+1)}{2}\right]$ when $k$ is odd: As $x$ goes from $\frac{\pi(2k-1)}{2}$ to $\frac{\pi(2k+1)}{2}$, $f(x)$ goes from $-1$ to $1$. Thus $(g\circ f)$ goes from $\frac{\pi(2k-1)}{2}$ to $\frac{\pi(2k+1)}{2}$, forming a positively sloped line.

2.) On the intervals $\left[\frac{\pi(2k-1)}{2},\frac{\pi\cdot(2k+1)}{2}\right]$ when $k$ is even: As $x$ goes from $\frac{\pi(2k-1)}{2}$ to $\frac{\pi(2k+1)}{2}$, $f(x)$ goes from $1$ to $-1$. Thus $(g\circ f)$ goes from $\frac{\pi(2k+1)}{2}$ to $\frac{\pi(2k-1)}{2}$, forming a negatively sloped line.

This is why you see the jigsaw pattern when you graph $(g\circ f)$ and don't see any change in $(f\circ g)$.

For $\pi\ge x\ge \frac{1}{2}\pi$, we have $\sin[x]=\sin[\pi-x]$. Therefore we have $$ \arcsin(\sin[x])=\arcsin(\sin[\pi-x])=\pi-x $$ Similarly for $-\pi\le x\le -\frac{\pi}{2}$, we have $\sin[x]=\sin[-\pi-x]$. Therefore we have $$ \arcsin(\sin[x])=\arcsin(\sin[-\pi-x])=\arcsin(-\sin[\pi+x])=-\arcsin(\sin(\pi+x))=-\pi-x $$

I think the rest of the graph follows by the same argument. I attached a graph, I do not know if it helps.