$\arcsin(\sin x)$ explanation? First off, I know this is a duplicate of this question. I'm asking this because I still don't quite understand the answer given there. But first, some graphs!


*

*$y=\sin(x)$ 

*$y=\arcsin(x)$

*$y=\sin(\arcsin(x))$

*$y=\arcsin(\sin (x))$
What I don't understand is why $f(x) = \arcsin(\sin(x))$ looks like it does. As stated in the previously linked question, $f(x)=\arcsin(x)$ is only defined between $[-1,1]$. So, how does it effect the rest of the graph? Shouldn't the domain be only values of $\sin x$ within the domain of $\arcsin x$?
Thanks!
 A: 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.
A: 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 $(f\circ g)$.
A: 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. 

