How to define $\arcsin(\sin\theta)$ on picewise sub-intervals? We know that $\sin: [-\pi/2,\pi/2]\to [-1,1]$ is invertible and $\arcsin: [-1,1]\to  [-\pi/2,\pi/2]$.
But I want to define $\arcsin(\sin\theta)$.      
If $\theta\in[\pi/2,3\pi/2]$, i.e., $\pi/2\leq \theta\leq 3\pi/2$, then
  $-\pi/2\leq \theta-\pi\leq \pi/2$ and so
  $\arcsin(\sin\theta)=\theta-\pi$.    
Again, if   $\theta\in[(2n-1)\pi/2,(2n+1)\pi/2]$ for  $n\in\mathbf{Z}$, then 
$-\pi/2\leq \theta-n\pi\leq \pi/2$  and so
$\arcsin(\sin\theta)=\theta-n\pi$ for all integers $n$.      
I don't know whether my approach is true or not. 
 A: Prelude
$\arcsin(\theta)$ is not the inverse of $\sin(\theta)$, but rather 
it's the inverse of $\sin(\theta)$ restricted to the domain $[-\pi/2,\pi/2]$.
Thus, restricted $\sin\theta$ has 
domain = $[-\pi/2,\pi/2]$
range = $[-1,1]$
and its inverse, $\arcsin(\theta)$ has
domain = $[-1,1]$
range = $[-\pi/2,\pi/2]$
when $\sin\theta$ is used in an expression such
as $\arcsin(\sin(\theta))$, the $\sin\theta$ part is ordinary $\sin$,
not restricted $\sin$. The composition is well defined
since the range of $\sin(\theta)$ is contained in the domain
of $\arcsin(\theta)$.
FInally, let me remark that the functions
$$\sin(\arcsin\theta)$$
and
$$\arcsin(\sin\theta)$$
Are completely different.
The function $\sin(\arcsin(\theta))$ has
domain = $[-1,1]$
range = $[-1,1]$
and is identically equal to $\theta$ on its domain, hence its
graph is the portion of the graph $y = \theta$ for $-1 \leq \theta \leq 1$.
The function $\arcsin(\sin(\theta))$ has
domain = $[-\infty,\infty]$
range = $[-\pi/2,\pi/2]$
and is a continuous, piecewise-linear with period $2\pi$ 
whose graph has a "saw-tooth" shape. 
Really last final remark:
while its correct to simplify the expression
$\sin(\arcsin(\theta))$ to $\theta$
for all $\theta$ for which it's defined (that is $-1 \leq \theta \leq 1$),
it's incorrect (and is a common student error) to 
simplify 
$\arcsin(\sin(\theta))$ to $\theta$
unless it is known that $-\pi/2 \leq \theta \leq \pi/2$
