Analyze the function $f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$ The problem
The task is to find the intervals for which the function $f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$ is constant and for which it takes the form of:
$f \left( x \right) =a \cdot \arcsin \left( x \right) +b, \\a,b=const, a \neq 0$
Constants a,b are also to be calculated.
What I have done so far
I've defined the functions domain, calculated its derivative and the derivatives domain.
$D_{f} x \in \left\langle -1,1\right\rangle\\
f' \left( x \right) =\frac{\sqrt{x^{2}}+x}{\sqrt{x^{2}-x^{4}}}\\
D_{f'} x \in \left(-1,0\right) \cup \left(0,1\right)\\ f' \left( x \right) =0 \Leftrightarrow \sqrt{x^{2}}+x=0\\ \left|x\right|=-x \Leftrightarrow x \le 0$
Which indicates that our function is constant for x'es from the interval (-1,0).
That's where I currently am. I don't know how to go about the second part of the task, that is finding the intervals for which the function behaves as:
$f \left( x \right) =a \cdot \arcsin \left( x \right) +b$
and calculating a,b.
Any help, greatly appreciated.
 A: I do not know what you are allowed to use. From Abramowitz/Stegun 4.3.45 and 4.4.2
(or http://dlmf.nist.gov/4.16#T3 and http://dlmf.nist.gov/4.23.E11)
we have 
$$\arcsin(\sqrt{1-x^2}) = \arccos x, \quad 0 \le x \le 1$$
$$\arccos x  = \frac{\pi}{2} - \arcsin x$$
Combining these results you get for $0 \le x \le 1$
$$\arcsin x-\arcsin(\sqrt{1-x^2}) = \arcsin x -\left(\frac{\pi}{2} - \arcsin x\right) = 2\arcsin x -\frac{\pi}{2}$$
and therefore $a=2$ and $b=-\frac{\pi}{2}.$
A: Hint
Let us consider  $$f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$$ and $$g \left( x \right) =a \cdot \arcsin \left( x \right) +b$$ In order they be identical (and this can only happen in the range [$0<x<1$]), a minimum requirement is that there will be two points where their values should be equal. Let us take the two limiting points $x=0$ and $x=1$. So, $f(0)=-\frac{\pi }{2}$ and $g(0)=b$, $f(1)=\frac{\pi }{2}$ and $g(1)=a \frac{\pi }{2}+b$; you then have two very simple linear equations to solve for $a$ and $b$.   
I am sure that you can take from here.
A: As $f$ is defined on $[-1,\,1]$, you may, for any $x\in[-1,\,1]$ take a unique $\alpha\in[-\pi/2,\,\pi/2]$ such that $\sin\alpha=x$. This number is by definition called $\alpha=\arcsin x$. Now you know that $\sin^2\alpha+\cos^2\alpha=1$ and $\cos\alpha\geq0$. This gives $\cos^2\alpha=1-\sin^2\alpha=1-x^2$
and therefore $\cos\alpha=\sqrt{1-x^2}$. Since $\sin(\pi/2-\alpha)=\cos\alpha$, you can conclude that 
$$\arcsin(\cos\alpha)=\pi/2-\alpha \quad\text{if}\quad\pi/2-\alpha\in[-\pi/2,\,\pi/2],$$
but if $\pi/2-\alpha\notin[-\pi/2,\,\pi/2]$ this means that $\alpha<0$, and thus that $\alpha\in[-\pi/2,\,0)$ you should then use $\sin(\pi/2+\alpha)=\cos\alpha$ and get
$$\arcsin(\cos\alpha)=\pi/2+\alpha \quad\text{if}\quad\alpha\in[-\pi/2,\,0].$$
Altogether, you have, for $x\geq0$, $f(x)=\alpha-\pi/2+\alpha=2\alpha-\pi/2$
and for $x\leq0$ $f(x)=\alpha-\pi/2-\alpha=-\pi/2$. 
A: Starting from what the OP found for the derivative of $f(x)=\arcsin(x)-\arcsin(\sqrt{1-x^2})$, we have
$$f'(x)={\sqrt{x^2}+x\over\sqrt{x^2-x^4}}=
\begin{cases}
0&\text{for }x\lt0\\
\displaystyle{2\over\sqrt{1-x^2}}&\text{for }x\gt0
\end{cases}$$
It follows that $f(x)$, which is continuous on $[-1,1]$, is constant on the interval $[-1,0]$ and equal to $2\arcsin(x)$ plus that constant on $[0,1]$. Since $f(0)=\arcsin(0)-\arcsin(1)=-\arcsin(1)=-\pi/2$, we get
$$\arcsin(x)-\arcsin\left(\sqrt{1-x^2}\right)=
\begin{cases}
-\displaystyle{\pi\over2}&\text{for }x\le0\\
2\arcsin(x)-\displaystyle{\pi\over2}&\text{for }x\ge0
\end{cases}$$
