How do you prove that $\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x$? I have the task to prove that
$$
     \arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x ,\left|x\right|\le 1
$$

I do not have any ideas from where I should start.
Can anyone help me solve it?
 A: Let $\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-x}})=y$ 
$\implies-\dfrac\pi2\le y\le\dfrac\pi2$
and $\dfrac12\sqrt{2-\sqrt {2-x}}=\sin y$
$\implies2-\sqrt {2-x}=(2\sin y)^2$
Using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$\sqrt {2-x}=2-4\sin^2y=2\cos2y$
$\implies x=2-(2\cos2y)^2=-2\cos4y$
$\implies\cos4y=-\dfrac x2$
$\arccos\left(-\dfrac x2\right)=\begin{cases}2\pi+4y  &\mbox{if }  -2\pi\le4y<-\pi \\-4y &\mbox{if }  -\pi\le4y<0\\ 4y &\mbox{if } 0\le4y\le\pi \\
2\pi-4y & \mbox{if }\pi<4y\le2\pi \end{cases}$
Now $\arccos\left(-\dfrac x2\right)=\dfrac\pi2-\arcsin\left(-\dfrac x2\right)$
$\implies\arccos\left(-\dfrac x2\right)=\dfrac\pi2+\arcsin\dfrac x2$   as $\arcsin(-u)=-\arcsin u$
A: You may just observe that, for $x \in [0,1)$, we have
$$
\left(\arcsin\left( \frac{1}{2} \sqrt{2-\sqrt {2-2x}}\right)\right)'=\frac14\frac1{\sqrt{1-x^2}}
$$ and we have
$$
\left(\frac{\pi}{8} + \frac{1}{4} \arcsin x\right)'=\frac14\frac1{\sqrt{1-x^2}}
$$ giving 

$$
\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x, \quad x \in [0,1]
$$ 

since both functions take the same value at $x=0$ and at $x=1$.
