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?

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 $x=1$.
