Prove that $ x_{k}=2^{k} \cdot \sin \frac{\pi}{2^{k}}$ equals $ x_{1}'=2, x'_{2}=2 \sqrt{2}, x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}$ Prove that $ x_{k}=2^{k} \cdot  \sin \frac{\pi}{2^{k}}$ equals $ x_{1}'=2, x'_{2}=2 \sqrt{2}, x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}$.
This is what I've managed:
$x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}= 2^{k} \cdot  \sin \frac{\pi}{2^{k}} \sqrt{\frac{2^{k+1} \cdot  \sin \frac{\pi}{2^{k}}}{2^{k} \cdot  \sin \frac{\pi}{2^{k}}+2^{k-1} \cdot  \sin \frac{\pi}{2^{k}}}}=2^{k} \cdot  \sin \frac{\pi}{2^{k}} \sqrt{\frac{2}{1+\cos \frac{\pi}{2^{k}}}}$
And I don't see how to proceed....

You can use that
$$\cos (2x) = \cos^2 x - \sin^2 x,$$
and hence
$$1 - \cos \frac{\pi}{2^k} = 1 - \cos^2 \frac{\pi}{2^{k+1}} + \sin^2 \frac{\pi}{2^{k+1}} = 2\sin^2 \frac{\pi}{2^{k+1}}.$$
Then write
$$\sqrt{\frac{2}{1+\cos \frac{\pi}{2^k}}} = \sqrt{\frac{2(1 - \cos \frac{\pi}{2^k})}{1 - \cos^2 \frac{\pi}{2^{k}}}} = \sqrt{\frac{4\sin^2 \frac{\pi}{2^{k+1}}}{\sin^2 \frac{\pi}{2^k}}} = \frac{2\sin \frac{\pi}{2^{k+1}}}{\sin \frac{\pi}{2^k}},$$
since all the sines involved are non-negative.
