Prove $\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$ using the half angle identities. Given is the sequence $x_1=0,\; x_{n+1}=\sqrt{2+x_n}$. Prove: $$\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$$
Hint:
Use the following formulas:
$$\cos\left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos x}{2}}$$
$$\sin\left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos x}{2}}$$
Any idea how to solve this problem?

By mathematical induction it can be proved that $x_n=2\cos(\frac{\pi}{2^n})$.
for base of induction $x_0=0=2\cos(\frac{\pi}{2})$. Also we have $x_{n+1}=\sqrt{2+x_n}=\sqrt{2+2\cos(\frac{\pi}{2^n})}=2\sqrt{\frac{1+\cos(\frac{\pi}{2^n})}{2}}=2\cos(\frac{\pi}{2^{n+1}})$. Therefore $\sqrt{2-x_n}=2\sqrt{\frac{1-\cos(\frac{\pi}{2^n})}{2}}=2\sin(\frac{\pi}{2^{n+1}})$, so $\lim_{n\rightarrow \infty}2^n\sqrt{2-x_n}=\lim_{n\rightarrow \infty}2^{n+1}\sin(\frac{\pi}{2^{n+1}})=\lim_{n\rightarrow \infty}\pi( \frac{\sin(\frac{\pi}{2^{n+1}})}{\frac{\pi}{2^{n+1}}})=\pi$
