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?

Note that as $x_1 \le 2$, $x_n \le 2 \forall n$. This can be shown inductively.
Thus, we may write $x_n = 2 \cos \theta_n$ for some $\theta_n \in [0, \pi/2]$, with $\theta_1 = \pi/2$
Now, $2 \cos \theta_{n+1} =   \sqrt{2(1+ \cos \theta_n)} = 2 \cos \frac{\theta_n}2 \implies \theta_{n+1} = \frac{\theta_n}{2} = \frac{\theta_1}{2^n} = \frac{\pi}{2^{n+1}}$. 
This implies that $2^n\sqrt{2 - x_n} = 2^{n+1} \sin \frac{\pi}{2^{n+1}}$. Finish off by using $\frac{\sin  x}{x} \to 1$ as $x \to 0$.
