Find $\lim_{n\to\infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}$. 
Find $\displaystyle \lim_{n\to\infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}$.

By geometry method, I know that this is $\pi$. But is there algebraic method to find this
? 
Thank you.
 A: Let $2=4\cos^{2}{\theta}$
we get
$$\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}=4\left|\sin{\frac{\theta}{2^n}}\right|$$
And as $n \longrightarrow \infty$ ,$\frac{\theta}{2^n} \longrightarrow 0$. Hence $\displaystyle \lim_{n \to \infty}2^n\sin{\frac{\theta}{2^n}}=\theta$
$\text{Limit}=|4\theta|$
As $\cos{\theta}=\pm \frac{1}{\sqrt{2}}$ we have $\theta=\pm \frac{\pi}{4}+n\pi$ $\quad ,n \in \mathrm{I}$
Here we take $\theta =\pm \frac{\pi}{4}$ (To know why please refer the note at the bottom) .
Hence the value of limit is $\pi$

Note: the reason why $\theta=\pm \frac{\pi}{4}$ only.  
We know that by the definition of square-root function $\sqrt{x^2}=|x|$ using the same reasoning. We have $\sqrt{2+2\cos{x}}=2\cos{\frac{x}{2}}$ iff $\cos{\frac{x}{2}}\ge 0 \Rightarrow \frac{x}{2}\in \left[2n\pi,\frac{\left(4n+1\right)\pi}{2}\right] \cup \left[\frac{\left(4n-1\right)\pi}{2},2n\pi\right] ,n\in \mathbb{I}$.
As we applied the process for $n$ times for every whole number $j \le n$ we need to have the following inequality $\cos{\frac{x}{2^j}}\ge 0 \Rightarrow \frac{x}{2^j}\in \left[2N\pi,\frac{\left(4N+1\right)\pi}{2}\right] \cup \left[\frac{\left(4N-1\right)\pi}{2},2N\pi\right] ,N\in \mathbb{I}$.
Now it is easy to reason out that our $\theta$ must belong to the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
