How is $\sqrt {2+\sqrt {2+\sqrt {2+}}} ... n $ times = $2\cos( π/2^{n+1})$? 
How is $\sqrt {2+\sqrt {2+\sqrt {2+}}} ... n $ times = $2\cos( π/2^{n+1})$?

No idea. Please help. I found this identity in a solution of a problem related to limits. Also if any more identities like this then please let me know in the answer column.
 A: Recall the identity
$$2\cos(\theta) = \sqrt{2+2\cos(2\theta)}$$
This means
$$2\cos(\theta) = \sqrt{2+\sqrt{2+2\cos(4\theta)}} = \sqrt{2+\sqrt{2+\sqrt{2+2\cos(8\theta)}}} \text{  and so on }$$
Setting $\theta = \dfrac{\pi}{2^{n+1}}$, and going on for $n$ terms, we see that we end up with $\cos(\pi/2)$, which is indeed zero.
EDIT All the square roots come with a positive sign, since each of the cosine term is of the form $\cos(\pi/2^k)$, where $k \geq 1$, i.e., $\pi/2^k \in [0,\pi/]$, where the cosine is non-negative.
A: $$2 \cos{\left (\frac{\pi}{2^{n+1}} \right )}  = 2 \sqrt{\frac{1+\cos{\left (\frac{\pi}{2^{n}} \right )}}{2}} = \sqrt{2 + 2 \cos{\left (\frac{\pi}{2^{n}} \right )}}$$
Keep on going...until you reach $n=2$.
A: one way is to use recurrence,
if $n=0$, then $0=0$... OK
suppose now that the relation, say $H_n$, is true and prove that $H_{n+1}$ is verified.
squaring the two sides of $H_{n+1}$
${(\sqrt{2+\sqrt{2+...}})}^2$ n+1 times = 2+$\sqrt{2+\sqrt{2+...}}$ n times
$= 2+2cos(\dfrac{\pi}{2^{n+1}})$
$= 4cos(\dfrac{\pi}{2^{n+2}})$ (according to Ron Gordon below in his hint)
...
Then $H_{n+1}$ is true and finally, you have thus your relation.
