Infinitely nested radical formulas for $\pi$ Has anyone come by the infinitely nested radical formula:
$$\Bigg\{\pi=\frac{12}{5}\cdot \lim\limits_{n \to \infty}  2^{n}\cdot\frac{1}{2^{\frac{2^{n}+1}{2^n}}} \sqrt{2^{\frac{2^{n-1}+1}{2^{n-1}}} -\cdot\cdot\cdot\cdot\sqrt{ 2^{\frac{3}{2}} +\sqrt{ 2^2 + (\sqrt{ 6}- \sqrt{ 2})}}}\Bigg\}$$ And can anyone prove it?
 A: From here:

If there exists an infinite iterative sequence $A=\{a_1, a_2,\cdots a_n\}$, and a sequence $B=\{c_1, c_2,\cdots c_n\}$ obtainable from $A$ such that for all $b, a_n =\cos b$, $a_{n+1} =\cos b/2, c_{n+1}=\sin b/2$. Then $$a_{n+1} =\sqrt{\frac{1+a_n}{2}} \text{  and  } c_n =\sqrt{\frac{1-a_n}{2}}$$

If $a_1 =\frac{\sqrt{6}-\sqrt{2}}{4}$, then $$c_2 = \frac{1}{2^{3/2}}\sqrt{2^2-(\sqrt{6}-\sqrt{2})}, c_3 = \frac{1}{2^{5/4}}\sqrt{2^{3/2}-\sqrt{2^2-(\sqrt{6}-\sqrt{2})}}$$ And thus $$c_n = \frac{1}{2^{\frac{2^n+1}{2^n}}} \sqrt{2^{\frac{2^{n-1}+1}{2^{n-1}}}-\cdots \sqrt{2^{5/4}+\sqrt{2^{3/2}+\sqrt{2^2 +(\sqrt{6}-\sqrt{2}})}}}$$ If $k = \frac{12}{5}\lim_{n \to \infty} 2^{n-1} c_n$, we get the required nested radical. Hope it helps.
A: I appreciate your work and finding. We may derive $\pi$ In infinite number of ways with the help of nested square roots
For explaining your derivation we have to think infinite nested radicals from the end.
Let us start from chord inside unit circle and value of lengths bisecting the chord
The length of chord is $2\sin\frac{75^\circ}{2}$ or 2 $\sin\frac{5\pi}{12}$$\frac{5\pi}{12}$ or $75^\circ$" />
The length of centre line bisecting the arc till the chord from centre is $\cos\frac{75^\circ}{2}$ as it bisects the angle, is equal to $\frac{(\sqrt6-\sqrt2)}{4}$ which can be simplified to $\frac{\sqrt{2-\sqrt3}}{2}$.
Length of bisected chord will be $2\sin\frac{75^\circ}{2}$ = $2\cdot\frac{\sqrt{2+\sqrt3}}{2}$
Now by bisecting the chord into halves leads to calculation of length from centre to midpoint of the chord which becomes smaller and smaller approaching towards unit of radius
Here you have applied 1/2 angle formula leading to nested square roots as follows
$\cos(\frac{75^\circ}{2^2})$ as $\frac{{\sqrt{2+{\sqrt{2-{\sqrt{3}}}}}}}{2}$, number of chords is 4
Now the number of chords within $75^\circ$ becomes 2^2
Sum of lengths of 4 chords will be $4\sin\frac{75^\circ}{2^2}$ $4\cdot\frac{{\sqrt{2-{\sqrt{2-{\sqrt{3}}}}}}}{2}$

Like this next iteration will be
$2^3\cdot \frac{{\sqrt{2-{\sqrt{2+{\sqrt{2-{\sqrt{3}}}}}}}}}{2}$
When we do this repeatedly we get $\cos(\frac{75^\circ}{2^n})$ as $\frac{{\sqrt{2+{\sqrt{2+...(n-1) times{\sqrt{2-{\sqrt{3}}}}}}}}}{2}$
Finally when we take sine value of the infinitesimal small angle $\frac{75^\circ}{2^n}$ it will be $$\frac{{\sqrt{2-{\sqrt{2+...(n-2) times{\sqrt{2-{\sqrt{3}}}}}}}}}{2}$$
Number of chords will be $2^n$ times
Chord length in radians is $r\theta$ = $\frac{5\pi}{12}$ which is nothing but $2^n\cdot \frac{{\sqrt{2-{\sqrt{2+...(n-1) times{\sqrt{2-{\sqrt{3}}}}}}}}}{2}$
Therefore $$\pi = \frac{12}{5}\cdot\lim_{n\to\infty} 2^n\cdot {\sqrt{2-{\sqrt{2+...(n-1) times{\sqrt{2-{\sqrt{3}}}}}}}}$$
Please forgive me for such a lengthy answer.
Please forgive for mistakes if you find any.
Thank you
My work 
