A formula for $\sin(\pi/2^n)$ May be this a duplicate, but I did not find any question related.
I found the following formula, but there was no proof of it:
$$2\sin\left(\frac{\pi}{2^{n+1}}\right)=\sqrt{2_1-\sqrt{2_2+\sqrt{2_3+\sqrt{2_4+\cdots\sqrt{2_n}}}}}$$
where
$$2_k=\underbrace{222\cdots222}_{k\text { times}}.$$
(The number $22$ is twenty-two for instance, and not $2\times 2=4$.)

Do you know a proof of this result? Do you know any references?


I think one way to prove it would be to deal with regular polygons inside a circle and play the angles and trigonometry. 
Do you think it would work?
Is there a different way to proceed?
 A: It's not true
The following Mathematica code defines the formulas left and right of the equals sign and shows the approximate values for $n$ from 2 to 10:
r2[n_] := FromDigits[ConstantArray[2, n]]
f1[n_] := 2 Sin[Pi/2^(n + 1)]
f2[n_] := Sqrt[2 - Sqrt[Fold[Sqrt[#1] + #2 &, Table[r2[k], {k, n, 2, -1}]]]]
Table[{n, f1[n] // N, f2[n] // N}, {n, 2, 10}] // Grid

The result is below. Note that the right formula even gives a complex result.
n    left        right
2    0.765367    1.64025i
3    0.390181    2.01854i
4    0.196034    2.04871i
5    0.0981353   2.04964i
6    0.0490825   2.04965i
7    0.0245431   2.04965i
8    0.0122718   2.04965i
9    0.00613591  2.04965i
10   0.00306796  2.04965i

A: Here's my repeated half-angle approach (I know, this is definitely not a great way to deal with it, but still am posting it here. This is my first answer, here in this website, so please bear with me..):
We know
$2\cos^2 \theta =1+\cos 2\theta\implies \cos \theta =\sqrt{\frac{1+\cos 2\theta}{2}}.$
Taking positive sign because I am going to take $\theta=\frac{π}{2^n}, n\ge 2.$
So $2\cos \theta =\sqrt{2+2\cos 2\theta}.$
Let $\theta=\frac{π}{2^n}, n\ge 2.$ Then
\begin{align}
2\cos \left(\frac{π}{2^n}\right)& =\sqrt{2+2\cos \left(\frac{π}{2^{n-1}}\right)} \;\;(1 \text{ radical}) \\\\
&=\sqrt{2+\sqrt{2+2\cos \left(\frac{π}{2^{n-2}}\right) } }\;\;(2\text{ radicals})\\\\
&\vdots\\\\
&=\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\cos \frac{π}{2}}}}\;\;(n-1\text{ radicals}) \\\\
&=\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}\;\;(n-1 \text{ radicals})\\\\
&=A_{n-1},\text{ say}.
\end{align}
Therefore, 
$2\cos \left(\frac{2π}{2^{n+1}}\right) =A_{n-1}$
$\implies 2\left[1-2\sin^2 \left(\frac π{2^{n+1}}\right) \right]=A_{n-1}$
$\implies 4\sin^2 \left(\frac π{2^{n+1}}\right) =2-A_{n-1}$
$\implies 2\sin \left(\frac{π}{2^{n+1}}\right) =\sqrt{2-A_{n-1}}$
Thus, 
$\sin \left(\frac{π}{2^{n+1}}\right) =\frac 12 \sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}\;\;(n\text{ radicals}), \forall n\ge 2.$
As for example, 
$\sin \frac{π}{8}=\sin \left(\frac{π}{2^{2+1}}\right)=\frac 12 \sqrt{2-\sqrt{2}}.$
