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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?

enter image description here

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?

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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
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  • $\begingroup$ You are right... I must have misunderstand something. $\endgroup$ – E. Joseph Feb 19 '17 at 14:53

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