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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?

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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?

Takahiro Waki
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E. Joseph
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2 Answers2

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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, $\sin \frac{π}{8}=\sin \left(\frac{π}{2^{2+1}}\right)=\frac 12 \sqrt{2-\sqrt{2}}.$

Dhrubajyoti Bhattacharjee
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2

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
Paul
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