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so I stumbled upon this equation/formula, and I have no idea how to prove it. I don't know how should I approach it: $$ \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{\vphantom{\large A}2\,}\,}\,}\,}\ =\ 2\cos\left(\vphantom{\Large A}\pi \over 2^{n + 1}\right) $$

where $n\in\mathbb N$ and the square root sign appears $n$-times.

I thought about using sequences and limits, to express the LHS as a recurrence relation but I didn't get anywhere.

edit: Solved, thanks for your answers and comments.

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    $\begingroup$ This may help: Try taking $n=1$, $n=2$, $n=3$, ... in each case; see if you can prove for small $n$ first. You may be able to use proof by induction... $\endgroup$ – Pixel Jan 24 '14 at 14:14
  • $\begingroup$ @TooOldForMath Could you please elaborate on your thought there, I am not sure how this equation could be interpreted through polygons. $\endgroup$ – David Jan 24 '14 at 14:31
  • $\begingroup$ Can we prove this without using induction? $\endgroup$ – BAYMAX Jun 2 '18 at 7:57
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Hint:

Use induction and the half-angle formula for cosine.

Solution:

For $n=1$, the claim is true, since $\cos(\pi/4)=\sqrt{2}/2$. By the half-angle formula $$2\cos(x/2)=\sqrt{2+2\cos(x)}$$ Therefore $$\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}=\sqrt{2+2\cos\left(\frac{\pi}{2^n}\right)}=2\cos\left(\frac{\pi}{2^{n+1}}\right)$$ where in the left square root expressions there are $n$ square roots and in the first equality we have used the induction hypothesis that the claim holds for $n-1$.

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  • $\begingroup$ I may be wrong, but wasn't this Vieta's beautiful result? $\endgroup$ – imranfat Jan 24 '14 at 14:48
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    $\begingroup$ No but they are related. Vietas result is something like $\frac{2}{\pi}=\prod_{k=1}^\infty \frac{1}{2}\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}$ where there are $k$ square roots. If I'm not wrong, one can use this result here and a formula for $\pi$ involving $2^n$-gons to prove Vieta's formula. $\endgroup$ – J.R. Jan 24 '14 at 14:49

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