Proving $\cos(\pi/2^n) =\frac{1}{2}\sqrt{2+\sqrt{2+\cdots\sqrt{2}}}$

Question

WolphramAlpha gives the following identities for the cosines of $\frac{\pi}{2^n}$:

$$ \cos\left(\frac{\pi}{8}\right)=\frac{1}{2}\sqrt{2+\sqrt{2}} $$ $$ \cos\left(\frac{\pi}{16}\right)=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}} $$ $$ \cos\left(\frac{\pi}{32}\right)=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} $$ and in general the following identity seems to be true $\forall n\in\mathbb{N}$:

$$ \cos\left(\frac{\pi}{2^n}\right)=\frac{1}{2}\sqrt{2+\sqrt{2+\ldots\sqrt{2}}} $$ With $n-1$ nested roots. Obviously for $n=2$ we have $\cos{\frac{\pi}{4}}$.

I have no idea where to start to prove this. Do you have any suggestions?

Hint: Apply the cosine half angle formula $2\cos(x/2)=\sqrt{2+2\cos(x)}$. — Christian E. Ramirez, Mar 21 '23 at 13:52
Just use induction, and the identity $\cos(2x)=2\cos^2(x)-1$. — Mark, Mar 21 '23 at 13:52

score 3 · Accepted Answer · answered Mar 21 '23 at 13:52

3

In essence, you want to show for positive integer $k$

$$\cos (\pi/2^k)=\sqrt{1+\cos(\pi/2^{k-1})\over 2},$$

which follows from the half-angle identity.

answered Mar 21 '23 at 13:52

Golden_Ratio

12,591

This was a duplicate. – Anne Bauval Mar 21 '23 at 13:53
1

@AnneBauval oups, merci de l'avoir signalé, Anne. – Golden_Ratio Mar 21 '23 at 13:55
Donc je pense que tu devrais détruire ta réponse. – Anne Bauval Mar 21 '23 at 14:13
@AnneBauval J'ai essayé, mais OP doit supprimer l'acceptation pour cette réponse – Golden_Ratio Mar 21 '23 at 14:23
Ah, d'accord! @FrankSoll can you please unaccept this answer, so that its author can delete it (as a duplicate)? – Anne Bauval Mar 21 '23 at 14:25

Proving $\cos(\pi/2^n) =\frac{1}{2}\sqrt{2+\sqrt{2+\cdots\sqrt{2}}}$

1 Answers1