Representing $\cos(\frac{π}{11})$ as cyclic infinite nested square roots of $2$

Question

How can we represent $2\cos(\frac{\pi}{11})$ and $2\cos(\frac{\pi}{13})$ as cyclic infinite nested square roots of 2

I have partially answered below for $2\cos(\frac{\pi}{11})$ which I derived it accidentally.

Currently I have figured out for $q$ as denominator with patterns of $2^{n\pm1}$here

Is there any way to figure out pattern of nested radicals for other rational number $p \over q$ in $2\cos(\frac{p}{q})\pi$.

P.S. I am able to figure out that the rational number $p \over q$ must be as follows $1\over4$ < $p\over q$ < $ 1 \over 2$ as $1<$ $\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm...}}}$ $<2$

Answer 1

Let $2\cos(\frac{\pi}{11})$ represented with the help of half angle cosine formula as follows

$2\cos(\frac{\pi}{11}) = \sqrt{2 + 2\cos(\frac{2\pi}{11})}=\sqrt{2 + \sqrt{2+ 2\cos(\frac{4\pi}{11})}}$

$\sqrt{2 + \sqrt{2+\sqrt{2+ 2\cos(\frac{8\pi}{11})}}}$

Now $\frac{8\pi}{11}$ is more than $\frac{\pi}{2}$ which makes $2\cos(\frac{8\pi}{11})$ to $-2\sin(\frac{5\pi}{22})$ and in turn this is $-2\cos(\frac{3\pi}{11})$

Next step plugging in leads to $\sqrt{2 + \sqrt{2+\sqrt{2- 2\cos(\frac{3\pi}{11})}}}$

Further expansion is $\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2+ 2\cos(\frac{6\pi}{11})}}}}$ and this is represented as $\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2- 2\sin(\frac{\pi}{22})}}}}$ which is equal to $\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2-2\cos(\frac{5\pi}{11})}}}}$

Final expansion is beautiful to observe $\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+2\cos(\frac{10\pi}{11})}}}}}$ which is equal to

$\sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2\cos(\frac{\pi}{11})}}}}}$

$\therefore 2\cos(\frac{\pi}{11}) = \sqrt{2 + \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2\cos(\frac{\pi}{11})}}}}}$

As these steps get repeated for ever we can conclude that $$2\cos(\frac{\pi}{11})$$ can be represented as cyclic infinite nested square roots of 2 as follows and single cycle is $$\sqrt{2 +\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-}}}}}...$$ which repeats infinitely. This is represented simply as $$2\cos\frac{\pi}{11}=cin\sqrt2[2+3-]$$ [Refer here for reverse way to derive cosine angle from cyclic infinite nested square roots of 2][1]

Answer 2

For your solution

$ 2 \cos \left(\frac{\pi }{11}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{11}\right)}}}}} $

I have add it (with your name) to Page 3 of my Notebook:

http://eslpower.org/Notebook.htm

I found the same result independently on July,2021. Here is my method:

Since $(2^5+1) \bmod 33=0$, we have

$ 2 \cos \left(\frac{\pi }{33}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{33}\right)}}}}} $

and

$ 2 \cos \left(\frac{3\pi }{33}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{3\pi }{33}\right)}}}}} $

then we have

$ 2 \cos \left(\frac{\pi }{11}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{11}\right)}}}}} $

For $ 2 \cos \left(\frac{\pi }{13}\right) $, we may use the same method:

Since $(2^6+1) \bmod 13=0$, we have

$ 2 \cos \left(\frac{\pi }{65}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{65}\right)}}}}}} $

and

$ 2 \cos \left(\frac{5 \pi }{65}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2 \cos \left(\frac{5 \pi }{65}\right)}}}}}} $

then we have

$ 2 \cos \left(\frac{\pi }{13}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{13}\right)}}}}}} $

For more of my new results, I have post as my answers to your another question: https://math.stackexchange.com/a/4232525/954936