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Is there nice radical expression for $$\cos\left(\frac{\pi}{2^k+1}\right)?$$
Example: $\cos\left(\dfrac{\pi}{5}\right)=\dfrac{\sqrt{5}+1}{4}$.
Please provide some concrete examples.
Also please provide a general procedure.
I would like to handle $$\cos\left(\frac{a\pi}{2^k+1}\right)$$ as well for $a\in\{0,1,2,\dots,k-1\}$.
For the form $p=2^k+1$, you can find an expression in real (not complex) radicals for,
$$\cos\Big(\frac{n\pi}{p}\Big)$$
and $n$ a positive integer only when $p$ is a Fermat prime, hence $p=3,5,17,257,65537$. For higher $p$ there can be several ways to express it.
I. For $p=17$:
$$\cos\frac{2\pi}{17}=\frac{1}{4}\Bigl(\frac{1}{x}+\sqrt{x}\,(17+4\sqrt{17})^{1/4}\Bigr),\quad x =\frac{1}{4}\Bigl(1-\sqrt{17}+\sqrt{\bigl(1-\sqrt{17}\bigr)^2+16}\Bigr)$$
or, from this post,
$$\cos\frac{2\pi}{17}=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}$$
II. For $p=257$:
See this post.
You can use the multiple angle formulas. For example you can compute sin5$\theta$ which will give you a polynomial p of the 5th degree in sin$\theta$ and cos$\theta$, or perhaps in cos$\theta$ alone if you are lucky. If you let $\theta = \pi/5$ then sin5$\theta$ = 0= p(cos$\theta$). If you can solve it, you've got an expression for sin($\pi/5$). From that you can deduce an expression for cos($\pi/5$) and it will be what you have above.
To create the multiple angle forumlas start with sinx = $\frac{e^{ix}-e^{-ix)}}{2i}$. For example sin5x =$\frac{e^{5ix}-e^{-5ix)}}{2i}$ . If d = $e^{ix}$ then this expression can be written as $ \frac {d^5-d^{-5}}{2i}$ = 1/2i$(cosx + isinx)^5$ - 1/2i$(cosx - isinx)^5$ Expand these with the binomial theorem and you will have your polynomial.
You can save some trouble by looking up the multiple angle formulas; and most likely someone has solved the polynomials up to some value of n.
The nested radicals are closely related to trigonometry and binary numbers. In these frameworks, such expressions are possible and generalize easily. The example below goes beyond the requirements of this question but is indicative of the possibilities that exist.
$2 \cos (\frac {4 \pi} {2 ^ 4 + 1}) = \sqrt {2_1 + \sqrt {2_2 - \sqrt {2_3 + \sqrt {2_4 + 2 \cos (\frac {4 \pi} {2 ^ 4 + 1})}}}}$
This expression can be generalized by introducing a variable $n$, the value of which increases when an analogous number of radicals with positive signs interpose between the terms $2_2$ and $2_{n - 2}$:
$2 \cos (\frac {4 \pi} {2 ^ n + 1}) = \sqrt {2_1 + \sqrt {2_2 + \ldots + \sqrt {2_{n - 2} - \sqrt {2_{n - 1} + \sqrt {2_n + 2 \cos (\frac {4 \pi} {2 ^ n + 1})}}}}}$
example with $n = 5$:
$2 \cos (\frac {4 \pi} {2 ^ 5 + 1}) = \sqrt {2_1 + \sqrt {2_2 + \sqrt {2_3 - \sqrt {2_4 + \sqrt {2_5 + 2 \cos (\frac {4 \pi} {2 ^ 5 + 1})}}}}}$
To learn more download the pdf file titled "Nested radicals - relation to trigonometry and binary numbers" found in https://independent.academia.edu/%CE%93%CE%99%CE%A9%CE%A1%CE%93%CE%9F%CE%A3%CE%A0%CE%9B%CE%9F%CE%A5%CE%A3%CE%9F%CE%A3 - four pages with elemental but not trivial results. It is up to you to take advantage of them, as this space is virtually unexplored.
Some general solution can be arrived by combining Trigonometry and infinite nested square roots of 2
My answer is very long. Please bear with me. But it is basic
Basic Trigonometric formula
$\cos2\theta = 2\cos^2\theta -1$ ==> $2\cos2\theta =4\cos^2\theta -2$ ==> $4\cos^2\theta = 2 + 2\cos2\theta $ $$2\cos\theta =\sqrt{2+2\cos2\theta} ...... \text Equation 1$$
$\cos2\theta = 1 - 2\sin^2\theta$ ==> $2\cos2\theta =2 - 4\sin^2\theta$ ==> $4\sin^2\theta = 2 - 2\cos2\theta $ $$2\sin\theta =\sqrt{2-2\cos2\theta} ...... \text Equation 2$$
First example
Now solving infinite nested radicals having only 2 having '-'(minus signs) throughout
$\sqrt{2-\sqrt{2-\sqrt{2-...}}}$ as follows $x = \sqrt{2-x}$ Now substituting $x = 2\cos\theta$ $2\cos\theta = \sqrt{2-2\cos\theta} = 2\sin\frac{\theta}{2}= 2\cos(\frac{\pi}{2}- \frac{\theta}{2})$
Now $\cos\theta = \cos(\frac{\pi}{2}- \frac{\theta}{2})$
$\theta = \frac{\pi}{2}- \frac{\theta}{2}$ ==>$ 2\theta = \pi-\theta ==> 3\theta = \pi$ ==> Therefore $ \theta = \frac{\pi}{3}$ = $\frac{2^0}{2^1+1}\cdot\pi$
Now easy to understand that $2cos\frac{\pi}{3} = 2\cdot\frac{1}{2} =1 $ which is the solution for infinite nested square roots of 2 having all '-'signs
2nd example
$$\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+...}}}}$$ ... Repetition of alternate '+' & '-' infinitely
Here $x = \sqrt{2-\sqrt{2+x}}$ Again substitution of $ x = 2\cos\theta$
$2\cos\theta = \sqrt{2-\sqrt2+2\cos\theta}$ ==> $2\cos\theta = \sqrt{2-2\cos\frac{\theta}{2}} $ ==>$ 2\sin\frac{\theta}{4}$ ==> $2\cos(\frac{\pi}{2}- \frac{\theta}{4})$
Now $$\theta = \frac{\pi}{2} - \frac{\theta}{4}$$ Solving this will be $\theta = \frac{2\pi}{5}$ =$\frac{2^1}{2^2+1}\cdot\pi$
3rd example
$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+...}}}}}}$ as $x = \sqrt{2-\sqrt{2+\sqrt{2+x}}}$ $2\cos\theta = \sqrt{2-\sqrt{2+\sqrt{2+2\cos\theta}}}$ Solving will be $= 2\sin\frac{\theta}{8} = 2\cos(\frac{\pi}{2}-\frac{\theta}{8})$
Therefore $\theta + \frac{\theta}{8} = \frac{\pi}{2}$ ==> $\theta = \frac{4}{9}\cdot\pi$ and is $\frac{2^2}{2^3+1}\cdot\pi$ which is 80°
$$2\cos80° = \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+...}}}}}}$$
By generalizing above method, we can get
$$ 2\cos\theta = \sqrt{2-\sqrt{2+\sqrt{2+... \text{k times} +2\cos\theta}}} $$
By simplifying we get
$$ \theta = \frac{\pi}{2} - \frac{\theta}{2^{k+1}}$$
$$ \theta + \frac{\theta}{2^{k+1}} = \frac{\pi}{2}$$
Further simplification will lead to relationship between cyclic infinite nested square roots of 2 and cosine values for angles as
$$ (\sqrt{2-\sqrt{2+\sqrt{2+... \text{k times}}}})...= 2\cos(\frac{2^k}{2^{k+1}+1}\cdot\pi)$$ where $ k\in \mathbb{N}\cup\{0\} $
Addendum : Please remember within brackets is one cycle of nested square roots of 2 which repeats infinitely (LHS) (named as cyclic infinite nested square roots of 2) converging to cosine values(RHS)
As a continuation of above method
$$\lim_{k\to\infty}(\sqrt{2-\sqrt{2+\sqrt{2+... \text{k times}}}})...$$ converges to $\sqrt{2-2} = 0$ as $\sqrt{2+\sqrt{2+\sqrt{2+...}}} = 2$ and proves $2cos\frac{\pi}{2}=0$
Interestingly substitution of $k= -1$ leads to $2\cos(\frac{2^{-1}}{2^0+1})$ = $2\cos\frac{\pi}{4} = \sqrt2$
