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We have the following identities:

$\sin(\frac{\pi}{1})=0$

$\sin(\frac{\pi}{2})=1$

$\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{\sqrt{4}}$

$\sin(\frac{\pi}{4})=\frac{1}{2}$

$\sin(\frac{\pi}{5})=\frac{\sqrt{5-\sqrt{5}}}{\sqrt{8}}$

On the other hand it is known that for integer $n$ , $\sin(\frac{\pi}{n}) $ can not always be expressed by real radicals.

!!! With expressed by real radicals, I mean expressible with sums and products and integers and n'th roots of such positive expressions. So no stuff like $(-1)^{1/7}$ or $i$ allowed.
!!!

And it is known that for integer $n$ , $\sin(\frac{\pi}{n}) $ can not always be expressed by $\sin(\frac{\pi}{q_i}) $ for a collection of $i$ integers $q_i$ relatively prime to $n$.

So combining those ideas, I wonder

When is $\sin(\frac{\pi}{n}) $ not expressible by real radicals and $\sin(\frac{\pi}{q_i}) $ ?

I assume the factorization of $n$ matters and it is only possible when it is possible for all primes that divide $n$

Im not completely sure though but this lead me to consider the main question :

For a prime $p$ :

When is $\sin(\frac{\pi}{p}) $ not expressible by real radicals and $\sin(\frac{\pi}{q_i}) $ ?

See also :

Algebraic numbers expressible in terms of real-valued radicals

mick
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  • $y=\sin x$ then let $w=\sqrt{1-y^2}+iy.$ Then when $|x|<\frac{\pi}2,$ $$\sin (x/n)=\frac1{2i}\left(\sqrt[n] w-\sqrt[n]{\overline w}\right)$$

    The question of "solving with radicals" usually allows the multivalued complex $n$th roots.

     – Thomas Andrews Feb 15 '23 at 23:49
  • For other $x,$ we can still get $\sin(x/n),$ too. For example, when $x=\pi,$ we choose $w=-1$ and $$\sin(\pi/n)=\frac1{2i}\left(\sqrt [n]{-1}-\frac1{\sqrt[n]{-1}}\right)$$ – Thomas Andrews Feb 15 '23 at 23:57
  • I don't see what you mean with "$\sin(\pi /p)$ cannot be solved by radicals", it is clearly wrong with the standard definition of solvable by radicals. Maybe you mean allowing only $n$-th roots of positive real numbers? If so this is not "solvable by radicals". – reuns Feb 16 '23 at 00:22
  • @ThomasAndrews You can embed $\Bbb{Q}(\zeta_m)$ into a tower $F_n/F_0$ such that $F_0=\Bbb{Q}$ and $F_n\cong F_{n-1}[x]/(x^{k_n}-a_n)$ with $a_n\in F_{n-1}$, no need of ambiguous things such as $(-1)^{1/3}$. – reuns Feb 16 '23 at 00:29
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    See https://math.stackexchange.com/a/2611466/276986 for solving the (totally) real radical problem – reuns Feb 16 '23 at 00:45
  • Thanks , I clarified and edited. vote to reopen. – mick Feb 16 '23 at 12:16
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    You should delete and reask, making clear what is real radical and including a link to https://math.stackexchange.com/a/2611466/276986 which does solve the problem when the $\sin(\pi /q_i)$ are not included. For now your question is messy, you can give some examples of real radical expressions but only at the end. – reuns Feb 16 '23 at 12:24

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