Closed form of $\cos(\frac{\pi}{7})$ I saw someone derive the closed form for $\cos(\frac{\pi}{5})=\frac{\varphi}{2}$, and got inspired to try to find a closed form expression for $\cos(\frac{\pi}{7})$ using the same method. In doing this, you get that $\cos(\frac{\pi}{7})$ is one of the solutions to the following polynomial.
$$8x^3-4x^2-4x+1$$
If there is an accessible closed form for the roots of this polynomial, I cant seem to find it. I have tried factoring, guessing and also looked at wolfram alpha, which only gives massive closed forms involving imaginary numbers.
So if anyone can find a closed form or show that none is accessible, that would be great!
 A: I've got sad news: we cannot express the solution to this equation using sums, products, and roots. The reasoning is strange to say the least: we can only find the solution to $\cos(\pi/p)$ where $p$ is on odd prime when $p$ is a Fermat prime. This means that we can assign $p$ to be something like $3, 5, 17, ...$, but $7$ is not an option. That's why we can find the solution to $\cos(\pi/5)$ but not for $\pi/7$. As an aside, this means we can find a solution for $\cos(\pi/17)$, which is really weird.
Part of why this is not doable is that this cubic equation has three roots, and when a cubic has three roots we might not have an approach to finding them. Cardano's method finds a root if the cubic has only one root, but we can employ his method with complex numbers in an attempt to find the remaining roots. Once you employ complex numbers you will need to find the cosine of some strange angles, and after performing some simplifications, you will need the value of...you guessed it, $\cos(\pi/7)$. Thus solving this problem is circular and there is nothing we can do about it.
We can find the solution in terms of complex numbers if interested, but the solution is not particularly helpful.
Hope this helps!
A: Maple says
$$
\frac{1}{12}\,\sqrt [3]{-28+84\,i\sqrt {3}}+\frac{7}{3}\,{\frac {1}{\sqrt [3]{-28+84\,
i\sqrt {3}}}}+\frac16
$$
despite that $i$ in there, the result is a real number.  Of course complex cube roots may be done trigonometrically, but this whole question is to avoid the trigonometry in $\cos\frac{\pi}{7}$.
