Closed algebraic form of $\cos(\frac{\pi}{7})$ We all know that $\cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$, $\cos(\frac{\pi}{4})=\frac{\sqrt 2}{2}$ and $\cos(\frac{\pi}{3})=\frac{1}{2}$. One can also prove  that $\cos(\frac{\pi}{5})=\frac{\sqrt 5+1}{4}$. But it seems that $\cos(\frac{\pi}{7})$ cannot be put in a closed form algebraic expression. Is there a proof of such a claim? I feel like this has to do with the Abel–Ruffini theorem.
The same for $\cos(\frac{\pi}{8})$ and $\cos(\frac{\pi}{9})$. But $\cos(\frac{\pi}{10})=\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}}$.

For any $n\geq 3$, $\cos\left(\frac{2\pi}{n}\right)$ is al algebraic number over $\mathbb{Q}$ with degree $\frac{\varphi(n)}{2}$, since its minimal polynomial can be computed from $\Phi_n(x)$ in a straightforward way. Our case is given by $n=14$, where the minimal polynomial of $\cos\left(\frac{\pi}{7}\right)$ is given by $q(x)=8x^3-4x^2-4x+1$. By letting $\omega=\frac{-1+\sqrt{-27}}{2}$, through the cubic formula we get
$$ \cos\left(\frac{\pi}{7}\right) = \frac{1}{6}\left(1+\frac{7^{2/3}}{\omega^{1/3}}+7^{1/3}\omega^{1/3}\right). $$
It might be interesting to notice that the LHS is very close to $\frac{9}{10}$.
