Closed algebraic form of $\cos(\frac{\pi}{7})$

Question

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 $\cos(\frac{\pi}{10})=\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}}$.

Answer 1

Define Euler's $\phi$ function: $$\phi(1) = 1\\ \phi(p^e) = (p-1)\cdot p^{e-1},~ \mathrm{p~ prime} \\ \phi(m\cdot n) = \phi(m)\cdot \phi(n)$$ Then Theorem (Lehmer, A Note on Trigonometric Algebraic Numbers)

The degree of $\cos(\frac{2\pi k}{n}),~ n > 3,~ \gcd(k,n)=1$, is $\frac{\phi(n)}{2}$.

In this case $n=7\cdot 2= 14$ and $k=1$. Then the degree of $\cos(\frac{\pi}{7})$ is $\frac{\phi(2\cdot 7)}{2} = \frac{1\cdot 6}{2} = 3$. Since there is a cubic formula, then indeed this value can be found. But as mentioned in the comments it requires roots of negative numbers. So we continue.

As you are aware, not all algebraic numbers can be finitely expressed in terms of arithmetic operations and $n$th roots, there is in fact a way to know if the cosine of an angle has such a representation.

Theorem

The numbers $\cos(\theta)$ and $\sin(\theta)$ can be written using only rational numbers, addition, subtraction, multiplication, division and roots of positive numbers if and only if $\phi(n)$ is a power of 2.

Emphasis original.

The proof is outlined in "Somewhat More than Governors Need to Know about Trigonometry" by Garibaldi. The extension to include negative root-taking will lead you, eventually, to Gauss's Disquisitiones Arithmeticae and constructable numbers. The upshot is that the Theorem 2 is extended to be "... is a power of two and a product of distinct Fermat primes", with the understanding that the empty product is $1$.

Answer 2

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 $\frac{9}{10}$.