TITLE: How do I prove that $\cos{\frac{2\pi}{7}}\notin\mathbb{Q}$?
QUESTION [16 upvotes]: How do I prove that $\cos{\frac{2\pi}{7}}\notin\mathbb{Q}$?
Should I use some geometrical approach or apagoge?

REPLY [9 votes]: Here is a slightly different way of phrasing Chris Eagle's argument:
If $\cos(2\pi/7)$ were rational, $X^2-2\cos(2\pi/7)\,X+1$ would be a factor of $$X^6+X^5+X^4+X^3+X^2+X+1$$ in $\mathbb Q[X]$, contradiction. 
EDIT. Let $n$ be an integer $\ge3$. Then the distinct conjugates of $\cos(2\pi/n)$ are the $\cos(2k\pi/n)$, where $k$ runs over the integers prime to $n$ lying between $1$ and $n/2$. 
In particular the degree of $\cos(2\pi/n)$ over $\mathbb Q$ is $\varphi(n)/2$, where $\varphi$ is Euler's totient function. 
This results immediately from the following well known facts. 
If $\zeta$ is a primitive $n$-th root of $1$ in $\mathbb C$, then $\mathbb Q(\zeta)/\mathbb Q$ is Galois of degree $\varphi(n)$. 
Let $G$ be the Galois group and $k$ an integer prime to $n$. Then there is a unique $\sigma_k$ in $G$ satisfying $\sigma_k(\zeta)=\zeta^k$. 
In particular $k\mapsto\sigma_k$ induces an isomorphism from the multiplicative group of $\mathbb Z/n\mathbb Z$ onto $G$.