How do I prove that $\cos{\frac{2\pi}{7}}\notin\mathbb{Q}$? How do I prove that $\cos{\frac{2\pi}{7}}\notin\mathbb{Q}$?
Should I use some geometrical approach or apagoge?

Since Chebyshev polynomials have the property $T_n\left( \cos \theta \right) = \cos(n \theta)$, 
we can use this property with $n=7$ and $\theta = \frac{2 \pi}{7}$.
Let $z = \cos\left( \frac{2\pi}{7} \right)$. Then $T_7(z) = 1$
Let $p(x) = T_7(x)-1 = 64 x^7 - 112 x^5 +56 x^3 - 7x - 1 = (x-1)( 8 x^3 + 4 x^2 - 4 x -1 )^2.$    
So $z$ must be a root of $q(x) = (2 x)^3 + (2 x)^2 - 2 (2 x) - 1$. Observe that $2z$ is a root of a monic polynomial, and thus is an algebraic integer. If $z$ were rational, then $2z$ must be an integer. Since $-2 \le 2 \cos \theta \le 2$, there are only 5 possibilities to consider, which are ruled out because $q(1) = 7$, $q(-1) = q(0) = q(1/2) = -1$, $q(-1/2) = 1$.
