See http://www.problem-solving.be/pen/viewtopic.php?t=213 Alternatively, it's not hard to find the minimal polynomial, it's then standard to check that it is irreducible (it has degree $3$, so it suffices to show that there are no roots).

Just as addendum the elementary way: the minimal polynomial is $8x^{3}-4x^{2}-4x+1$, we see that it is one by setting $2 \cos \frac{\pi}{7} = -\zeta_7-\zeta_7^{-1}$ (here $\zeta_7=e^{\frac{8 \pi i}7}$ which results in $-\zeta_7^3-\zeta_7^2-\zeta_7-1-\zeta_7^{-1}-\zeta_7^{-2}-\zeta_7^{-3}$, being zero by $0=\zeta_7^7-1=(\zeta_7-1)(\zeta_7^6+\zeta_7^5+\zeta_7^4+\zeta_7^3+\zeta_7^2+\zeta_7+1)$. The polynomial is irreducible because if we set $x=y-1$ we get $8y^{3}-28y^{2}+28y-7$, being irreducible by Eisenstein.