TITLE: How to prove that $\cos (2\pi/n)$ is algebraic?
QUESTION [7 upvotes]: Prove that for all integers $n$, $\cos (2\pi/n)$ is an algebraic number.

REPLY [9 votes]: We have $\zeta_n = e^{2\pi i/n}$ being an algebraic integer for all $n$ simply because it is a root of $x^n -1$. Now notice that $\zeta_n + \zeta_n^{-1} =2\cos(\frac{2\pi i}{n})$ and hence $\frac{1}{2}(\zeta_n + \zeta_n^{-1})$ is contained in the finite algebraic extension $\Bbb{Q}(\zeta_n)$ from which it follows that it itself is algebraic over $\Bbb{Q}$. This comes from the equivalence of the following facts:
Let $E/F$ be a field extension and $\alpha \in E$. TFAE:

$\alpha$ is algebraic over $F$
$\dim_F F[\alpha] < \infty$
$F[\alpha]$ is a field
$F[\alpha] = F(\alpha)$.

Lastly I should add that $\frac{1}{2}(\zeta_n + \zeta_n^{-1})$ is in fact an algebraic integer, meaning to say that its minimal polynomial has coefficients in $\Bbb{Z}$.