Proving that $\cos(2\pi/n)$ is algebraic

I want to prove this without using any of the properties about the field of algebraic numbers (specifically that it is one). Essentially I just want to find a polynomial for which $\cos\frac{2\pi}{n}$ is a root.

I know roots of unity and De Moivre's theorem is clearly going to be important here but I just can't see how to actually construct the polynomial from these facts.

Solution 1:

We know $\cos(2\pi/n)+i\sin(2\pi/n)$ is algebraic ($n$th root of unity). And the complex conjugate of algebraic is algebraic (in fact it's just another $n$th root of unity), so add it to its conjugate because also the sum of two algebraics is algebraic. Then divide by two.

Solution 2:

Let $\theta=\frac{2\pi}{n}$. By De Moivre's formula we have $$ (\cos \theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta = 1 $$

Expand the left-hand side using the binomial theorem!

Every other term of the expansion contains a $\pm i$ and an odd power of $\sin\theta$; we know these terms cancel out because the imaginary part of the right-hand side is zero.

If we remove all those, what is left is real terms with only even powers of $\sin \theta$. Substitute $(\sin\theta)^{2n} = (1-\cos^2\theta)^n$. Now you have a polynomial identity with integer coefficients in $\cos \theta$ -- in other words, an integer polynomial with $\cos\theta$ as root. Thus $\cos\theta$ is algebraic.

[This is actually one way to derive the Chebyshev polynomials].