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.

Euler's Formula implies
$$
\left[\cos\left(\frac{2\pi}n\right)+i\sin\left(\frac{2\pi}n\right)\right]^n=1\tag{1}
$$
The Binomial Theorem says
$$
\begin{align}
1
&=\sum_{k=0}^ni^k\binom{n}{k}\cos^{n-k}\!\left(\frac{2\pi}n\right)\sin^k\!\left(\frac{2\pi}n\right)\\
&=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k}\!\left(\frac{2\pi}n\right)\sin^{2k}\!\left(\frac{2\pi}n\right)\\
&+i\sum_{k=0}^{\lfloor(n-1)/2\rfloor}(-1)^k\binom{n}{2k+1}\cos^{n-2k-1}\!\left(\frac{2\pi}n\right)\sin^{2k+1}\!\left(\frac{2\pi}n\right)\tag{2}
\end{align}
$$
Looking at the real part of $(2)$ yields
$$
\begin{align}
1
&=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k}\!\left(\frac{2\pi}n\right)\sin^{2k}\!\left(\frac{2\pi}n\right)\\
&=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k}\!\left(\frac{2\pi}n\right)\left[1-\cos^2\!\left(\frac{2\pi}n\right)\right]^k\\
&=\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^m\cos^{n-2m}\!\left(\frac{2\pi}n\right)\sum_{k=m}^{\lfloor n/2\rfloor}\binom{n}{2k}\binom{k}{m}\tag{3}
\end{align}
$$
Equation $(3)$ gives a polynomial equation whose root is $\cos\!\left(\frac{2\pi}n\right)$. Furthermore, the coefficient of $\cos^n\!\left(\frac{2\pi}n\right)$ is $2^{n-1}$, which implies that $2\cos\!\left(\frac{2\pi}n\right)$ is an algebraic integer. This polynomial also has a root of $1$, so it is reducible.

Example: Using $n=7$ in $(3)$, we have that $\cos\!\left(\frac{2\pi}7\right)$ satisfies
$$
64x^7-112x^5+56x^3-7x-1=0\tag{4}
$$
Note that $1$ is also a root of $(4)$.
