How prove $\cos 20^{\circ}$ is not rational? How prove $\cos 20^{\circ}$ is not rational ?
I tried something to make a proof .Let me show you my work .
$$\cos 60^{\circ}=4\cos ^320^{\circ}-3\cos 20 ^{\circ}\\
\frac{1}{2}=4\cos ^320^{\circ}-3\cos 20 ^{\circ}\\\cos 20^{\circ}=x\\8x^3-6x-1=0$$ possible rational roots are $\in\{\pm 1,\pm \frac{1}2,\pm \frac 14,\pm\frac18\}$ but no one of them work in the equation ,so the equation has no rational root(s) ,hence $x \in \mathbb{Q^c}$
Is my trial true  ?
Is there other idea(s) to show $\cos 20^{\circ}$ is not rational ?
 A: Let $n\in\mathbb{N}$ with $n\geq 3$. 

Claim. $\cos\frac{2\pi}{n}$ is an algebraic number over $\mathbb{Q}$ with degree $\frac{\varphi(n)}{2}$.

Proof. It is simple to construct from $\Phi_n(x)$ a polynomial with integer coefficients and degree $\frac{\varphi(n)}{2}$ which vanishes at $x=\cos\frac{2\pi}{n}$. For instance, in the case $n=18$ we have that the palindromic polynomial $\Phi_{18}(x) = x^6-x^3+1$ vanishes at $x=\exp\left(\frac{2\pi i}{18}\right)$. Since
$$ \frac{\Phi_{18}(x)}{x^3} = \left(x^3+\frac{1}{x^3}\right)-1 = \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)-1$$
we have that $p(x)=8x^3-6x-1$ vanishes at $x=\cos\frac{2\pi}{18}$. The general theory grants that $\Phi_n(x)$ is an irreducible polynomial, hence $\exp\left(\frac{2\pi i}{n}\right)$ is an algebraic number over $\mathbb{Q}$ with degree $\varphi(n)$. De Moivre's formula $e^{i\theta}=\cos\theta+i\sqrt{1-\cos^2\theta}$ then grants that the polynomial found by the above procedure is the minimal polynomial of $\cos\frac{2\pi}{n}$.
Corollary #1. Since $20^\circ=\frac{2\pi}{18}$ and $\frac{\varphi(18)}{2}>1$, $\cos 20^\circ\not\in\mathbb{Q}$.
Corollary #2. We cannot construct a regular $9$-agon with straightedge and compass alone.
