Prove $\cos{\frac{\pi}{n}}\notin\mathbb Q$ for any postive integer $n\ge 4$ Let $n\ge 4$ postive integer,show that
$$\cos{\dfrac{\pi}{n}}\notin\mathbb Q$$
Now I have solve for a case:
Assmue that
$$\cos{\dfrac{\pi}{n}}=\dfrac{q}{p},(p,q)=1,p,q\in N^{+}$$ use Chebyshev polynomials?
$$T_{n}(\cos{x})=\cos{(nx)}$$
so we have
$$T_{n}\left(\dfrac{q}{p}\right)=-1$$
then we have
$$2^{n-1}\left(\dfrac{q}{p}\right)^n+a_{n-1}\left(\dfrac{q}{p}\right)^{n-1}+\cdots+a_{0}+1=0$$
so we have
$$q|\le a_{0}+1$$
since
$|a_{0}|=0,1,-1$,

(1):$a_{0}\neq -1$,so we have
$$\dfrac{q}{p}\le\dfrac{2}{3}$$,but
$$\cos{\dfrac{\pi}{n}}\ge\cos{\dfrac{\pi}{4}}=\dfrac{\sqrt{2}}{2}>\dfrac{2}{3}$$
contradiction。

(2):But for $a_{0}=-1$, I can't prove it

An elementary proof. From deMoivre's Theorem, if $k\in \Bbb N$ then $\cos kx=\sum_{(0\le 2j\le k)}\binom {k}{2j}(-1)^j(\cos x)^{k-2j}(1-\cos^2 x)^j.$ So if $\cos (\pi/n)\in \Bbb Q$ and $m|n$ then $\cos (\pi/m)\in \Bbb Q.$ So it suffices to show $\cos (\pi/m)\not \in \Bbb Q$ when (i)$\,m=4,$ or (ii) $\,m=6,$ or (iii) $\,m=9,$ or (iv) $\,m \ge 5$ is prime.
For $m=9,$ let $y=2\cos (\pi/9).$  Then $4(y/2)^3-3(y/2)=\cos (\pi/3)=1/2,$ so $y^3-3y-1=0.$ If $y=A/B$ with $A,B\in \Bbb N$ and $\gcd (A,B)=1$ then $A^3-3AB^2-B^3=0 ,$ so  we have $$1\equiv 1+A^3-3AB^2-B^3\equiv 1+A+AB+B\equiv (1+A)(1+B) \pmod 2$$ implying $1+A$ and $1+B$ are odd, implying $A$ and $B$ are even, contrary to $\gcd(A,B)=1.$
For prime $m\ge 5,$ if $\cos \pi/m=A/B$ with $A,B\in \Bbb N$ and $\gcd(A,B)=1$ then by deMoivre's Theorem we have $$(\bullet)\quad -B^m=\sum_{j=0}^{(m-1)/2}\binom {m}{2j}A^{m-2j}(B^2-A^2)^j(-1)^j.$$ The RHS of $(\bullet)$ is divisible by $A$ so $A|(-B^m).$ But $\gcd(A,B)=1$ so we must have $A=1.$
$\quad$ Now $B^m\equiv B \mod m,$ as $m$ is prime. And if $m$ is an odd  prime and $0< 2j<m$ then $m$ divides $\binom {m}{2j}.$ So from $(\bullet)$ we have $$-B\equiv -B^m\equiv A^m\equiv 1 \pmod m.$$ So $B\ge m-1.$
But now $1/2=\cos (\pi/3)<\cos (\pi/m)=A/B=1/B\le 1/(m-1)\le 1/4,$  a paradox.
