Let $n\ge 4$ postive integer,show that $$\cos{\dfrac{\pi}{n}}\notin Q$$
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 $$T_{n}(\dfrac{q}{p})=-1$$ 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