I was asked to prove that there is no rational $r$ that satisfies the identity $\cos(\pi r) = \frac{1}{\sqrt 3}$.
I tried to approach the problem by contradiction using chebyshev polynomials using $$T_n(\cos(\theta)) := \cos( n \theta)$$
by choosing $\theta = \pi r$ for some rational $r=\frac pq $ then taking special values for $n$ and using the rational root test but I didn't get any contradiction so far.
Any advice to continue or an alternative would be appreciated. thanks.
Suppose $\cos\pi r=1/\sqrt3=\sqrt3/3$. Then $\sin\pi r=\pm\sqrt6/3$. We may assume, without loss of generality, $\sin\pi r=\sqrt6/3$. Now assume $r$ is rational, so $r=p/q$ for some integers $p$ and $q$, $q>0$, $p/q$ in lowest terms. Then $${\sqrt3\over3}+i{\sqrt6\over3}=\cos\pi r+i\sin\pi r=2e^{\pi ip/q}$$ Raising to the power $q$, we get $$\left({\sqrt3\over3}+i{\sqrt6\over3}\right)^q=\pm2^q$$ Expanding by the Binomial Theorem and equating real parts, we get $$\left({\sqrt3\over3}\right)^q-{q\choose2}\left({\sqrt3\over3}\right)^{q-2}{2\over3}+{q\choose4}\left({\sqrt3\over3}\right)^{q-4}\left({2\over3}\right)^2-\cdots=\pm2^q$$ We see that $q$ must be even, for otherwise the left side is a rational times the irrational number $\sqrt3$. If you clear fractions by multiplying through by $3^q$, then the right side is even, and so is every term on the left side except the first. That's a contradiction, so $r$ can't be rational.