How to prove that $\cos(\frac{\pi}{13})$ is algebraic and not rational?

Question

How to prove that $\cos(\frac{\pi}{13})$ is algebraic and not rational?

My Try: $\cos(\frac{\pi}{13})+i\sin(\frac{\pi}{13})$ and $\cos(\frac{\pi}{13})-i\sin(\frac{\pi}{13})$ are roots of $x^{26} -1$. So $x^2 - 2 \cos(\frac{\pi}{13})+1$ is a divisor of $x^{26} -1$. But I can not proceed further. Will this help at all?

Can anyone help me?

Answer 1

Here's an approach that will work, albeit only with a lot of calculation:

If $x=\cos(\pi/13)$, then $\sqrt{1-x^2}=\sin(\pi/13)$, so that $(x+i\sqrt{1-x^2})^{13}=-1$. If you grind through the binomial expansion on the left, you'll get polynomials $P$ and $Q$ with integer coefficients such that

$$\left(x+i\sqrt{1-x^2}\right)^{13}=P(x)+iQ(x)\sqrt{1-x^2}$$

This actually gives $x$ as the root of two polynomial expressions, $P(x)+1=0$ and $Q(x)=0$. Pick either one (or play around with them first to see if you can get something simpler) and show it has no rational roots.

Answer 2

Let $x=\cos\pi/13$, then we have $-1=\cos\pi=T_{13}(x)$ (Chebyshev polynomial of first kind), i.e. $T_{13}(x)+1=0$. That's a polynomial equation with integer coefficients, the highest being $2^{12}$, the lowest $1$. If the rational number $x=p/q$ with integer, coprime $p,q$ were a solution, we'd have $p|1$ and $q|2^{12}$ (that's the Rational root theorem ), so $x$ would be $\pm1/2^k$. But that's impossible, because $\cos x$ is monotone decreasing for $x\in[0,\pi/2]$ and thus $1>\cos\pi/13>\cos\pi/3=1/2$.