Proving the irrationality of the number $\cos\frac\pi9$ The question is if there exists a standard way of proving that 
$$ \cos{ π \over 9}$$ is an irrational number.
 A: Observe that $\cos(\frac{\pi}{3}) = \dfrac{1}{2}$. Thus using the identity: $\cos (3x) = 4\cos^3 x - 3\cos x$, $\cos(\frac{\pi}{9})$ is a zero of the equation: $4x^3 - 3x = \dfrac{1}{2}$,and using the irrational root test, one concludes that this equation has no rational root, hence $\cos(\frac{\pi}{9})$ must be irrational.
A: Using either trig identities, or plug-and-simplify of $\frac{1}{2} (e^{\pi i/9}+e^{-\pi i /9})$, we verify that $\cos\frac{\pi}{9}$ is a root of $f(x)=8x^3-6x-1$.  By the Rational Root Theorem, the only possible rational roots of this polynomial are $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4},\pm \frac{1}{8}$, and none of these eight are correct.  
A: Hint: Use $\frac{1}{2}=\cos\frac{\pi}{3}=4\cos^3\frac{\pi}{9}-3\cos\frac{\pi}{9}$, then prove that $4x^3-3x=\frac{1}{2}$ does not have rational roots
A: Using Galois theory: Let $n\in\mathbb N$. If $2\cos(2\pi/n)=e^{2\pi i/n}+e^{-2\pi i/n}$ were rational, it would be fixed under the automorphism of $\mathbb Q[e^{2\pi i/n}]$ that sends $e^{2\pi i/n}$ to $e^{k2\pi i/n}$ for $\gcd(k,n)=1$, i.e., we would have
$$\cos(2\pi/n)=\cos(2k\pi/n)$$
if $\gcd(k,n)=1$. If $n>4$ there are such angles with different cosine (to be seen e.g. by drawing a unit circle, or take $k=2$ or $n/2-1$ according to as $n$ is odd or even).
We conclude that $\cos(2\pi/n)$ is irrational for $n>4$.
