Proving the irrationality of the number $\cos\frac\pi9$

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

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.

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.