How to prove $\frac{\arccos\frac{\sqrt{3}}{6}}{\pi}$ is not a rational number?

Question

The number: $$\frac{\arccos\frac{\sqrt{3}}{6}}{\pi}$$ How to prove it's not a rational number?

Answer 1

Let $$r=\dfrac{\arccos\dfrac{\sqrt3}6}\pi,$$ then $$\pi r=\arccos\left(\dfrac13\cos\dfrac\pi6\right),$$ $$3\cos\pi r=\cos\dfrac\pi6,$$ $$\dfrac\pi6=2\pi n\pm\arccos(3\cos\pi r),$$ $$\dfrac\pi6=2\pi n\pm(4\pi^3r^3-3\pi r),$$ $$1=12n\pm(24\pi^2r^3-18r),$$ $$\pi^2=\frac{18r\pm(1-12n)}{24r^3},\text{ where }n\in\mathbb N.$$

$\pi^2$ isn't rational, so $r$ isn't rational.