How do i prove that $\frac{1}{\pi} \arccos(1/3)$ is irrational?

Let $\theta = \arccos\dfrac13$ so that $\cos\theta=\dfrac13$.

If $\theta$ is a rational multiple of $\pi$, say $\theta=\dfrac mn \pi$, then $\cos(n\theta)=\pm1$. Now $\cos(n\theta)=T_n(\cos\theta)$, where $T_n$ is the $n$th-degree Chebyshev polynomial. Using mathematical induction and some trigonometric identities, you can show that the leading coefficient in the $n$th-degree Chebyshev polynomial is $2^{n-1}$ for $n\ge2$. We have $$ \pm1=T_n\left(\frac 13\right) = 2^{n-1}\left(\frac13\right)^n+\text{lower-degree terms}. $$ Multiplying both sides by $3^n$, we get $$ \pm3^n = 2^{n-1} + \text{terms divisible by $3$}. $$ And that says a positive power of $2$ is a multiple of $3$, which violates uniqueness of prime factorizations.

Let’s call this angle $\Theta$. If it were rational, then there would be an $N$ such that $\cos(N\Theta)=1$. That would say that $\cos\Theta$ was a root of $T_N-1$, where $T_N$ is the appropriate Čebyšev polynomial, which it can’t be ’cause it’s transcendental.