How do i prove that $\frac{1}{\pi} \arccos(1/3)$ is irrational? How do i prove that $\frac{1}{\pi} \arccos(1/3)$
 is irrational?
 A: 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.
A: 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.
A: I try to use induction method without using  Chebyshev polynomial. We want to show that for $\cos\pi\alpha=\frac{1}{3}$, then $\alpha$ is irrational.
Let $\alpha=\frac{r}{s}$ where $r,s\in \mathbb Z$ and $s>0$, hence amoung $\cos(n\pi\alpha)$ we have at most $2s$ numbers. Since $\cos\pi\alpha=\frac{1}{3}$, hence by $\cos2\theta=2\cos^2\theta-1$ we arrive to $\cos(2^m\pi\alpha)=\frac{t}{3^{2^{m-1}}}$ which $3\nmid t$. and $t\in \mathbb Z$. So we have infinity numbers of $\cos(2^m\pi\alpha)$ which are different together.  Hence $\alpha$ must be irrational.
A: $$\alpha :=\arccos\left ( \frac{1}{3}\right )$$
$$\frac{1}{\pi}\alpha = q \stackrel{?}{\in} \mathbb{Q}$$
$$\left (e^{i\alpha}\right )^2 = e ^{i\cdot2\alpha}=\cos{2\alpha}+i\underbrace{{}\sin{2\alpha}}_{\neq 0} \notin \mathbb{R}$$
$$\left (e^{i\alpha}\right )^2 =(\underbrace{e^{i\cdot2\pi}}_1 )^q =1\in \mathbb{R}$$
