Rational or Not

Let $a,b\in\mathbb N$ , $d,m\in\mathbb Z$ , $n$ not be a power of $2$ nor a square, $\cos(\theta)=\frac1{\sqrt n}$ , $T_b(x)$ be Chebyshev polynomials, and say $\frac{\theta}{\pi}=\frac ab$ , then
$\begin{aligned} \pm1&=\cos(b\theta)\\ &=T_b(cos(\theta))\\ &=T_b\left(\frac1{\sqrt n}\right)\\ &=2^{b-1}\left(\frac1{\sqrt n}\right)^b+d\left(\frac1{\sqrt n}\right)^{b-1}+m\left(\frac1{\sqrt n}\right)^{b-2}\\ \pm(\sqrt n)^b&=2^{b-1}+d\sqrt n+mn\\ \end{aligned}$
If $b$ is even, then $2^{b-1}=n(\pm n^{\frac{b-2}2}-m)$ , which is a contradiction since $n$ has a prime factor that isn't $2$ .
If $b$ is odd, then $2^{b-1}=-mn$ , which is a contradiction.
For the case where $n$ is a square but not a power of $2$ , we have $2^{b-1}=\sqrt n(\pm(\sqrt n)^{b-1}-d-m\sqrt n)$ which is a contradiction since $\sqrt n$ has a prime factor that isn't $2$ .
Therefore, $\frac1{\pi}\arccos\left(\frac1{\sqrt n}\right)$ is irrational when $n$ is not a power of $2$ .