Is $arcsin(1/4) / pi$ irrational?

This is a partial case of the classical result.

https://en.wikipedia.org/wiki/Niven%27s_theorem

Let $theta= arcsin(1/4)$. Assume $theta$ is a rational multiple of $pi$.

Then, there exists some $n$ such that $sin(ntheta)=0$. This gives $cos(n theta)= pm 1$.

Set $z=cos(theta)+i sin(theta)$, then $z^n= pm 1$ and $frac{1}{z^n}=pm 1$.

This gives that $z$ and $frac{1}{z}$ are algebraic integers, and hence so is $$2 isin(theta)=z- frac{1}{z}$$

Therefore, $frac{i}{2}$ is an algebraic integer, which is a contradiction.

Yes, $arcsin(frac14)/pi$ is irrational.

Suppose $arcsin(frac14)/pi = m/n$, where $m$ and $n$ are integers.

Then $sin(n arcsin(frac14))=sin(m pi)=0$.

We analyze this usng the formulas from Browmich as cited on Mathworld:

$$frac{sin(narcsin(x))}{n}=x-frac{(n^2-1^2)x^3}{3!} + frac{(n^2-1^2)(n^2-3^2)x^5}{5!} + cdots$$ $$frac{sin(narcsin(x))}{n cos(arcsin(x))}=x-frac{(n^2-2^2)x^3}{3!} + frac{(n^2-2^2)(n^2-4^2)x^5}{5!} + cdots$$ for $n$ odd or even respectively.

So the right-hand sides must be 0 for $x=frac14$.

However, when we multiply the terms on the right-hand sides by $2^nn$ (if $n$ is odd) or $2^{n-2}n$ (if $n$ is even), we find that all the terms are integers, except that the last non-zero term is $pmfrac12$.

So the right-hand side can't be 0, the left-hand side can't be 0, and $arcsin(frac14)/pi$ must be irrational.