Is $\arcsin(1/4) / \pi$ rational? An approximation given by a calculator seem to suggest that it isn't, but I found no proof. Thanks in advance!

1$\begingroup$ You are asking in the wrong forum. You may get the reason for this if you ask in math.stackexchange.com ... but even there you may be required to do some of your own research before asking. $\endgroup$ – Gerald Edgar Jul 23 at 13:49

1$\begingroup$ What is the difference between the two sites? $\endgroup$ – ikp Jul 23 at 13:51

$\begingroup$ Look at the "?" at the top right of the window. That describes this forum. Then do the same for the other one. $\endgroup$ – Gerald Edgar Jul 23 at 13:58

1$\begingroup$ OK, thanks! I think this question suits any of the two sites. I'm a mathematician but not an expert on analysis or number theory, and I thought that this question might be wellknown to experts. $\endgroup$ – ikp Jul 23 at 14:05

2$\begingroup$ See the essentially identical question math.stackexchange.com/q/3617176/442 already at math.stackexchange.com . $\endgroup$ – Gerald Edgar Jul 23 at 14:14
This is a partial case of the classical result.

$\begingroup$ That's what I thought at first glance (reusing Niven's integrals and so forth), I didn't know the precise, general reference though. Thanks ! $\endgroup$ – Loïc Teyssier Jul 23 at 17:01

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:
$$\frac{\sin(n\arcsin(x))}{n}=x\frac{(n^21^2)x^3}{3!} + \frac{(n^21^2)(n^23^2)x^5}{5!} + \cdots$$ $$\frac{\sin(n\arcsin(x))}{n \cos(\arcsin(x))}=x\frac{(n^22^2)x^3}{3!} + \frac{(n^22^2)(n^24^2)x^5}{5!} + \cdots$$ for $n$ odd or even respectively.
So the righthand sides must be 0 for $x=\frac14$.
However, when we multiply the terms on the righthand sides by $2^nn$ (if $n$ is odd) or $2^{n2}n$ (if $n$ is even), we find that all the terms are integers, except that the last nonzero term is $\pm\frac12$.
So the righthand side can't be 0, the lefthand side can't be 0, and $\arcsin(\frac14)/\pi$ must be irrational.

1$\begingroup$ Can you assume WLOG that $n$ is even, to simplify this somewhat? I don't think you use the fact that $m/n$ is in lowest terms anywhere, so you can replace $m$ and $n$ by $2m$ and $2n$, respectively, so that you don't have to split into cases later. $\endgroup$ – Nathaniel Johnston Jul 23 at 15:42
Let $\theta= \arcsin(1/4)$. Assume $\theta$ is a rational multiple of $\pi$.
Then, there exists some $n$ such that $\sin(n\theta)=0$. 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 $$2 i\sin(\theta)=z \frac{1}{z}$$
Therefore, $\frac{i}{2}$ is an algebraic integer, which is a contradiction.

2$\begingroup$ There's a typo, you want theta without pi the denominator, and the rest follows correctly. Nice proof! It's essentially the same as Matt F's, but replaces the explicit formulas with closed ringness of algebraic integers. Pretty neat. $\endgroup$ – Dror Speiser Jul 23 at 16:54


$\begingroup$ The typo had been fixed already by someone else. You don't want $\pi$ in the defn of $\theta$. I rolled back to the correct version. $\endgroup$ – Gabe Conant Jul 23 at 17:04

$\begingroup$ @GabeConant Ups, Ty. This is what happens when I do not drink enough coffee :) $\endgroup$ – Nick S Jul 23 at 17:09
