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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!

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    $\begingroup$ See the essentially identical question math.stackexchange.com/q/3617176/442 already at math.stackexchange.com . $\endgroup$ Jul 23 '20 at 14:14
  • $\begingroup$ yes, seems very relevant, I'll look. Thanks a lot! $\endgroup$
    – ikp
    Jul 23 '20 at 14:21
  • $\begingroup$ "An approximation given by a calculator seem to suggest that it isn't [rational]": how come you can see (ir)rationality from a finite decimal representation? (If I got what you meant right.)(This applies to the MSE post as well.) $\endgroup$ Jul 23 '20 at 16:56
  • $\begingroup$ Of course, "seem to suggest" is not very precise. I meant that I couldn't see any pattern immerging from the finitely many digits. $\endgroup$
    – ikp
    Jul 23 '20 at 17:48
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This is a partial case of the classical result.

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

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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:

$$\frac{\sin(n\arcsin(x))}{n}=x-\frac{(n^2-1^2)x^3}{3!} + \frac{(n^2-1^2)(n^2-3^2)x^5}{5!} + \cdots$$ $$\frac{\sin(n\arcsin(x))}{n \cos(\arcsin(x))}=x-\frac{(n^2-2^2)x^3}{3!} + \frac{(n^2-2^2)(n^2-4^2)x^5}{5!} + \cdots$$ 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 $\pm\frac12$.

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

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    $\begingroup$ 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. $\endgroup$ Jul 23 '20 at 15:42
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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.

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    $\begingroup$ Nice proof! It's essentially the same as Matt F's, but replaces the explicit formulas with closed ringness of algebraic integers. Pretty neat. $\endgroup$ Jul 23 '20 at 16:54

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