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

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!

• See the essentially identical question math.stackexchange.com/q/3617176/442 already at math.stackexchange.com . Jul 23, 2020 at 14:14
• yes, seems very relevant, I'll look. Thanks a lot!
– ikp
Jul 23, 2020 at 14:21
• "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.) Jul 23, 2020 at 16:56
• Of course, "seem to suggest" is not very precise. I meant that I couldn't see any pattern immerging from the finitely many digits.
– ikp
Jul 23, 2020 at 17:48

This is a partial case of the classical result.

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

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.

• 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. Jul 23, 2020 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.

• Nice proof! It's essentially the same as Matt F's, but replaces the explicit formulas with closed ringness of algebraic integers. Pretty neat. Jul 23, 2020 at 16:54