# $\sin(\alpha) = \frac{\sqrt{n}}{k}$, where $n$ and $k$ are integers and $\alpha$ is a rational multiple of $\pi$

It is well known that the solutions of the equation

$$\sin\left(\frac\pi x\right)= \frac{\sqrt3}{2}$$

are

$$x=\frac{3}{6n+2}, n\in\mathbb{Z}$$

and

$$x=\frac{3}{6n+1}, n\in\mathbb{Z}.$$

Are there any other known values $$\alpha$$ such that $$\sin(\alpha) = \frac{\sqrt{n}}{k}$$, where $$k$$ and $$n$$ are positive integers and $$\alpha$$ is a rational multiple of $$\pi$$?

• You meant $\sqrt{n}/k$, right? – J.G. Apr 23 at 19:29
• How $3=\frac3 {6\cdot 3+1}$ or $3=\frac3 {6\cdot 3+2}$? – user Apr 23 at 19:32
• When $k,n$ are positive integers, $k\sqrt{n}>1$ unless $k,n=1.$ – Thomas Andrews Apr 23 at 19:33
• I think OP means $x=\frac{3}{6n+2}, \frac{3}{6n+1}$ for integer values of $n.$ @user – Thomas Andrews Apr 23 at 19:34
• Don't forget $\sin(\pi/4) = \sqrt{2}/2$ and $\sin(\pi/2) = \sqrt{4}/2$. There's also $\sin 0 = \sqrt{0}/2$, but writing this argument as $\pi/x$ is problematic. (I realize you asked for $k$ to be positive. Still, why not just write $\sin(\pi x)$?) That said, you can take any integer $k$ and $n$ you like, with $k\leq n^2$, and the corresponding $\alpha$ is simply $\arcsin(\sqrt{k}/n)$; that's rarely a nice number, though. Are you specifically interested in $\alpha$ being a rational multiple of $\pi$? (This question asks something similar.) – Blue Apr 24 at 9:13

Since sine is a continuous function, it will take any value $$\frac{\sqrt{n}}{k}$$ such as $$-1\le\frac{\sqrt{n}}{k}\le 1$$. (I assumed that you ment $$\frac{\sqrt{n}}{k}$$ based on your example).
• @Klangen: Well, we have $\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$ but for any value $a$ that falls into $[-1,1]$ interval you can just take $\sin^{-1} a$ – Vasya Apr 24 at 12:34