It is well known that the solutions of the equation

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


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


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

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 $\pi$?

  • 3
    $\begingroup$ You meant $\sqrt{n}/k$, right? $\endgroup$ – J.G. Apr 23 at 19:29
  • $\begingroup$ How $3=\frac3 {6\cdot 3+1}$ or $3=\frac3 {6\cdot 3+2}$? $\endgroup$ – user Apr 23 at 19:32
  • 1
    $\begingroup$ When $k,n$ are positive integers, $k\sqrt{n}>1$ unless $k,n=1.$ $\endgroup$ – Thomas Andrews Apr 23 at 19:33
  • 2
    $\begingroup$ I think OP means $x=\frac{3}{6n+2}, \frac{3}{6n+1}$ for integer values of $n.$ @user $\endgroup$ – Thomas Andrews Apr 23 at 19:34
  • 1
    $\begingroup$ 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.) $\endgroup$ – 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).

  • $\begingroup$ Can you give any specific values? That is what I'm looking for $\endgroup$ – Klangen Apr 24 at 6:11
  • $\begingroup$ @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 $\sin^{-1} a$ $\endgroup$ – Vasya Apr 24 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.