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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 $\pi$?

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    $\begingroup$ 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
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    $\begingroup$ When $k,n$ are positive integers, $k\sqrt{n}>1$ unless $k,n=1.$ $\endgroup$ – Thomas Andrews Apr 23 at 19:33
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    $\begingroup$ 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
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    $\begingroup$ 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
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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).

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  • $\begingroup$ 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

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