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Do there exist rational numbers $q \in (0,1) \cap \mathbb Q$ such that $$\sin\left(\frac{\pi}{2}q\right) \in \mathbb Q\;?$$

Clearly if $q \in \mathbb Z$, yes. But what about the case $0 < q < 1$?

As $\sin(\pi/6) = 1/2$ we have $q = 1/3$ is a solution. Are there any others?

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    $\begingroup$ This is similar to this question. Some of the answers there will bear on this question. In particular, see Niven's Theorem. $\endgroup$ – robjohn Jan 27 '15 at 22:14
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    $\begingroup$ "Clearly if $q\in\Bbb{Z}$, yes." It is clear because it is a vacuous statement. There are no integers in the open interval $(0,1)$, so there is no $q\in(0,1)\cap\Bbb{Q}$ where $q\in\Bbb{Z}$. $\endgroup$ – KSmarts Jan 27 '15 at 22:43
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    $\begingroup$ Yes, I was relaxing the hypothesis in that sentence to make the point that if $q$ is integer then (of course) $\sin(\pi q/2)$ is rational. The intent was to underline the focus of the question to those $q$ in the open interval $(0,1)$. $\endgroup$ – Simon S Jan 27 '15 at 23:33
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The only rationals $r$ such that $\sin(\pi r)$ is rational are those for which $ \sin(\pi r)$ is in $\{-1,-1/2,0,1/2,1\}$. This is because $2 \sin(\pi r)$ is an algebraic integer, and algebraic integers that are rational are ordinary integers.

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  • $\begingroup$ Following @robjohn's link above, I've found that this result is called Niven's theorem. I am guessing I should take it that the clue you are giving here is the heart of the proof of Niven. If so, could you elaborate a little more on why $2\sin(\pi r)$ is an algebraic integer? $\endgroup$ – Simon S Jan 27 '15 at 22:20
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    $\begingroup$ $2 \sin(\pi r) = -i \exp(i\pi r) + i \exp(-i \pi r)$. $\pm i$ and $\exp(\pm i \pi r)$ are algebraic integers, being roots of unity. The sum or product of algebraic integers is an algebraic integer. $\endgroup$ – Robert Israel Jan 27 '15 at 22:23
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$$q=\frac{1}{3}\Rightarrow \sin \left( \frac{\pi}{2} \frac{1}{3}\right) =\frac{1}{2}$$

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  • $\begingroup$ Thanks and apologies, I meant to include this one in the question. $\endgroup$ – Simon S Jan 27 '15 at 22:13

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