Possible Duplicate:
Sine values being rational

I'm guessing that if I look in Ivan Niven's elementary book on irrational numbers, I'll find the answer to this quickly, but I'm posting it here in case people find it useful.

For what rational values of $x/\pi$ is $\sin x$ rational?

Obviously $\sin 0$, $\sin (\pi/6)$, $\sin (\pi/2)$ and their counterparts in the other quadrants will do it. I believe I've seen it asserted by someone who should know, that those are the only ones. How is that proved?


marked as duplicate by GEdgar, Ross Millikan, Srivatsan, t.b., lhf Dec 2 '11 at 19:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This link has an elementary argument. $\endgroup$ – André Nicolas Dec 2 '11 at 17:18
  • $\begingroup$ Maybe the difference is that this question explicitly asks for a proof of that claim. (But the one to defend that this is not a duplicate should be, first of all, the OP.) $\endgroup$ – Martin Sleziak Dec 2 '11 at 17:18
  • $\begingroup$ It's beginning to look as if this should be closed as a (nearly?) exact duplicate. I was moved to ask this by an earlier similar question posted today, but the answers getting posted there make it appear that it's being treated as if the questioner meant just what I asked here. $\endgroup$ – Michael Hardy Dec 2 '11 at 19:08

Those are the only ones. I'll answer the question for cosine instead: $2 \cos \frac{2 \pi k}{n}$ is the sum of two algebraic integers $\zeta_n + \zeta_n^{-1} = e^{ \frac{2 \pi i k}{n} } + e^{ - \frac{2 \pi i k}{n} }$, hence an algebraic integer, so it is rational if and only if it is an integer. Hence $\cos \frac{2 \pi k}{n} = 0, \pm \frac{1}{2}, \pm 1$.

In fact, more can be said. $\mathbb{Q}(\cos \frac{2 \pi k}{n})$ is the real subfield of $\mathbb{Q}(\zeta_n)$, hence has degree $\frac{\varphi(n)}{2}$ over $\mathbb{Q}$.

  • 2
    $\begingroup$ This is a nice counterpart to my answer in another thread. :-) $\endgroup$ – robjohn Dec 2 '11 at 19:37

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