0
$\begingroup$

For $0 \le \theta \le \pi/2$, when are both $\theta/\pi$ and $\sqrt2\sin\theta$ rational?

I think $\theta=0, \pi/4$ is the only cases. This problem seems to be related to Niven's theorem, but I cannot prove this.

$\endgroup$
  • 1
    $\begingroup$ Hint: sine is periodic $\endgroup$ – Sorfosh Dec 5 '18 at 14:57
  • $\begingroup$ In addition to the hint provided, you can also look at the other quadrants clearly. $\endgroup$ – KM101 Dec 5 '18 at 15:05
  • $\begingroup$ Sorry for unclear statement. The range of $\theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned. $\endgroup$ – Jeongu Kim Dec 5 '18 at 15:10
  • 1
    $\begingroup$ Hint: $2\sin^2(\theta)=1-\cos(2\theta)$ $\endgroup$ – Ingix Dec 5 '18 at 15:45
  • $\begingroup$ Wow, great idea! $\endgroup$ – Jeongu Kim Dec 5 '18 at 15:54
0
$\begingroup$

Let us assume that $\theta\in\pi\mathbb{Q}$ and $\sqrt{2}\sin\theta\in\mathbb{Q}$. If $\sin\frac{\pi p}{q}=\cos\left(\frac{\pi q}{2q}-\frac{2\pi p}{2q}\right)=\cos\left(\frac{2\pi|q-2p|}{4q}\right)$ is an algebraic number of degree $2$ over $\mathbb{Q}$, then we must have $\frac{1}{2}\varphi(4q)=2$ or $\varphi(4q)=8$, so $q\in\{2,3,4,5,6\}$. Now a manual inspection completes the job, or $\cos\frac{\pi}{5}\in\mathbb{Q}(\sqrt{5})\setminus\mathbb{Q}$ and $\cos\frac{\pi}{6}\in\mathbb{Q}(\sqrt{3})\setminus\mathbb{Q}$.

$\endgroup$

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.