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

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.

• Hint: sine is periodic – Sorfosh Dec 5 '18 at 14:57
• In addition to the hint provided, you can also look at the other quadrants clearly. – KM101 Dec 5 '18 at 15:05
• 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. – Jeongu Kim Dec 5 '18 at 15:10
• Hint: $2\sin^2(\theta)=1-\cos(2\theta)$ – Ingix Dec 5 '18 at 15:45
• Wow, great idea! – Jeongu Kim Dec 5 '18 at 15:54

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 $$\cos\frac{\pi}{6}\in\mathbb{Q}(\sqrt{3})\setminus\mathbb{Q}$$.