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

Question

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.

Answer 1

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