Product of two algebraic numbers to be rational

Question

Given two algebraic numbers $\alpha,\beta\in\mathbb{A}\setminus\mathbb{Q}$ not rational themselves (so they basically have to have roots etc.), when is their product $\alpha\beta$ a rational? I know that in the case of $\alpha$ and $\beta$ being of the form $$ \alpha=a+b\sqrt{c},\quad\beta=a-b\sqrt{c}\quad\implies\quad\alpha\beta=a^2-b^2c\in\mathbb{Q} $$ they are rational. Also when we have $\alpha=\sqrt{a+b}$ and $\beta=\sqrt{a-b}$ and $a^2-b^2$ is square, the product is rational. But are there any special conditions on $\alpha$ and $\beta$?

As for context, I want to show that $\cos\left(\frac{\pi}{N}n\right)-\cos\left(\frac{\pi}{N}k\right)\not\in\mathbb{Q}$ for $n\neq\frac{N}{5}$ and $k\neq\frac{2N}{5}$ (both are smaller than $N$). This is equivalent to $\sin\left(\frac{\pi}{2N}(n-k)\right)\sin\left(\frac{\pi}{2N}(n+k)\right)\not\in\mathbb{Q}$ for $n\neq\frac{N}{5}$ and $k\neq\frac{2N}{5}$. As trigonometric numbers are algebraic and irrational (except in the case $0,\frac{\pi}{6},\frac{\pi}{2}$ because of Niven's theorem), I only have to show that a product of algebraic irrational numbers is non-rational.

Edit: I'm not entirely sure, but this paper could be the solution, atleast for the trigonometric case.

Answer 1

$\alpha\beta\in\mathbb{Q}\setminus 0$ implies that the fields $K(\alpha)$ and $K(\beta)$ are equal. In particular $\beta=p(\alpha)$ with a polynomial $p\in\mathbb{Q}[X]$ of degree smaller than the degree of the minimal polynomial of $\alpha$. In concrete cases this is most likely not very helpful...