# Product of two algebraic numbers to be rational

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.

$$\alpha\beta\in\mathbb{Q}\setminus 0$$ 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...