Minimal polynomial for trig function

Question

I am trying to find the minimal polynomial of $\tan \theta $ over $\mathbb{Q}(\tan 2\theta$). I'm assuming that $0 <2\theta < \frac{\pi}{2}.$ If I assume that $\tan \theta \notin \mathbb{Q}(\tan 2\theta),$ then the minimal polynomial is $ x^2 + \frac {2}{\tan 2\theta}x -1=0.$ I can see that for particlar choice of $\theta$, for example, $\theta = \frac{\pi}{6},$ I have $\tan \theta \in \mathbb{Q}(\tan 2\theta).$ Could you help me find some values of $\theta$ such that $\tan \theta \notin \mathbb{Q}(\tan 2\theta)$? Thank so much. Any help/suggestions would be apprecited.

Answer 1

Hint

Show that $\tan(\frac{\pi}{8}) \notin \mathbb Q$.

Answer 2

The hint by N.S. gives the solution. Let me add more details. Let me denote $\tan\theta$ by $x$ and $\tan2\theta$ by $a$. The usual formula tells us $$a=\frac{2x}{1-x^2}$$ This can be rewritten as a quadratic relation between $a$ and $x$ as $$ax^2+2x-a=0$$ which has discriminant $4(1+a^2)$.

As the range of tan function is whole of real numbers $a$ can take any integer value. For all integers $a$ we can see $(1+a^2)$ (being an integer next to a square) is not a square and hence the discriminant is not a square of an integer. (not can it be a square of a rational number). So the root $x=\tan\theta$ is irrational whenever $\tan2\theta$ is a non-zero integer.