TITLE: Three angles are linearly independent over $\mathbb{Q}$?
QUESTION [7 upvotes]: If$$\tan \alpha = 1, \text{ }\tan \beta = {3\over 2}, \text{ }\tan \gamma = 2,$$then does it follow that $\alpha$, $\beta$, $\gamma$ are linearly independent over $\mathbb{Q}$?
It is possible to test combinations $m\alpha+n\beta+\ell\gamma$ with some small integer coefficients $m,n,\ell$. The tool for doing that is the sum formula for tangents of two angles with known tangents:
$$
\tan(x\pm y)=\frac{\tan x\pm\tan y}{1\mp\tan x\tan y}.
$$
For example, judging from a picture $\beta+2\gamma$ is relatively close to $\pi=4\alpha$, but the calculations:
$$
\tan 2\gamma=\frac{2+2}{1-2\cdot2}=-\frac43,
$$
$$
\tan(\beta+2\gamma)=\frac{3/2-4/3}{1+(3/2)(4/3)}=\frac1{18}
$$
show that it is not a match.

REPLY [15 votes]: The angles are linearly independent.
Since $\alpha$ is a rational multiple of $\pi$, the question is whether, letting $\beta = \arctan 3/2$ and $\gamma = \arctan 2$, we have 
$$m\beta + n\gamma \equiv 0 \pmod{\pi}$$
for some integers $m$ and $n$, not both zero.
If this were the case, we would have 
$$1 = (e^{2i\beta})^m (e^{2i\gamma})^n = \left( \frac{2+3i}{2-3i}\right)^m \left( \frac{1+2i}{1-2i}\right)^n.$$
But since $2+3i$, $2-3i$, $1+2i$, $1-2i$ are all non-associated irreducible elements in $\mathbf{Z}[i]$, which is a unique factorization domain, this is absurd.