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.

  • 2
    $\begingroup$ @Narasimham what context is there in the version on which the comment was written: math.stackexchange.com/revisions/1712645/1 $\endgroup$
    – quid
    Mar 25, 2016 at 18:56
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    $\begingroup$ @JyrkiLahtonen there are some interesting related items such as $$ \arctan 1 + \arctan 2 + \arctan 3 = \pi, $$ which I have always liked $\endgroup$
    – Will Jagy
    Mar 25, 2016 at 22:18
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    $\begingroup$ Me, too @Will. I learned about that on this site! And have used it as an exercise in freshman calculus a few times. $\endgroup$ Mar 25, 2016 at 22:20

1 Answer 1


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.


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