[Random]
Leaky's mention of polynomial division in the previous conversations on solving $n+10$ divides $n^3-100$ once again trigger this attempt again:
$\sin a = \frac{b}{c}, \cos a = \frac{d}{f}, \tan a = \frac{bf}{cd}$
$ce^{2ia}-c=2b,fe^{2ia}+f=2d,(bf-icd)e^{2ia}+(bf+icd)=0$
Suppose $e^{2ia}=q\in \Bbb{Q}$. Then
$c(q-1)=2b,f(q+1)=2d,(bf-icd)q+(bf+icd)=0\implies bf(q+1)+cd(1-q)i=0 \implies bf(q+1)=0 \wedge cd(1-q)=0$
Equation 1 and 2 may have countably many solutions as the constraints are $q=\frac{2b}{c}+1, q=\frac{2d}{f}-1$