Prove that $\tan 3a = \tan 3b = \tan 3c$ if $\sin a + \sin b + \sin c = 0$ and $\cos a + \cos b + \cos c = 0$ Let $a, b$ and $c$ be three angles such that:
$$\sin a + \sin b + \sin c = 0$$
$$\cos a + \cos b + \cos c = 0$$
Prove that $\tan 3a = \tan 3b = \tan 3c$.
I haven't done anything meaningful yet on this problem because I have no idea how I should start.
Thank you in advance!

We have
$sin(a)=-sin(b)-sin(c)$
and
$cos(a)=-cos(b)-cos(c)$
thus
$1=2+2sin(b)sin(c)+2cos(b)cos(c)$
$\implies$
$cos(b-c)=-\frac 12=cos(a-b)=cos(a-c)$
$\implies$
$b-c=\pm\frac{2\pi}{3}+2k_1\pi$
and
$a-b=\pm\frac{2\pi}{3}+2k_2\pi$
$\implies$
$\tan(3a)=\tan(3b)=\tan(3c)$.
