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!
By geometry:
$(\cos a,\sin a),(\cos b,\sin b),(\cos c,\sin c)$ are three unit vectors which sum to zero.
Then they must be the vertices of an equilateral triangle inscribed in the trigonometric circle, at angles $$k\frac{2\pi}3+\phi$$ for $k=0,1,2$.
Obviously,
$$\tan\left(3\left(k\frac{2\pi}3+\phi\right)\right)=\tan3\phi$$ for all $k$.
From this finding, we can derive a more analytical solution.
Combining the two initial equations, we must have
$$(\sin a+\sin b)^2+(\cos a+\cos b)^2=(-\sin c)^2+(-\cos c)^2=1$$ or after simplification $$2\sin a\sin b+2\cos a\cos b=2\cos(a-b)=-1$$ giving $$a-b=\pm\frac{2\pi}3.$$
By symmetry, this holds for $b-c$ and $c-a$ and leads us to
$$b=a\pm\frac{2\pi}3,c=a\mp\frac{2\pi}3,$$ then $$3b=3a\pm2\pi,3c=3a\mp2\pi.$$
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
$a-b=\pm\frac{2\pi}{3}+2k_2\pi$
$\implies$
$\tan(3a)=\tan(3b)=\tan(3c)$.
