Equality to be proved using Demoivre's theorem If $\sin (A)+\sin (B)+\sin (C)=\cos (A)+\cos (B)+\cos (C)=0$  then prove that $\sin (2A)+\sin (2B)+\sin (2C)=0$ using Demoivre's theorem.  $$\text {Attempt} $$ From above information we know that  $e^{iA}+e^{iB}+e^{iC}=0$ squaring we get an equation. But since we are concerned about inaginary part only we compare both sides to get $i (\sin (2A)+\sin (2B)+\sin (2C)+2 (\sin (A+B)+\sin (B+C)+\sin (A+C))=0$ but now how to continue from here. It isnt given that A,B,C are angles of triangle so we cant use $A+B=\pi-C $ or anything like that. Thanks
 A: Consider the triangle $T\subset S^2$ formed by the three points $e^{i\alpha}$, $e^{i\beta}$, $e^{i\gamma}$. The assumptions say that the centroid of $T$ coincides with its circumcenter $0$. From this we may conclude that $T$ is in fact equilateral. This implies $\beta=\alpha+120^\circ$, $\gamma=\alpha+240^\circ$. It follows that $2\beta=2\alpha+240^\circ$, $2\gamma=2\alpha+480^\circ\sim2\alpha+120^\circ$. The three points $e^{2i\alpha}$, $e^{2i\beta}$, $e^{2i\gamma}$ therefore again form an equilateral triangle on $S^1$, so that the stated claim holds true.
Now the proof using complex numbers: It is assumed that
$z:=e^{i\alpha}+e^{i\beta}+e^{i\gamma}=0$. It follows that
$$z^2=e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma}+2\bigl(e^{i(\alpha+\beta)}+e^{i(\beta+\gamma)}+e^{i(\gamma+\alpha)}\bigr)=0\tag{1}$$
as well. But
$$e^{i(\alpha+\beta)}+e^{i(\beta+\gamma)}+e^{i(\gamma+\alpha)}=e^{i(\alpha+\beta+\gamma)}\bigl(e^{-i\alpha}+e^{-i\beta}+e^{-i\gamma}\bigr)=e^{i(\alpha+\beta+\gamma)}\bar z=0\ ,$$
so that $(1)$ implies $e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma}=0$.
