If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$, 
If $$\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0,$$ prove that
  $$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C).$$


My solution:
From the given, 
$$\cos^3A+\cos^3B+\cos^3C=3\cos A\cos B\cos C$$
Now, 
L.H.S$$=\cos3A+\cos3B+\cos3C$$
$$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C-3\cos C$$
$$=12\cos A \cos B \cos C.$$
My solution ends here. How should I complete?
NOTE: I am not allowed to use complex numbers at my level. So, please help me solve this problem without using complex numbers.
 A: we have $$sinA+sinB=-sinC$$ and $$cosA+cosB=-cosC$$ squaring and adding both we get
$$2+2cos(A-B)=1$$ that is
$$cos(A-B)=\frac{-1}{2}$$ $\implies$
$$A-B=\frac{2\pi}{3}$$ similarly
$$B-C=\frac{-4\pi}{3}$$ and
$$C-A=\frac{2\pi}{3}$$
Now $$cos3A=cos\left(3\left(B+\frac{2\pi}{3}\right)\right)=cos(3B+2\pi)=cos3B$$
similarly $$cos3C=cos3B$$
Hence
$$cos3A+cos3B+cos3C=3cos3A=3cos(A+A+A)=3cos\left(A+B+\frac{2\pi}{3}+C-\frac{2\pi}{3}\right)=3cos(A+B+C)$$
A: $\cos (A+B+C)=\cos (A+B) \cos C- \sin (A+B) \sin C \\ = (\cos A \cos B -\sin A \sin B) \cos C -( \sin A \cos B +\cos A \sin B) \sin C \\ =\cos A \cos B \cos C- \sum_{cyc} \sin A \sin B \cos C $
$( \sin A+ \sin B)^2= (-\sin C)^2 , (\cos A +\cos B )^2=(-\cos C)^2  \\ \implies 2+ 2\sin A \sin B +2 \cos A \cos B =1 \\ \implies  \sin A \sin B = -\cos A \cos B -\dfrac{1}{2} \\ \implies \sum_{cyc} \sin A \sin B \cos C =\sum_{cyc} (-\cos A \cos B -\dfrac{1}{2})\cos C  \\ =-3 \cos A \cos B \cos C -\dfrac{1}{2} (\cos A + \cos B + \cos C)=-3 \cos A \cos B \cos C$
A: Trigonometry is not needed here. 
You have three unit vectors with sum $0$, which can only happen if the vectors are the sides of an (oriented) equilateral triangle in some order.  Geometrically this is the statement that one unit circle centered on another unit circle intersects it at points $\pm 60$ degrees from the center of the first circle.
But in that case $A,B,C$ are (in some order) equal to  $\theta, \theta - 120, \theta + 120$ for some angle $\theta$, and we have the stronger result that $3A = 3B = 3C = (A+B+C) = 3\theta$ as angles.  Of course the sines and cosines of those angles will also be equal and this implies the given formula.  
A: Hint: Anyway this is a solution with complex numbers. The solution without the complex numbers can be deduced from here, use the formula $e^{ix}=cos(x)+isin(x)$  and compute the real part at it stage of the following method.
$cos(A)+cos(B)+cos(C)=sin(A)+sin(B)+sin(C)=0$ is equivalent to $e^{iA}+e^{iB}+e^{iC}=0$.
$e^{iA}+e^{iB}+e^{iC}=0$ implies $(e^{iA}+e^{iB}+e^{iC})^3=0$.
We use the identity:
$(a+b+c)^3 = a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2+b^3+3b^2c+3bc^2+c^3$
Thus we have:
$e^{3iA}+ 3e^{2iA}e^{iB}+3e^{2iA}e^{iC}+3e^{iA}e^{2iB}+6e^{i(A+B+C)}+3e^{iA}e^{2iC}+e^{3iB}+3e^{2iB}e^{iC}+3e^{iB}e^{2iC}=0$
$=e^{3iA}+e^{3iB}+e^{3iC}+3e^{iA}e^{iB}(e^{iA}+e^{iB})+3e^{iA}e^{iC}(e^{iA}+e^{iC})+3e^{iB}e^{iC}(e^{iB}+e^{iC})+6e^{i(A+B+C)}=0$
We use here $e^{iA}+e^{iB}+e^{iC}=0$ which implies $e^{iA}e^{iB}(e^{iA}+e^{iB})=-e^{iA}e^{iB}e^{iC}$.
We deduce
$e^{3iA}+e^{3iB}+e^{3iC} -3e^{iA}e^{iB}e^{iC}-3e^{iA}e^{iB}e^{iC}-3e^{iA}e^{iB}e^{iC}+6e^{i(A+B+C)}=0$.
$e^{3iA}+e^{3iB}+e^{3iC} = 3(e^{iA}+e^{iB}+e^{iC})$
thus $cos(3A)+cos(3B)+cos(3C)=3(cos(A)+cos(B)+cos(C))$.
