Trigonometry proof based on if $\cos A+$.... If $$\cos A+\cos B+\cos C=\sin A.+\sin B+\sin C=0$$ then prove that: $$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C)$$
My Attempt;
Here,
$$\cos A+\cos B+\cos C=0$$
$$\cos A+\cos B=-\cos C$$
squaring on both sides,
$$\cos^2A+2\cos A.\cos B+\cos^2B=\cos^2C$$
Again,
$$\sin A+\sin B+\sin C=0$$
$$\sin A+\sin B=-\sin C$$
squaring on both sides
$$\sin^2A+2\sin A.\sin B+\sin^2B=\sin^2C$$
Now,
$$L.H.S=\cos 3A+\cos 3B+\cos 3B$$
$$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C-3\cos C$$
$$=4(\cos^3A+\cos^3B+\cos^3C)-3(\cos A+\cos B+\cos C)$$
$$=4(\cos^3A+\cos^3B+\cos^3C)-3\times 0$$
$$=4(\cos^3A+\cos^3B+\cos^3C$$
Now, please help me to continue from.here.
NOTE: PLEASE DO NOT USE COMPLEX NUMBERS IN THE SOLUTION
 A: HINT:
We have $$\sum(\cos A+i\sin A)=0$$
Using If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$., 
$$\sum(\cos A+i\sin A)^3=3\prod(\cos A+i\sin A)$$
Can you take it from here?
See also: Clarification regarding a question
A: Maybe you could try a complex approach (which is simpler ;-))
Let $z,u,v$ be three complex numbers:
$z=\cos(A)+i\sin(A)$, $u=\cos(B)+i\sin(B)$, $v=\cos(C)+i\sin(C)$.
then by hypothesis $z+u+w=0$. Hence 
$$0=(z+u+v)^3=z^3+u^3+v^3+3uv(u+v)+3uz(u+z)+3zv(z+v)+6zuv\\=
z^3+u^3+v^3+3uv(-z)+3uz(-v)+3zv(-u)+6zuv=z^3+u^3+v^3-3zuv.$$
Therefore $z^3+u^3+v^3=3zuv$
which implies, after separating real and imaginary parts,
$$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C)$$
and
$$\sin 3A+\sin 3B+\sin 3C=3\sin(A+B+C).$$
So you get two (real) trigonometric relations at one stroke!
A: Take a unit vector starting from the origin of a planar Cartesian coord. system at angle $A$.
Where is a geometric value of $\cos A$...?
Now add the next unit vector, starting at the end of the former one, at angle $B$ now.
Can you see what happens next? ...the sum of three cosines equal zero means the three-segment chain ends back at the $OY$ axis.
Similary, the sum of sines equal zero means the chain ends at the $OX$ axis.
Put together that means the chain returns to the origin...
And a closed 3-segment chain of equal-lengths segments is an equilateral triangle.
So each angle differs from the previous one by $1/3$ of a full turn:
$$B = A + \frac 23 \pi,\ C = A + \frac 43 \pi$$
or
$$B = A - \frac 23 \pi,\ C = A - \frac 43 \pi$$
Now you can substitute those B and C to your equality, expand functions, then reduce and see if you get an identity.
...or you can see that
$$3B = 3A + 2\pi \text{ and } 3C = 3A + 4\pi$$
(or both pluses replaced with minuses) so $$\cos 3A + \cos 3B + \cos 3C = 3\cos 3A$$
and $$3\cos(A + B + C) = 3\cos(3A \pm 2\pi) = 3\cos 3A$$
which immediately implies the desired equality.
