<s>[INST] 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
 [/INST] 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!
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