$\cos{(A-2B)}+\cos{(B-2C)}+\cos{(C-2A)}=\cos{(2A-B)}+\cos{(2B-C)}+\cos{(2C-A)}$ Let $A,B,C\in \mathbb{R}$ with $\sin{A}+\sin{B}+\sin{C}=0$. Prove that
$$\cos{(A-2B)}+\cos{(B-2C)}+\cos{(C-2A)}=\cos{(2A-B)}+\cos{(2B-C)}+\cos{(2C-A)}$$
 A: This is not so much an answer, but a caveat that you cannot simply resolve the $LHS$ of math110's proposed identity (1) into the $RHS$. WolframAlpha gives the complete version as,
$$\cos{(A-2B)}+\cos{(B-2C)}+\cos{(C-2A)}-\big(\cos{(2A-B)}+\cos{(2B-C)}+\cos{(2C-A)}\big) = 8\sin{\left(\frac{A-B}{2}\right)}\sin{\left(\frac{A-C}{2}\right)}\sin{\left(\frac{B-C}{2}\right)}\big( \sin A +\sin B +\sin C\big)$$
Hence, to prove (1), you have to show that the $LHS$ of this new equation is identically equal to the $RHS$. 
A: The first idea that comes to mind it to express everything in terms of $\sin A,\ \sin B,\ \sin C$ (and the cosines). It turns out that it works:
Begin with $\cos (A - 2B ) = \cos A \cos 2B + \sin A \sin 2B $ (by the formula for $\cos (x+y)$). Now, get rid of the double angles: $\cos 2B = 1 - 2 \sin^2 B$ and $\sin 2B = 2 \sin B \cos B$. This gives you:
$$\cos (A - 2B ) = \cos A - 2 \cos A \sin^2B + 2 \sin A \sin B \cos B $$
Now, the expression on the left side is (using cyclic summation):
$$ L = \sum_{\text{cyc}} \cos A - 2\sum_{\text{cyc}} \cos A \sin^2B + 2 \sum_{\text{cyc}}\sin A \sin B \cos B $$
Note that it follows from the assumption $\sin A + \sin B + \sin C = 0$ (by moving $\sin B$ to the other side, and multiplying by $\sin B$) that $- \sin^2 B = \sin A \sin B + \sin C \sin B$, so this is:
$$ L = \sum_{\text{cyc}} \cos A + 2\sum_{\text{cyc}} \cos A \sin B \sin A + 2\sum_{\text{cyc}} \cos A \sin B \sin C + 2 \sum_{\text{cyc}}\sin A \sin B \cos B $$
This has grown a little messy, but if you shift the summation in the last sum so that it becomes $ \sum_{\text{cyc}}\sin C \sin A \cos A$, you can rearrange it into something nicer:
$$ L = \sum_{\text{cyc}} \cos A + 2\sum_{\text{cyc}} \cos A (\sin A \sin B + \sin B \sin C + \sin C \sin A ) \\
= (1 + 2\sin A \sin B + 2\sin B \sin C + 2\sin C \sin A )\sum_{\text{cyc}} \cos A 
$$
Now, look at the right side. Of course, we might redo the computations, and it would come out the same. There is a nicer way to see this: right side differs from the left side just by ordering of $A,B,C$ (in the sense that if you denote the expression on the left by $L(A,B,C)$, and the one on the right by $R(A,B,C)$, then $R(A,B,C) = L(C,B,A)$). But the formula we arrived at does not depend on the order of $A,B,C$! So $R(A,B,C) = L(C,B,A) = L(A,B,C)$, which is what we wanted.

To make sure the proposed solution is more readable, let me explain what I mean by $\sum_{\text{cyc}}$. If $f(A,B,C)$ is any expression involving $A,B,C$, by $\sum_{\text{cyc}} f(A,B,C)$ I mean the expression $f(A,B,C) + f(B,C,A) + f(C,A,B)$. For example, $\sum_{\text{cyc}} \sin A \cos B = \sin A \cos B  + \sin B \cos C  + \sin C \cos A$. The reasoning strongly relies on the fat that the problem does not change under cyclic rearrangement of $A,B,C$, so if I do some computation for, say, $\sin^2 B$, roughly the same can be done for $\sin^2 A $ and $\sin^2 C$. Note that $\sum_{\text{cyc}} f(A,B,C) = \sum_{\text{cyc}} f(B,C,A)$. Also, $\sum_{\text{cyc}} f(A,B,C)g(A,B,C) = g(A,B,C) \sum_{\text{cyc}} f(A,B,C)$ provided that $g(A,B,C)= g(B,C,A)=g(C,A,B)$
