$\cos{(A-2B)}+\cos{(B-2C)}+\cos{(C-2A)}=\cos{(2A-B)}+\cos{(2B-C)}+\cos{(2C-A)}$ Let $A,B,C\in \mathbb{R}$ with $\sin{A}+\sin{B}+\sin{C}=0$. Prove that
$$\cos{(A-2B)}+\cos{(B-2C)}+\cos{(C-2A)}=\cos{(2A-B)}+\cos{(2B-C)}+\cos{(2C-A)}$$

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 $RHS$. 
