Trignometric Equality Proof 
Prove that $$\frac{\cos 3A + Cos 3B}{2\cos(A-B) -1}  =(\cos A + \cos B)\cos(A+B)-(\sin A+\sin B)\sin(A+B)$$

I used $\cos 3A=4{\cos ^3A}-3\cos A$, but it is getting more and more complicated.

Using Prosthaphaeresis Formulas,
$$(\cos A+\cos B)\cos(A+B)-(\sin A+\sin B)\sin(A+B)$$
$$=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2\cos(A+B)-2\sin\dfrac{A+B}2\cos\dfrac{A-B}2\sin(A+B)$$
$$=2\cos\dfrac{A-B}2\cos\left(\dfrac{A+B}2+A+B\right)$$
Now apply Prosthaphaeresis Formulas on $\cos3A+\cos3B$
Finally use $\cos3x=\cos x(4\cos^2x-3)=\cos x\{2(1+\cos2x)-3\}=\cos x(2\cos2x-1)$  for $2x=A-B$
