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.
 A: $$(\cos\alpha+\cos\beta)\cos(\alpha+\beta)-(\sin\alpha+\sin\beta)\sin(\alpha+\beta)=$$
$$\cos\alpha\cos(\alpha+\beta)-\sin\alpha\sin(\alpha+\beta)+\cos\beta\cos(\alpha+\beta)-\sin\beta\sin(\alpha+\beta)=$$
$$=\cos(2\alpha+\beta)+\cos(2\beta+\alpha)=$$
$$=2\cos\frac{3\alpha+3\beta}{2}\cos\frac{\alpha-\beta}{2}$$ and
$$\frac{\cos3\alpha+\cos3\beta}{2\cos(\alpha-\beta)-1}=\frac{2\cos\frac{3\alpha+3\beta}{2}\cos\frac{3\alpha-3\beta}{2}}{2\cos(\alpha-\beta)-1}.$$
Thus, it remains to prove that
$$\cos\frac{3\alpha-3\beta}{2}=\cos\frac{\alpha-\beta}{2}\left(2\cos(\alpha-\beta)-1\right),$$
which is true because
$$\cos\frac{3\alpha-3\beta}{2}=4\cos^3\frac{\alpha-\beta}{2}-3\cos\frac{\alpha-\beta}{2}=$$
$$=\cos\frac{\alpha-\beta}{2}\left(2(1+\cos(\alpha-\beta))-3\right)=$$
$$=\cos\frac{\alpha-\beta}{2}\left(2\cos(\alpha-\beta)-1\right).$$
Done!
A: 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 $2x=A-B$
