$\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ if........... Help please:                                                                    If $\sin\alpha+\sin\beta= \sqrt{3} (\cos\beta-\cos\alpha)$ then show that $\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$                                       please tell me how can I approach 

Apart from the Prosthaphaeresis Formulas already mentioned with the unmentioned assumption that $\displaystyle \sin\frac{\alpha+\beta}2\ne0$
we can try as follows :
Rearranging we have  $\displaystyle\sin\alpha+\sqrt3\cos\alpha=\sqrt3\cos\beta-\sin\beta$
As $\displaystyle 1^2+(\sqrt3)^2=4,$ we write $1=2\sin30^\circ,\sqrt3=2\cos30^\circ$ to get 
$\displaystyle2\sin30^\circ\sin\alpha+2\cos30^\circ\cos\alpha=2\cos30^\circ\cos\beta-2\sin30^\circ\sin\beta$
$\displaystyle\implies \cos(\alpha-30^\circ)=\cos(30^\circ+\beta)$
$\displaystyle\implies \alpha-30^\circ=n360^\circ\pm(30^\circ+\beta)$
Taking the '-' sign, $\displaystyle\implies \alpha-30^\circ=n360^\circ-(30^\circ-\beta)\implies \alpha=n360^\circ-\beta$ 
This makes both sides equal for all $\alpha,$ so no solution available  from here
Taking the '+' sign, $\displaystyle\implies \alpha-30^\circ=n360^\circ+(30^\circ-\beta)\implies \alpha-\beta=n360^\circ+60^\circ$ 
Now, use  $\displaystyle2\cos^2A=1+\cos2A$ as $\displaystyle\cos2A=2\cos^2A-1$ for $\displaystyle A=\frac{\alpha-\beta}2$
