What is the value of $ \sin (a+ \frac{b}{2}(2n+1) ) - \sin (a+ \frac{b}{2} ) $? What is the value of $$ \sin \bigg(a+\displaystyle \frac{b}{2}(2n+1) \bigg) - \sin \bigg(a+\displaystyle \frac{b}{2} \bigg) ? $$
I already tried the sum-to-product rules and did not go so far with that. It is supposed to equal $$ 2 \sin \bigg( \frac{bn}{2} \bigg) \cos \bigg( \frac{2a+bn+b}{2} \bigg) .$$

Take d =  $ \frac{bn}{2}$ and c = $\frac{2a+bn+b}{2}$
So your equation reduces to 
 $$ \sin \big(c+d \big) - \sin \big(c-d \big) $$
which is $$ 2\cos (c)\sin(d) $$ = $$ 2\cos (\frac{2a+bn+b}{2})\sin(\frac{bn}{2}) $$
QED
