I am reading this quesiton and accepted answer.
Question is about proof.
$S = \sin{(a)} + \sin{(a+d)} + \cdots + \sin{(a+nd)}$
$S \times \sin\Bigl(\frac{d}{2}\Bigr) = \sin{(a)}\sin\Bigl(\frac{d}{2}\Bigr) + \sin{(a+d)}\cdot\sin\Bigl(\frac{d}{2}\Bigr) + \cdots + \sin{(a+nd)}\cdot\sin\Bigl(\frac{d}{2}\Bigr) $
After cancellation I get $S * \sin(\frac{d}{2})=\frac{1}{2}(\cos(a-\frac{d}{2})-\cos(a+\frac{d(2n+1)}{2}))$
So I need to prove that $\frac{1}{2}(\cos(a-\frac{d}{2})-\cos(a+\frac{d(2n+1)}{2}))$ = ${\sin(n \times \frac{d}{2})} \times \sin\biggl( \frac{2 a + (n-1)\cdot d}{2}\biggr)$
This is where I don't know how to prove
