Find a closed form for ${\displaystyle \sin(a)+\sin(a+b)+\sin(a+2b)+\cdots +\sin(a+(n-1)b)}$ .
Method 1Edit
To sum the series
${\displaystyle \sin(a)+\sin(a+b)+\sin(a+2b)+\cdots +\sin(a+(n-1)b)=S}$ .
Multiply each term by (this is my question)
${\displaystyle 2\sin \left({\tfrac {b}{2}}\right)}$. (Why multiply with this term.)
Then we have
${\displaystyle 2\sin(a)\sin \left({\tfrac {b}{2}}\right)=\cos \left(a-{\tfrac {b}{2}}\right)-\cos \left(a+{\tfrac {b}{2}}\right)}$
and similarly for all terms to
${\displaystyle 2\sin {\bigl (}a+(n-1)b{\bigr )}\sin \left({\tfrac {b}{2}}\right)=\cos \left(a+{\tfrac {(2n-3)b}{2}}\right)-\cos \left(a+{\tfrac {(2n-1)b}{2}}\right)}$ .
Summing, we find that nearly all the terms cancel out and we are left with
${\displaystyle 2S\sin \left({\tfrac {b}{2}}\right)=\cos \left(a-{\tfrac {b}{2}}\right)-\cos \left(a+{\tfrac {(2n-1)b}{2}}\right)=2\sin \left(a+{\tfrac {(n-1)b}{2}}\right)\sin \left({\tfrac {nb}{2}}\right)}$ .
Hence
${\displaystyle S=\sin \left(a+{\tfrac {(n-1)b}{2}}\right){\frac {\sin \left({\tfrac {nb}{2}}\right)}{\sin \left({\tfrac {b}{2}}\right)}}}$ .
Similarly, if
${\displaystyle C=\cos(a)+\cos(a+b)+\cos(a+2b)+\cdots +\cos {\bigl (}a+(n-1)b{\bigr )}}$
then
${\displaystyle C=\cos \left(a+{\tfrac {(n-1)b}{2}}\right)\cdot {\frac {\sin \left({\tfrac {nb}{2}}\right)}{\sin \left({\tfrac {b}{2}}\right)}}}$
Why multiple both terms with 2sin b/2
