Prove that $\displaystyle \sum$sin(k$\displaystyle \theta$)= [sin(n$\displaystyle \theta$/2)sin((n+1)$\displaystyle \theta$/2)]/sin$\displaystyle \theta$/2)
Follow Math Help Forum on Facebook and Google+
Originally Posted by LCopper2010 Prove that $\displaystyle \sum$sin(k$\displaystyle \theta$)= [sin(n$\displaystyle \theta$/2)sin((n+1)$\displaystyle \theta$/2)]/sin$\displaystyle \theta$/2) Easy way? $\displaystyle \sum_{k}\sin(k\theta)=\sum_{k}\Im\left(e^{ik\theta }\right)$
I still don't really understand...could you elaborate a bit? Thanks!
Originally Posted by LCopper2010 I still don't really understand...could you elaborate a bit? Thanks! I'm not going to tell you the answer, but maybe a helpful hint is in order. Remember the geometric sum formula.
View Tag Cloud