Prove that

$\displaystyle \sum$sin(k$\displaystyle \theta$)= [sin(n$\displaystyle \theta$/2)sin((n+1)$\displaystyle \theta$/2)]/sin$\displaystyle \theta$/2)

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- Mar 2nd 2010, 04:23 PMLCopper2010complex variables
Prove that

$\displaystyle \sum$sin(k$\displaystyle \theta$)= [sin(n$\displaystyle \theta$/2)sin((n+1)$\displaystyle \theta$/2)]/sin$\displaystyle \theta$/2) - Mar 2nd 2010, 04:52 PMDrexel28
- Mar 3rd 2010, 10:09 AMLCopper2010more help
I still don't really understand...could you elaborate a bit? Thanks!

- Mar 3rd 2010, 12:09 PMDrexel28