Trigonometry/The summation of finite series

      To sum the series

      \displaystyle \sin(A) + \sin(A+B) + \sin(A+2B) + ... + \sin(A+(n-1)B) = S.

      Multiply each term by

      \displaystyle 2\sin \left( \frac{B}{2} \right).

      Then we have

      \displaystyle 2\sin(A) \sin \left( \frac{B}{2} \right) = \cos \left(A - \frac{B}{2} \right) - \cos \left(A + \frac{B}{2} \right)

      and similarly for all terms to

      \displaystyle 2\sin(A+(n-1)B) \sin \left( \frac{B}{2} \right) = \cos \left(A + \frac{(2n-3)B}{2} \right) - \cos \left(A + \frac{(2n-1)B}{2} \right).

      Summing, we find that nearly all the terms cancel out and we are left with

      \displaystyle 2S \sin \left( \frac{B}{2} \right) = \cos \left(A - \frac{B}{2} \right) - \cos \left(A + \frac{(2n-1)B}{2} \right) = 2\sin \left ( A + \frac{n-1}{2}B \right ) \sin \left ( \frac{nB}{2} \right ).

      Hence

      \displaystyle S = \sin \left ( A + \frac{n-1}{2}B \right ) \frac{\sin \left ( \frac{nB}{2} \right )}{\sin \left( \frac{B}{2} \right)}.

      Similarly, if

      \displaystyle C = \cos(A) + \cos(A+B) + \cos(A+2B) + ... + \cos(A+(n-1)B) then
      \displaystyle C = \cos \left ( A + \frac{n-1}{2}B \right ) \frac{\sin \left ( \frac{nB}{2} \right )}{\sin \left( \frac{B}{2} \right)}.
      Last modified on 14 May 2011, at 18:28