Trigonometry/The summation of finite series

Problem StatementEdit

Find a closed form for
\displaystyle \sin(A) + \sin(A+B) + \sin(A+2B) + ... + \sin(A+(n-1)B).

Note: A 'closed form' is not mathematically defined, but just means a simplified formula which does not involve '...', or a summation sign. In our problem, we should look for a formula that only involves variables A, B, n, and known operations like the four operations, radicals, exponents, logarithm, and trigonometric functions.

Method 1Edit

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)}.

Method 2Edit

Consider the following sum

\displaystyle s = e^{iA} + e^{i(A+B)} + + ... + e^{i(A+(n-1)B)}.

Since s is a geometric series with common ratio e^{iB}, we get

\displaystyle s = \frac{e^{iA}(e^{inB}-1)}{e^{iB}-1} = \frac{e^{iA}(e^{inB}-1)}{e^{iB}-1} = \frac{e^{iA}e^{inB/2}(e^{inB/2}-e^{-inB/2})}{e^{iB/2}(e^{iB/2}-e^{-iB/2})}
 = e^{i(A+(n-1)B/2)} \frac{\sin \left(\frac{nB}{2}\right)}{\sin \left(\frac{B}{2}\right)}

Therefore,

\displaystyle \sin(A) + \sin(A+B) + \sin(A+2B) + ... + \sin(A+(n-1)B) = Im(s) = \sin \left( A+\frac{(n-1)B}{2} \right) \frac{\sin \left(\frac{nB}{2}\right)}{\sin \left(\frac{B}{2}\right)}
\displaystyle \cos(A) + \cos(A+B) + \cos(A+2B) + ... + \cos(A+(n-1)B) = Re(s) = \cos \left( A+\frac{(n-1)B}{2} \right) \frac{\sin \left(\frac{nB}{2}\right)}{\sin \left(\frac{B}{2}\right)}
Last modified on 22 August 2013, at 15:05