# 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,

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