Voofie | How to sum sin(n q) and cos(n q)?

How to sum sin(n q) and cos(n q)?

02 Sep 10

This is a short how to about how to sum sin(n q) and cos(n q).

Content

1 The Problem
2 Change to exponential
3 Special case

The Problem

Find a close form for the sum:

$\f$\displaystyle \sum _{n=n_1}^{n_2} \sin (n \theta )\f$$ and $\f$\displaystyle \sum _{n=n_1}^{n_2} \cos (n \theta )\f$$

Change to exponential

If you write:

$\f[e^{i \theta }=\cos \theta + i \sin \theta \f]$ The problem can be solved easily since:

$\f[\sum _{n=n_1}^{n_2} e^{i n \theta }=\sum _{n=n_1}^{n_2} \cos n \theta + i\sum _{n=n_1}^{n_2} \sin n \theta \f]$

$\f$\sum _{n=n_1}^{n_2} e^{i n \theta }\f$$ can be summed easily, since it is just a geometry sum.

Now,

$\f$\begin{align*} \sum _{n=n_1}^{n_2} e^{i n \theta } &= e^{i n_1\theta }\sum _{n=0}^{n_2-n_1} e^{i n \theta }\\ &= \frac{e^{i n_1\theta }\left(1-e^{i\left(n_2-n_1+1\right)\theta }\right)}{1-e^{i \theta }}\\ &= \frac{e^{i n_1\theta }-e^{i\left(n_2+1\right)\theta }}{1-e^{i \theta }} \end{align*}\f$$

Therefore, the desire cosine and sine sum is just the real and imaginary part of the expression.

$\f[\sum _{n=n_1}^{n_2} \sin (n \theta )= \frac{1}{2i}\left(\frac{e^{i n_1\theta }-e^{i\left(n_2+1\right)\theta }}{1-e^{i \theta }}-\frac{e^{-i n_1\theta }-e^{-i\left(n_2+1\right)\theta }}{1-e^{-i \theta }}\right) \f]$ $\f[\sum _{n=n_1}^{n_2} \cos (n \theta )=\frac{1}{2}\left(\frac{e^{i n_1\theta }-e^{i\left(n_2+1\right)\theta }}{1-e^{i \theta }}+\frac{e^{-i n_1\theta }-e^{-i\left(n_2+1\right)\theta }}{1-e^{-i \theta }}\right)\f]$