# Fourier Series/Half-Range Sine Series of Cosine over 0 to Pi

## Theorem

On the interval $\left({0 \,.\,.\, \pi}\right)$:

$\cos x = \displaystyle \frac 8 \pi \sum_{m \mathop = 1}^\infty \frac {m \sin \left({2 m x}\right)} {4 m^2 - 1}$

## Proof

Let $f \left({x}\right)$ be the function defined as:

$\forall x \in \left({0 \,.\,.\, \pi}\right): f \left({x}\right) = \cos x$

Let $f$ be expressed by a half-range Fourier sine series:

$\displaystyle f \left({x}\right) \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} l$

where for all $n \in \Z_{> 0}$:

$b_n = \displaystyle \frac 2 l \int_0^l \cos x \sin \frac {n \pi x} l \, \mathrm d x$

In this context, $l = \pi$ and so this can be expressed more simply as:

$\displaystyle f \left({x}\right) \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$

where for all $n \in \Z_{> 0}$:

$b_n = \displaystyle \frac 2 \pi \int_0^\pi \cos x \sin n x \, \mathrm d x$

First the case when $n = 1$:

 $\displaystyle b_1$ $=$ $\displaystyle \frac 2 \pi \int_0^\pi \cos x \sin x \, \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac 2 \pi \left[{\frac {\sin^2 x} 2}\right]_0^\pi$ Primitive of $\cos x \sin x$ $\displaystyle$ $=$ $\displaystyle \frac 2 \pi \left({\frac {\sin^2 \pi} 2 - \frac {\sin^2 0} 2 }\right)$ $\displaystyle$ $=$ $\displaystyle \frac 2 \pi \left({0 - 0}\right)$ Sine of Multiple of Pi $\displaystyle$ $=$ $\displaystyle 0$

When $n \ne 1$:

 $\displaystyle b_n$ $=$ $\displaystyle \frac 2 \pi \int_0^\pi \cos x \sin n x \, \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac 2 \pi \left[{\frac {-\cos \left({n - 1}\right) x} {2 \left({n - 1}\right)} - \frac {\cos \left({n + 1}\right) x} {2 \left({n + 1}\right)} }\right]_0^\pi$ Primitive of $\cos x \sin n x$ $\displaystyle$ $=$ $\displaystyle \frac 2 \pi \left({\left({\frac {-\cos \left({n - 1}\right) \pi} {2 \left({n - 1}\right)} - \frac {\cos \left({n + 1}\right) \pi} {2 \left({n + 1}\right)} }\right) - \left({\frac {-\cos 0} {2 \left({n - 1}\right)} - \frac {\cos 0} {2 \left({n + 1}\right)} }\right)}\right)$ $\displaystyle$ $=$ $\displaystyle \frac 2 \pi \left({\left({\frac {-\cos \left({n - 1}\right) \pi} {2 \left({n - 1}\right)} - \frac {\cos \left({n + 1}\right) \pi} {2 \left({n + 1}\right)} }\right) + \frac 1 {2 \left({n - 1}\right)} + \frac 1 {2 \left({n + 1}\right)} }\right)$ Cosine of Zero is One and simplifying $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \left({\frac {1 - \cos \left({n - 1}\right) \pi} {\left({n - 1}\right)} + \frac {1 - \cos \left({n + 1}\right) \pi} {\left({n + 1}\right)} }\right)$ Further simplification

Thus for $n = 2 m$ for $m \in \Z$:

 $\displaystyle b_n$ $=$ $\displaystyle \frac 1 \pi \left({\frac {1 - \cos \left({2 m - 1}\right) \pi} {\left({2 m - 1}\right)} + \frac {1 - \cos \left({2 m + 1}\right) \pi} {\left({2 m + 1}\right)} }\right)$ $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \left({\frac {1 - \left({- 1}\right)} {\left({2 m - 1}\right)} + \frac {1 - \left({- 1}\right)} {\left({2 m + 1}\right)} }\right)$ Cosine of Multiple of Pi $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \left({\frac 2 {\left({2 m - 1}\right)} + \frac 2 {\left({2 m + 1}\right)} }\right)$ $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \left({\frac {2 \left({2 m + 1}\right) + 2 \left({2 m - 1}\right)} {\left({2 m + 1}\right) \left({2 m - 1}\right)} }\right)$ $\displaystyle$ $=$ $\displaystyle \frac {4 m + 2 + 4 m - 2} {\pi \left({2 m + 1}\right) \left({2 m - 1}\right)}$ $\displaystyle$ $=$ $\displaystyle \frac {8 m} {\pi \left({4 m^2 - 1}\right)}$

and for $n = 2 m + 1$ for $m \in \Z$:

 $\displaystyle b_n$ $=$ $\displaystyle \frac 1 \pi \left({\frac {1 - \cos \left({2 m}\right) \pi} {\left({2 m - 1}\right)} + \frac {1 - \cos \left({2 m + 2}\right) \pi} {\left({2 m + 1}\right)} }\right)$ $\displaystyle$ $=$ $\displaystyle \frac 1 \pi \left({\frac {1 - 1} {\left({2 m - 1}\right)} + \frac {1 - 1} {\left({2 m + 1}\right)} }\right)$ Cosine of Multiple of Pi $\displaystyle$ $=$ $\displaystyle 0$

Thus we have:

 $\displaystyle b_{2 m}$ $=$ $\displaystyle \frac {8 m} {\pi \left({4 m^2 - 1}\right)}$ $\displaystyle b_{2 m + 1}$ $=$ $\displaystyle 0$

and so over the given interval:

 $\displaystyle \cos x$ $=$ $\displaystyle \sum_{m \mathop = 1}^\infty \frac {8 m} {\pi \left({4 m^2 - 1}\right)} \sin \left({2 m x}\right)$ $\displaystyle$ $=$ $\displaystyle \frac 8 \pi \sum_{m \mathop = 1}^\infty \frac {m \sin \left({2 m x}\right) } {\left({4 m^2 - 1}\right)}$

$\blacksquare$