Q:

Trigonometric Identities//Fourier Series

Basically I have to find the value of a constant $M$ from this equation: $$l(x)=0=\sum M\Big(\frac{n\pi}{L}\Big)\sin(n\pi x) $$
using the Fourier Series. However the usual Fourier Series formula is:
$$l(x)=a_0 + \sum a_n \cos \Big(\frac {n\pi x}{L}\Big) +b_n \sin \Big(\frac {n\pi x}{L}\Big) $$
So is it possible to express $\sin(n\pi x)$ as $\sin \big(\frac {n\pi x}{L}\big)$ or is there any other way?

A:

Multiply both sides by $sin(m\pi x)$ and then integral over one period.
$$
\int _0^\frac{2}{m}l(x)\sin (m \pi x)dx = \int _0^\frac{2}{m} \sum_n M\frac{n\pi}{L}sin(n\pi x)sin(m \pi x)dx
$$
This integral on the right hand side is equal to zero for all $n$ except when $n=m$ where it is $\frac{M\pi}{L}$. So you have:
$$
M = \frac{L}{\pi}\int _0^\frac{2}{m}l(x)\sin (m \pi x)dx
$$

