Given that $\{\sin\left[\frac{(2n-1)\pi}{2L}x\right] : n\in\mathbb N\}$ is the complete set of eigenfunctions of a regular Sturm-Liouville with boundary points $0$ and $L$ and weight function $1$, and that $$f(x)=\sum_{n=1}^{\infty}A_{n}\sin\left[\frac{(2n-1)\pi}{2L}x\right],$$ then we know that $$A_{n} = \frac{\int_{0}^{L}f(x)\sin\left[\frac{(2n-1)\pi}{2L}x\right]dx}{\int_{0}^{L}\sin^{2}\left[\frac{(2n-1)\pi}{2L}x\right]dx}=\frac{2}{L}\int_{0}^{L}f(x)\sin\left[\frac{(2n-1)\pi}{2L}x\right]dx. $$
So it seems that the above generalised Fourier series is really nothing more than the classic Fourier since series, with $n$ replaced by $\frac{2n-1}2$.
Without knowing any Sturm-Liouville theory, how could I have immediately seen that the above series is in fact the classic Fourier sine series?
This is not the classical sine series because $$ \left. \sin\left[\frac{(2n-1)\pi}{2L}x\right]\right|_{x=L} = \sin\left((2n-1)\frac{\pi}{2}\right)=\left\{\begin{array}{cc}1, & n \mbox{ is odd}\\-1, & n \mbox{ is even}\end{array}\right. \;. $$ You can definitely tell from this fact alone that this is not the classical sine series. You can also tell because the frequency multiplier $(n-\frac{1}{2})\pi$ is not harmonic (i.e., integer multiples of a base frequency.)
This series arises from conditions $f(0)=0$ and $f'(L)=0$ because the derivatives of the eigenfucntions vanish at $L$.