I am tasked with finding the real form of the Fourier series where $\phi(x)=\sin(x)$ on $(-1, 1)$. So far I have the following:
Since $\sin(x)$ is an odd function, it has a Fourier sine series such that
$$\phi(x) = \sin(x) = \sum_{n=1}^{\infty} b_n \sin(nx) \textrm{ for } x \in (-1,1)$$ and $$b_n = \frac{2}{l} \int_{0}^{l} \phi(x)\sin(\frac{n\pi x}{l})dx$$
Substituting $\phi(x) = \sin(x)$ and $l=1$ gives
$$ \begin{align*} b_n &= 2 \int_{0}^{1} \sin(x)\sin(n\pi x)dx && \\ &= \frac{2\cos(1)\sin(\pi n) - 2\pi n\sin(1)\cos(\pi n)}{\pi^2 n^2-1} \\ &= -\frac{2 \pi n \sin(1)(-1)^n}{\pi^2n^2-1} && \textrm{ for } n^2 \neq \frac{1}{\pi^2} \end{align*} $$
$\therefore$ The real form of the Fourier series $\phi(x) = \sin(x)$ on $(-1, 1)$ is
$$ \begin{align*} \sin(x) &= 2\pi \sum_{n=1}^{\infty} \frac{n \sin(1)\sin(nx)(-1)^{n+1}}{\pi^2n^2-1} \end{align*} $$
But when I compute the value of the series for $x=\frac{1}{2}$ I get
$$\sin(1/2) \approx 0.15848388659160$$
Which is incorrect. What gives?