
Fourier series of sinx
Find the full series of sinx on (l,l) in its real and complex forms. Assume that l is not an integer multiple of $\displaystyle \pi$. Hint: First find the series for $\displaystyle e^{ix}$.
I don't see how finding the fourier series of $\displaystyle e^{ix}$ helps in finding the series for sinx.
THanks

The reason they said that is because:
$\displaystyle sin(x)=\frac{e^{ix}e^{ix}}{2i}$

So I simplified the complex version for the coefficient of the fourier series to
$\displaystyle \frac{1}{2i{\pi}n}(e^{in\pi+il}e^{in\piil})$
I'm not sure how to simplify it from here.
THanks

Simplifying…
To simplify $\displaystyle \frac{1}{2 \imath \pi n}(e^{\imath n \pi+\imath l}e^{\imath n\pi\imath l}) $ we can use the fact that $\displaystyle e^{\imath x} = \cos x + \imath\sin x$ to get that $\displaystyle \frac{e^{\imath x}  e^{\imath x}}{2i} = sin x$
Thus $\displaystyle \frac{1}{2 \imath \pi n} (e^{\imath n \pi+\imath l}e^{\imath n\pi \imath l})=\frac{1}{2\imath \pi n}(e^{\imath (n\pi l)}e^{\imath( n\pi l)}) $
$\displaystyle =\frac{1}{\pi n} \sin (n\pi l) $
Kevin C.