Express $$f(t) = \begin{cases} 0,& 0\leqslant t<\pi\\ -\sin t,& \pi\leqslant t\leqslant 2\pi. \end{cases} $$ as a full-range Fourier series.
I found $ C_0 $ to be $2/\pi$. But I'm having trouble expressing $C_n$. I've messed around with it for a long time but I can't figure out where to go. Any help? Thank you
We compute the $\cos$ coefficients: \begin{align} a_n &= \frac 2{2\pi} \int_\pi^{2\pi}-\sin t\cos\left(2\pi t\frac n{2\pi} \right)\ \mathsf dt\\ &= -\frac1\pi \int_\pi^{2\pi} \sin t\cos nt\ \mathsf dt\\ &= -\frac1\pi \left[\frac{n \sin (t) \sin (n t)+\cos (t) \cos (n t)}{n^2-1}\right]_\pi^{2\pi}\\ &= \frac{\cos (2 \pi n)}{n^2-1} + \frac{\cos (\pi n)}{n^2-1}, n\geqslant 2 \end{align} Since $\cos(2\pi n)=1$ and $\cos(\pi n) = (-1)^n$ for integers $n$, this reduces to $$ \frac{1-(-1)^{n+1}}{(1-n^2)\pi}. $$ Similarly, we compute the $\sin$ coefficients: $$ b_1 = -\frac1\pi\int_\pi^{2\pi}\sin^2t\ \mathsf dt = -\frac12, $$
\begin{align} b_n &= \frac 2{2\pi} \int_\pi^{2\pi}-\sin t\sin\left(2\pi t\frac n{2\pi} \right)\ \mathsf dt\\ &= -\frac1\pi\left[\frac{n \sin (t) \cos (n t)-\cos (t) \sin (n t)}{n^2-1}\right]_\pi^{2\pi}\\ &= -\frac1\pi\left(\frac{\sin (2 \pi n)}{n^2-1} +\frac{\sin (\pi n)}{n^2-1}\right), n\geqslant 2. \end{align} Since $\sin(2\pi n)=\sin(\pi n)=0$ for integers $n$, this reduces to zero. It follows that the Fourier series for $f$ is $$ \frac1\pi -\frac12\sin t +\sum_{n=2}^\infty \frac{1-(-1)^{n+1}}{(1-n^2)\pi}\cos nt. $$