Previous exercise
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. \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}
\begin{align} a_n &= \frac {1-(-1)^{n+1}}{(1-n^2)\pi} \end{align}
\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} \begin{align} b_n &=0\end{align}
I want to solve this equation by using result of the previous exercise
I dont know where to start, someone you can help me ?