Is $\frac{1}{n}\sin (\frac{n\pi}{2})-\frac{\pi}{2n}\cos (\frac{n\pi}{2})=\frac{(-1)^{n+1}}{(2n-1)^2}$, where $n \in \mathbb N$?
I am doing Fourier series, and my hand computed solution is the one on the left hand side, but the solution given is the one on the right hand side. I am always losing points because I do not "simplify" my solutions (i.e. make them look like the right hand side) but I don't know how. Am I missing any trig identities? The only ones I know of (that have helped me in Fourier series) are $\sin(n \pi)=0$ and $\cos(n \pi) = (-1)^n$.