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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 $\cos(n \pi) = (-1)^n$.

Chad
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  • Does it check out when $n = 2$? It seems that there is no way the $\pi$ in front of cosine will be cancelled, when $n$ is even. – pjs36 May 22 '16 at 05:20
  • What problem did you try to solve, obtaining the left hand side? It looks likely that you made a mistake at that solution. – Mark Fischler May 22 '16 at 05:26
  • @MarkFischler I was trying to find $b_n$ where $b_n=\frac{2}{\pi} \int_0^\pi f(x) \sin(nx)dx$ where $f(x) = \frac{2cx}{\pi}$ for $ 0 \le x \le \frac{\pi}{2}$ and $f(x) = \frac{2c(\pi-x)}{\pi}$ for $\frac{\pi}{2} \le x \le \pi$. My original solution was $\frac{8c}{\pi^2} [\frac{1}{n}\sin(\frac{n\pi}{2})-\frac{\pi}{2n}\cos(\frac{n\pi}{2})]$ and the book has $\frac{8c}{\pi^2}\frac{(-1)^{n+1}}{(2n-1)^2}$ – Chad May 22 '16 at 05:31

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