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I was reviewing the Fourier series for an exam and one of the practice exercises was to find the cosine and sine series of $f(x)=cos(x)$ in the interval $[0,\pi]$. When I calculate the cosine series of course the result is $a_0=0$ and $a_n=0$ for $n\neq 1$, so the final result is only $f(x)=cos(x)$.

Nevertheless, when I try to calculate the sine series, as $cos(x)$ is an even function I suppose the result should be $b_n=0$, but instead I get $b_n=\frac{4n}{\pi(n^2-1)}$ for $n$ even. Is there something wrong here? or the result is just this one and there's a mistake in my understanding of the Fourier series?

The calculations I made were: $$b_n=\frac{2}{\pi}\int_0^\pi cosx\space sin(nx) \space dx=\frac{1}{\pi}\int_0^\pi sin[(1+n)x]-sin[(1-n)x]\space dx=\frac{1}{\pi}\left[-\frac{cos[(1+n)x]}{1+n}-\frac{cos[(1-n)x]}{n-1}\right]_0^\pi=\frac{1}{\pi}\left[\frac{1}{n+1}+\frac{1}{n-1}+\frac{(-1)^n}{n+1}+\frac{(-1)^n}{n-1}\right]$$ For $n$ odd $b_n=0$.But for $n$ even: $$ b_n=\frac{1}{\pi}\left(\frac{4n}{n^2-1}\right) $$

Does this make sense? Thank you very much.

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When you want to represent $x\mapsto\cos x$ in the interval $[0,\pi]$ as a series of the form $$\cos x=\sum_{n=1}^\infty b_n\sin(nx)\qquad(0<x<\pi)\tag{1}$$ you have to be aware that series $$f(x)\sim{a_0\over2}+\sum_{n=1}^\infty\bigl(a_n\cos(nx)+b_n\sin(nx)\bigr)$$ refer to the interval $[-\pi,\pi]$. In $(1)$ nothing is required about the $x$-interval $[-\pi,0]$; on the other hand we want all $a_n=0$. This means that we should invent an $f$ which is odd, and is $\equiv\cos x$ on $[0,\pi]$. Such an $f$ is given by $$f(x):={\rm sgn}(x)\cos x\quad(0<|x|<\pi)\ ,$$ but the value of $f$ at $0$ and $\pm\pi$ is not quite as desired. For this $f$ all $a_n=0$, and we have $$b_n={1\over\pi}\int_{-\pi}^\pi{\rm sgn}(x)\cos x\sin(nx)\>dx={2\over\pi}\int_0^\pi\cos x\sin(nx)\>dx\ ,$$ as you had in your question. I'm taking your values $b_n$, and obtain the following figure. Note Gibbs' phenomenon at the jump points.

enter image description here

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