[Math] Fourier sine series of $f = \cos x$

Question

Let $f:(0,\pi) \to \mathbb{R}$ defined by $x \mapsto \cos x $

Show that the Fourier sine series of (odd extension) is given by

$$\sum\limits_{n=2}^\infty \frac{2n(1+(-1)^n)}{\pi(n^2-1)}$$

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So far, because it's an odd series, I used $\displaystyle b_n =\frac{2}{\pi}\int^\pi_0 \cos x \sin nx dx$

$$\begin{align} b_n &= \frac{2}{\pi}\int^\pi_0 \cos x \sin nx dx \\
&=\frac{2}{\pi}\int^\pi_0\sin x ' \sin nx dx \\
&= \frac{2}{\pi}{[-\sin x \sin nx ]^\pi_0+n\int^\pi_0 \sin x \cos nx dx} \\
&=\frac{2}{\pi}{[-\sin x \sin nx]^\pi_0+n\int^\pi_0-\cos x ' \cos nx dx}
\end{align}$$

but now I'm thinking I've gone down the wrong path.

Answer 1

Use trigonometric identities: $$ \sin a\cos b=\frac12(\sin(a+b)+\sin(a-b)). $$