cos(x) as a sine series

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cos(x) as a sine series

Differential Equations
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\[\qquad\text{\(f(x)\) as a sine series}\] \begin{equation*} f(x)=\cos(x),\qquad0
\[\begin{align*} % a_0,n a_0&=a_n=0\qquad\text{(odd function)}\\ \end{align*}\] \begin{align*} % b_n b_n&=\frac1\pi\int\limits_{-\pi}^\pi g(x)\sin(nx)\,\text dx\\ &=\frac2\pi\int\limits_{0}^\pi \cos(x)\sin(nx)\,\text dx\qquad\text{(even integrand)}\\ &=\frac1\pi\int\limits_{0}^\pi \sin\big((1+n)x\big)-\sin\big((1-n)x\big)\,\text dx\\ &=\frac1\pi\left[\frac{-\cos\big((1+n)x\big)}{1+n}+\frac{\cos\big((1-n)x\big)}{1-n}\Big|_0^\pi\right]\\ &=\frac1\pi\left[\frac{1-\cos\big((1+n)\pi\big)}{1+n}+\frac{\cos\big((1-n)\pi\big)-1}{1-n}\right]\\ &=\frac1\pi\left[\frac{1-(-1)^{n+1}}{1+n}+\frac{(-1)^{n+1}-1}{1-n}\right]\\ &=\frac1\pi\left[\frac{1+(-1)^{n}}{1+n}+\frac{(-1)^{n+1}-1}{1-n}\right]\\ &=\frac1\pi\left[\frac{(1-n)\big(1+(-1)^n\big)-(1+n)\big((-1)^{n}+1\big)}{1-n^2}\right]\\ &=\frac1\pi\left[\frac{-2n\big((-1)^{n}+1\big)}{1-n^2}\right]\\ &=\frac2\pi\left[\frac{n\big((-1)^{n}+1\big)}{n^2-1}\right]\\ \end{align*}
\[\begin{align*} % S(x) S(x)&=\frac2\pi\sum\limits_{n=1}^\infty\frac{n\big((-1)^{n}+1\big)}{n^2-1}\sin(nx)\\ &=\frac4\pi\sum\limits_{n=2,4,6,\dots}^\infty\frac{n\sin(nx)}{n^2-1}\\ &=\frac8\pi\sum\limits_{r=1}^\infty\frac{r\sin(2rx)}{4r^2-1}\\ \end{align*}\]

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for some reason in the back of my book it has \[\frac8\pi\sum\limits_{r=1}^\infty\frac{r\sin(2rx)}{4r^2+1}\] as the answer, which seams like a mistake to me , but maybe i have overlooked something
The math checks out.
He used the last trig identity on this page convert it. http://www.sosmath.com/trig/Trig5/trig5/trig5.html
Were you expecting bn as 0?
the only qualm i have on this question is the plus sign in the denominator of answer in the back of the book, i got a minus sign , which seams to graph the correct thing so im not sure .
I got minus as well.
i guess the back of my book is wrong then ,
yeah, even i get minus doing it another way also...
Thanks to both of you

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