a) Find the half range Fourier sine series of $\cos(x)$ on $\displaystyle 0 < x < \frac{\pi}{2}$.
b) Use this extension to show that $\displaystyle \sum_{m=0}^{\infty}\frac{(2m+1)}{4(2m+1)^{2}-1}(-1)^{m} = \frac{\pi}{8\sqrt2}$
For a) I have solved it by using:
$$b_n = \frac{2}{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} \cos(x)\sin(2nx) \mathrm{d}x $$
So, $$b_n = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}} \cos(x)\sin(2nx) \mathrm{d}x $$ I solved $b_n$ by using the formula for $\sin A \cos B = \frac{1}{2}(\sin(A+B)+\sin(A-B)$.
So I got, $$\frac{1}{2}\int_{0}^{\frac{\pi}{2}} \sin(2nx+x)\sin(2nx-x) dx $$
This gives: $$\frac{\sin(\pi n)+1}{2n+1}+\frac{1-\sin(\pi n)}{2n-1}$$
After simplifying, I got: $$b_n = \frac{4}{\pi}\frac{-\sin(\pi n)+2n}{4n^{2}-1} $$
This is where my problem is. I'm thinking that what I get from part a) should be similar to part b), but with my $n=2m+1$. From part b), it seems the answer should be: $$\frac{n\sin(\frac{\pi n}{2})}{4n^{2}-1}$$ Can someone tell me where I went wrong with part a)? Thanks.
$b_n = \dfrac{8n}{\pi(4n^{2}-1)} $ and $$\cos x=\sum_{n=1}^\infty \dfrac{8n}{\pi(4n^{2}-1)}\sin 2nx$$ set $x=\dfrac{\pi}{4}$: $$\cos \dfrac{\pi}{4}=\sum_{n=1}^\infty \dfrac{8n}{\pi(4n^{2}-1)}\sin n\dfrac{\pi}{2}$$ if $n=2m$ then $\sin n\dfrac{\pi}{2}=0$. if $n=2m+1$ then $\sin n\dfrac{\pi}{2}=(-1)^m$ thus $$\dfrac{1}{\sqrt{2}}=\sum_{m=0}^\infty \dfrac{8(2m+1)}{\pi(4(2m+1)^{2}-1)}(-1)^m$$