Cosine series for the Sine

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Problem 2 on page 153 of our text book asks us to show that the cosine series for $\sin(x)$ over $[0,\pi]$ is

$$\frac{2}{\pi} - \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\cos(2nx)}{4n^2-1}.$$

Using our online Fourier series calculator, we get a different result. How can we reconcile these?

AbS

According to our Fourier cosine series calculator,

$$\sin(x) \approx \frac{2}{\pi} + \sum_{n=1}^{\infty}\left(\begin{cases} \frac{2\left((-1)^n +1\right)}{\pi \left(n^2 - 1\right)} & \text{for $n\neq 1$} \\ 0 & \text{otherwise} \end{cases}\right)\cos(nx) = \frac{2}{\pi} - \frac{2}{\pi}\sum_{n=1}^{\infty}\left(\begin{cases} \frac{(-1)^n +1}{n^2 - 1} & \text{for $n\neq 1$} \\ 0 & \text{otherwise} \end{cases}\right)\cos(nx)$$ over $[0,\pi].$

So $a_n = \frac{2}{n^2-1}$ where $n$ is even, and $a_n = 0$ where $n$ is odd. Hence,

$$\sin(x) \approx \frac{2}{\pi} - \frac{4}{\pi}\sum_{n=2, 4, 6...}\frac{\cos(nx)}{n^2-1} = \frac{2}{\pi} - \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos(2nx)}{(2n)^2-1}$$ over $[0,\pi].$