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Find the Fourier sine series of $f(x)=1$ over $[0,1]$. Use this to show that
$$\frac{\pi}{4} = \sin(1) + \frac{1}{3}\sin(3) + \frac{1}{5}\sin(5) +\cdots.$$
The Fourier sine series of $f(x)$ over $[0,L]$ is
$$f(x) = \sum_{n=1}^{\infty} b_n\sin\left(\frac{n\pi x}{L}\right),$$
where
$$b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right).$$
For $f(x) = 1$ and $L = 1$,
$$b_n = 2\int_{0}^{1} \sin(n\pi x) = -\frac{2}{n\pi}\cos(n\pi x)\Bigr|_{0}^{1} =$$
$$= \frac{2}{n\pi}(1 - \cos(n\pi)) = \frac{2}{n\pi}\left(1 - (-1)^n\right).$$
Thus,
$$b_n =\begin{cases} 0 & \text{if $n$ is even} \\ \frac{4}{n\pi} & \text{if $n$ is odd} \end{cases}.$$
Hence,
$$f(x) = 1 = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{2n - 1}\sin((2n - 1)\pi x).$$
Multiplying both sides by $\frac{\pi}{4}$, we get
$$\frac{\pi}{4} = \sum_{n=1}^{\infty}\frac{1}{2n - 1}\sin((2n - 1)\pi x) = \sin(\pi x) + \frac{1}{3}\sin(3\pi x) + \frac{1}{5}\sin(5\pi x) +\cdots.$$
Comments
The Fourier sine series of $f(x)$ over $[0,L]$ is
$$f(x) = \sum_{n=1}^{\infty} b_n\sin\left(\frac{n\pi x}{L}\right),$$
where
$$b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right).$$
For $f(x) = 1$ and $L = 1$,
$$b_n = 2\int_{0}^{1} \sin(n\pi x) = -\frac{2}{n\pi}\cos(n\pi x)\Bigr|_{0}^{1} =$$
$$= \frac{2}{n\pi}(1 - \cos(n\pi)) = \frac{2}{n\pi}\left(1 - (-1)^n\right).$$
Thus,
$$b_n =\begin{cases} 0 & \text{if $n$ is even} \\ \frac{4}{n\pi} & \text{if $n$ is odd} \end{cases}.$$
Hence,
$$f(x) = 1 = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{2n - 1}\sin((2n - 1)\pi x).$$
Multiplying both sides by $\frac{\pi}{4}$, we get
$$\frac{\pi}{4} = \sum_{n=1}^{\infty}\frac{1}{2n - 1}\sin((2n - 1)\pi x) = \sin(\pi x) + \frac{1}{3}\sin(3\pi x) + \frac{1}{5}\sin(5\pi x) +\cdots.$$