$\displaystyle c+2l=\pi{},\ c=\ -\pi{},\ gives\ l=\pi{} \\[2ex] \displaystyle f\left(x\right)=\vert{}\sin{x\vert{}\ ;f\left(-x\right)=\vert{}sin⁡(-x)\vert{}=\vert{}-\sin{x\vert{}=\sin{x=f(x)}}}. \\[2ex] \displaystyle \therefore{}\vert{}\sin{x\vert{}\ is\ even\ function.} \\[2ex] \displaystyle \therefore{}bn=0 \\[2ex] \displaystyle \therefore{}f\left(x\right)=\vert{}\sin{x\vert{}} \\[2ex] $

$\displaystyle a_0=\frac{1}{l}\int_c^{c+l}f(x)dx=\frac{1}{\pi{}}\int_{-\pi{}}^{\pi{}}\vert{}\sin{x\vert{}\ dx} \\[3ex] \displaystyle \therefore{}a_0=\frac{2}{\pi{}}\int_0^{\pi{}}\vert{}\sin{x\vert{}\ dx}=\frac{2}{\pi{}}\int_0^{\pi{}}\sin{x\ dx} \\[3ex] \{\vert{}\sin x\vert{} \ is \ positive \ in \ the \ interval \ (0,\pi{})\} $

$\displaystyle a_0={-\frac{2}{\pi{}}[-\cos{x]}}_0^{\pi{}} \\[2ex] \displaystyle a_0=-\frac{2}{\pi{}}\left[-1-1\right]=\frac{4}{\pi{}} \\[2ex] \displaystyle a_0=\frac{1}{l}\int_c^{c+l}f(x)\cos \Bigg(\frac{n\pi{}x}{l}\Bigg)dx=\frac{1}{\pi{}}\int_{-\pi{}}^{\pi{}}\vert{}\sin{x\vert{}\cos \Bigg (\frac{n\pi{}x}{\pi{}} \Bigg) dx} \\[2ex] \displaystyle \therefore{}a_n=\frac{2}{\pi{}}\int_0^{\pi{}}\vert{}\sin{x\vert{}.\cos \Bigg( \frac{n\pi{}x}{\pi{}} \Bigg) dx} \\[2ex] $

$ \{both \vert{}\sin{x\vert{}\ \ and\ \ \cos⁡(\frac{n\pi{}x}{\pi{}})\ }are \ even functions\}\\[3ex] \displaystyle =\dfrac{1}{\pi{}}\int_0^{\pi{}}2\cos nx.\sin xdx\\[3ex] \displaystyle=\frac{1}{\pi{}}\int_0^{\pi{}}\sin{\left(nx+x\right)}-\sin{\left(nx-x\right)}dx…\left(1\right)\\[3ex] \displaystyle={\frac{1}{\pi{}}\left[-\frac{\cos{\left(n+1\right)}x}{n+1}+\frac{\cos{\left(n-1\right)}x}{n-1}\right]}_0^{\pi{}}\\[3ex] \displaystyle=\frac{1}{\pi{}}\left[\frac{{\left(-1\right)}^n}{n+1}-\frac{{\left(-1\right)}^n}{n-1}\right]-\left[-\frac{1}{n+1}-\frac{1}{n-1}\right]\\[3ex] \displaystyle=\frac{1}{\pi{}}\left[\frac{1}{n+1}-\frac{1}{n-1}\right].[{\left(-1\right)}^n+1]\\[3ex] \displaystyle=\frac{{\left(-1\right)}^n+1}{\pi{}}\left[\frac{n-1-n-1}{\left(n+1\right)\left(n-1\right)}\right]....(2)\\[3ex]Put\ n=1\ in\ (1)\\[2ex] \displaystyle\therefore{}a_1=\frac{1}{\pi{}}\int_0^{\pi{}}\sin{2x-0\ }dx=\frac{1}{\pi{}}{\left[\frac{-\cos{2x}}{2}\right]}_0^{\pi{}}\\[3ex] \displaystyle a_1 =\frac{1}{\pi{}}\times{}-\frac{1}{2}\left[1-1\right]=0 \\[4ex] By\ fourier\ series,\\[3ex] $

$\displaystyle f\left(x\right)=\frac{a_0}{2}+\sum_1^{\infty{}}a_n\cos{\frac{n\pi{}x}{l}}+\sum_1^{\infty{}}b_n\sin{\frac{n\pi{}x}{l}} \\[2ex] \displaystyle f\left(x\right)=\frac{2}{\pi{}}+0+\sum_2^{\infty{}}\frac{{\left(-1\right)}^n+1}{\pi{}}\left[\frac{n-1-n-1}{\left(n+1\right)\left(n-1\right)}\right]\cos nx \\[2ex] \displaystyle \vert{}\sin{x\vert{}=\frac{2}{\pi{}}+\frac{1}{\pi{}}\left[2\left(\frac{1}{3}-\frac{1}{1}\right)\cos{2x+0+2\left(\frac{1}{5}-\frac{1}{3}\right)\cos{4x}}\right]} \\[2ex] \displaystyle \vert{}\sin{x\vert{}=\frac{2}{\pi{}}-\frac{4}{\pi{}}\left[\frac{1}{3}\cos{2x+\frac{1}{15}\cos{4x}+…}\right]} \\[2ex] $