$\displaystyle For\ Half\ range\ cosine\ series,\ a_0=\frac{2}{l}\int_0^l\sin⁡(\frac{\pi{}x}{l})\ dx \[2ex]

\displaystyle ={\frac{2}{l}\left[-\cos{\left(\frac{\pi{}x}{l}\right)}.\frac{l}{\pi{}}\right]}_0^l \[2ex]

\displaystyle =\frac{2}{l}\times{}-\frac{l}{\pi{}}\ \left[-1-1\right]\[2ex] \displaystyle =-\frac{2}{\pi{}}\left(-2\right)\[2ex] \displaystyle =\frac{4}{\pi{}} \[2ex]

$ $\displaystyle Now,\ a_n=\frac{2}{l}\int_0^lf\left(x\right).\cos⁡(n\pi{}x/l)\ dx \[2ex]

\displaystyle =\frac{2}{l}\int_0^l\sin(\pi{}x/l).\cos⁡\left(\frac{n\pi{}x}{l}\right)\ dx \[2ex]

\displaystyle =\frac{1}{l}\int_0^l\left[\sin\left(\frac{n\pi{}x}{l}+\frac{\pi{}x}{l}\right)-\sin\left(\frac{n\pi{}x}{l}-\frac{\pi{}x}{l}\right)\right]\ dx…\left(1\right) \[2ex]

\displaystyle =\frac{1}{l}\times{}\frac{l}{\pi{}}\left{\frac{{\left(-1\right)}^n}{n+1}-\frac{{\left(-1\right)}^n}{n-1}+\frac{1}{n+1}-\frac{1}{n-1}\right} \[2ex]

\displaystyle =\frac{1}{\pi{}}\left{{\left(-1\right)}^n\left[\frac{1}{n+1}-\frac{1}{n-1}\right]+1\left[\frac{1}{n+1}-\frac{1}{n-1}\right]\right} \[2ex]

\displaystyle =\frac{1}{\pi{}}\left[\frac{1}{n+1}-\frac{1}{n-1}\right]\ \left[{\left(-1\right)}^n+1\right] \[2ex]

\displaystyle =-\frac{2\left[{\left(-1\right)}^n+1\right]}{\pi{}\left(n\bullet{}n-1\right)}…\left(n\not=1\right) \[2ex]

$ $ Put\ n=1\ in\ (1y)\[2ex] \displaystyle \therefore{}\ a_1=\frac{1}{l}\int_0^l\sin\left(\frac{2\pi{}x}{l}\right)-\ 0dx

\[2ex]

\displaystyle ={\frac{1}{l}\left[-\cos{\left(\frac{2\pi{}x}{l}\right).\frac{l}{2\pi{}}}\right]}_0^l \[2ex]

\displaystyle =0 \[2ex]

$ $\displaystyle By\ Fourier\ Series, \[2ex]

\displaystyle \therefore{}\ \sin{\left(\frac{\pi{}x}{l}\right)}=\frac{2}{\pi{}}+0-\frac{2}{\pi{}}\sum_2^{\infty{}}\frac{\left[{\left(-1\right)}^n+1\right]}{n^2}-1.\cos{\left(\frac{n\pi{}x}{l}\right)} \[2ex]

\displaystyle =\ \frac{2}{\pi{}}-\frac{4}{\pi{}}[\frac{1}{2^2-1}\cos{\left(\frac{2\pi{}x}{l}\right)}+\frac{1}{4^2-1}\cos{\left(\frac{4\pi{}x}{l}\right)}+… \[2ex] $