Using the delay theorem, we know that \(\displaystyle{L}{\left\lbrace{g{{\left({t}\right)}}}{s}{t}{e}{p}{\left({t}-{c}\right)}\right\rbrace}={c}^{{-{c}{s}}}{L}{\left\lbrace{g{{\left({t}+{c}\right)}}}\right\rbrace}\)

Thus:

\(\displaystyle{L}{\left\lbrace{\sin{{\left({t}-\pi\right)}}}{s}{t}{e}{p}{\left({t}-\pi\right)}\right\rbrace}={e}^{{-\pi{s}}}{L}{\left\lbrace{\sin{{\left({t}+\pi-\pi\right)}}}\right\rbrace}={e}^{{-\pi{s}}}{L}{\left\lbrace{\sin{{t}}}\right\rbrace}={e}^{{-\pi{s}}}\frac{{1}}{{{s}^{{2}}+{1}}}=\frac{{{c}^{{-\pi{s}}}}}{{{s}^{{2}}+{1}}}\)

Thus:

\(\displaystyle{L}{\left\lbrace{\sin{{\left({t}-\pi\right)}}}{s}{t}{e}{p}{\left({t}-\pi\right)}\right\rbrace}={e}^{{-\pi{s}}}{L}{\left\lbrace{\sin{{\left({t}+\pi-\pi\right)}}}\right\rbrace}={e}^{{-\pi{s}}}{L}{\left\lbrace{\sin{{t}}}\right\rbrace}={e}^{{-\pi{s}}}\frac{{1}}{{{s}^{{2}}+{1}}}=\frac{{{c}^{{-\pi{s}}}}}{{{s}^{{2}}+{1}}}\)