$${find}\:{S}\left(\theta\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{3}} \left({n}\theta\right)}{{n}!} \\$$
$${sin}^{\mathrm{2}} \left({n}\theta\right)=\frac{\mathrm{3}}{\mathrm{4}}{sin}\left({n}\theta\right)−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{n}\theta\right) \\$$$${Thus}, \\$$$${S}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{3}{sin}\left({n}\theta\right)−{sin}\left(\mathrm{3}{n}\theta\right)}{{n}!} \\$$$$=\frac{\mathrm{3}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\theta\right)}{{n}!}−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{sin}\left(\mathrm{3}{n}\theta\right)}{{n}!} \\$$$$=\frac{\mathrm{3}}{\mathrm{4}}{Im}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{{in}\theta} }{{n}!}\right)−\frac{\mathrm{1}}{\mathrm{4}}{Im}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{\mathrm{3}{in}\theta} }{{n}!}\right) \\$$$$=\frac{\mathrm{3}}{\mathrm{4}}{Im}\left({e}^{{e}^{{i}\theta} } \right)−\frac{\mathrm{1}}{\mathrm{4}}{Im}\left({e}^{{e}^{\mathrm{3}{i}\theta} } \right) \\$$$${Im}\left({e}^{{e}^{{i}\theta} } \right)={Im}\left({e}^{{cos}\theta} {e}^{{isin}\theta} \right)={e}^{{cos}\theta} {sin}\left({sin}\theta\right) \\$$$${Im}\left({e}^{{e}^{\mathrm{3}{i}\theta} } \right)={e}^{{cos}\mathrm{3}\theta} {sin}\left({sin}\left(\mathrm{3}\theta\right)\right) \\$$$${S}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}{e}^{{cos}\theta} {sin}\left({sin}\theta\right)−{e}^{{cos}\mathrm{3}\theta} {sin}\left({sin}\left(\mathrm{3}\theta\right)\right)\right) \\$$
$${thank}\:{you}\:{sir}. \\$$