If $\sin 5\theta =a{\cos }^4\theta \sin \theta +b{\cos }^2\theta {\sin }^3\theta +c{\sin }^5\theta$ then $abc=$

${(\cos \theta +i\sin \theta )}^5={\cos }^5\theta +{}^5{C}_1{\cos }^4\theta \left(i\sin \theta \right)$

$+{}^5{C}_2{\cos }^3\theta {\left(i\sin \theta \right)}^2+{}^5{C}_3{\cos }^2\theta {\left(i\sin \theta \right)}^3$

$+{}^5{C}_4\cos \theta {\left(i\sin \theta \right)}^4+{}^5{C}_5{\left(i\sin \theta \right)}^5$

$\cos 5\theta +i\sin 5\theta =({\cos }^5\theta -10{\cos }^3\theta {\sin }^2\theta -5\cos \theta {\sin }^4\theta )$

$+i(5{\cos }^4\theta \sin \theta -10{\cos }^2\theta {\sin }^3\theta +{\sin }^5\theta )$

$a=5,b=-10,c=1$