$4\sin \frac{5\mathrm{\theta }}{2}\cos \frac{3\mathrm{\theta }}{2}\cos 3\mathrm{\theta }=$

$\begin{array}{l}=2\left(2\sin \frac{5\mathrm{\theta }}{2}\cos \frac{3\mathrm{\theta }}{2}\right)\cos 3\mathrm{\theta }\\ =2(\sin 4\mathrm{\theta }+\sin \mathrm{\theta })\cos 3\mathrm{\theta }\\ =2\sin 4\mathrm{\theta cos}3\mathrm{\theta }+2\sin \mathrm{\theta cos}3\mathrm{\theta }\\ =\sin 7\mathrm{\theta }+\sin \mathrm{\theta }+\sin 4\mathrm{\theta }-\sin 2\mathrm{\theta }\\ =\sin 7\mathrm{\theta }+\sin 4\mathrm{\theta }-\sin 2\mathrm{\theta }+\sin \mathrm{\theta }\end{array}$