Answer:
The expression is given below as
[tex]3\sin (\frac{5\pi}{2})\cos (\frac{3\pi}{2})[/tex]
Concept:
The product to sum identity to be used is given below as
[tex]\begin{gathered} \sin A\cos B=\frac{1}{2}(\sin (A+B)+\sin (A-B) \\ A=\frac{5\pi}{2},B=\frac{3\pi}{2} \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} 3\sin (\frac{5\pi}{2})\cos (\frac{3\pi}{2})=3\times\frac{1}{2}(\sin (\frac{5\pi}{2}+\frac{3\pi}{2})+\sin (\frac{5\pi}{2}-\frac{3\pi}{2}) \\ 3\sin (\frac{5\pi}{2})\cos (\frac{3\pi}{2})=\frac{3}{2}(\sin (\frac{5\pi+3\pi}{2})+\sin (\frac{5\pi-3\pi}{2}) \\ 3\sin (\frac{5\pi}{2})\cos (\frac{3\pi}{2})=\frac{3}{2}(\sin (\frac{8\pi}{2})+\sin (\frac{2\pi}{2}) \\ 3\sin (\frac{5\pi}{2})\cos (\frac{3\pi}{2})=\frac{3}{2}(\sin (\frac{8\pi}{2})+\sin (\frac{2\pi}{2}) \\ 3\sin (\frac{5\pi}{2})\cos (\frac{3\pi}{2})=\frac{3}{2}(\sin (4\pi)+\sin (\pi) \end{gathered}[/tex]
Hence,
The final answer is
[tex]\Rightarrow\frac{3}{2}(\sin (4\pi)+\sin (\pi)[/tex]
