Answer:
Explanation:
Here, we want to use the power-reducing formula
We have that as follows:
[tex]\begin{gathered} sin^2(3x)cos^2(3x)\text{ = } \\ sin^2\text{ 3x = }\frac{1-cos\text{ 6x}}{2} \\ \\ cos^23x\text{ = }\frac{1+cos\text{ 6x}}{2} \end{gathered}[/tex]
Thus, we have the product as:
[tex][/tex]
