use the power - reducing formulas to rewrite the expression in terms of first powers of the cosines of…

Explanation:

Step1: Recall power - reducing formulas

The power - reducing formulas are $\sin^{2}\alpha=\frac{1 - \cos(2\alpha)}{2}$ and $\cos^{2}\alpha=\frac{1+\cos(2\alpha)}{2}$. First, rewrite $\sin^{4}(3x)$ as $(\sin^{2}(3x))^{2}$. $\sin^{2}(3x)=\frac{1 - \cos(6x)}{2}$, so $\sin^{4}(3x)=(\frac{1 - \cos(6x)}{2})^{2}=\frac{1 - 2\cos(6x)+\cos^{2}(6x)}{4}$. And $\cos^{2}(3x)=\frac{1+\cos(6x)}{2}$.

Step2: Multiply the two expressions

$\sin^{4}(3x)\cos^{2}(3x)=\frac{1 - 2\cos(6x)+\cos^{2}(6x)}{4}\times\frac{1+\cos(6x)}{2}$. Now, rewrite $\cos^{2}(6x)$ using the power - reducing formula $\cos^{2}(6x)=\frac{1+\cos(12x)}{2}$. So the expression becomes $\frac{1 - 2\cos(6x)+\frac{1+\cos(12x)}{2}}{4}\times\frac{1+\cos(6x)}{2}$. First, simplify the numerator of the first fraction: $1 - 2\cos(6x)+\frac{1+\cos(12x)}{2}=\frac{2-4\cos(6x)+1+\cos(12x)}{2}=\frac{3 - 4\cos(6x)+\cos(12x)}{2}$. Then the product is $\frac{\frac{3 - 4\cos(6x)+\cos(12x)}{2}}{4}\times\frac{1+\cos(6x)}{2}=\frac{3 - 4\cos(6x)+\cos(12x)}{16}\times(1 + \cos(6x))$. Expand the product: [ \begin{align*} &\frac{3 - 4\cos(6x)+\cos(12x)}{16}\times(1+\cos(6x))\ =&\frac{3+3\cos(6x)-4\cos(6x)-4\cos^{2}(6x)+\cos(12x)+\cos(6x)\cos(12x)}{16}\ \end{align*} ] Rewrite $\cos^{2}(6x)=\frac{1+\cos(12x)}{2}$ again: [ \begin{align*} =&\frac{3 - \cos(6x)-4\times\frac{1+\cos(12x)}{2}+\cos(12x)+\cos(6x)\cos(12x)}{16}\ =&\frac{3 - \cos(6x)-2 - 2\cos(12x)+\cos(12x)+\cos(6x)\cos(12x)}{16}\ =&\frac{1 - \cos(6x)-\cos(12x)+\cos(6x)\cos(12x)}{16} \end{align*} ] Using the product - to - sum formula $\cos A\cos B=\frac{1}{2}[\cos(A + B)+\cos(A - B)]$, where $A = 6x$ and $B=12x$, $\cos(6x)\cos(12x)=\frac{1}{2}[\cos(18x)+\cos(6x)]$. [ \begin{align*} &\frac{1 - \cos(6x)-\cos(12x)+\frac{1}{2}[\cos(18x)+\cos(6x)]}{16}\ =&\frac{1 - \cos(6x)-\cos(12x)+\frac{1}{2}\cos(18x)+\frac{1}{2}\cos(6x)}{16}\ =&\frac{1-\frac{1}{2}\cos(6x)-\cos(12x)+\frac{1}{2}\cos(18x)}{16}\ =&\frac{1}{16}-\frac{1}{32}\cos(6x)-\frac{1}{16}\cos(12x)+\frac{1}{32}\cos(18x) \end{align*} ]

Answer:

$\frac{1}{16}-\frac{1}{32}\cos(6x)-\frac{1}{16}\cos(12x)+\frac{1}{32}\cos(18x)$