use the power reduction formulas to rewrite the expression. (hint: your answer should not contain any…

Explanation:

Step1: Apply power - reduction formulas

The power - reduction formulas are $\sin^{2}\alpha=\frac{1 - \cos(2\alpha)}{2}$ and $\cos^{2}\alpha=\frac{1+\cos(2\alpha)}{2}$. For the given expression $7\sin^{2}(x)\cos^{2}(x)$, substitute the formulas: [ \begin{align*} 7\sin^{2}(x)\cos^{2}(x)&=7\times\frac{1 - \cos(2x)}{2}\times\frac{1+\cos(2x)}{2}\ \end{align*} ]

Step2: Use the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$

Here $a = 1$ and $b=\cos(2x)$, so we have: [ \begin{align*} 7\times\frac{1 - \cos(2x)}{2}\times\frac{1+\cos(2x)}{2}&=\frac{7}{4}(1-\cos^{2}(2x)) \end{align*} ]

Step3: Apply the power - reduction formula again for $\cos^{2}(2x)$

Since $\cos^{2}(2x)=\frac{1+\cos(4x)}{2}$, we get: [ \begin{align*} \frac{7}{4}(1-\cos^{2}(2x))&=\frac{7}{4}\left(1-\frac{1 + \cos(4x)}{2}\right)\ &=\frac{7}{4}\times\frac{2-(1 + \cos(4x))}{2}\ &=\frac{7}{8}(1-\cos(4x)) \end{align*} ]

Answer:

$\frac{7}{8}(1 - \cos(4x))$