Choose the correct answer from the drop-down menu. Rewrite \( 4 \sin ^{2} \theta \cos ^{2} \theta \) in terms with no power greater than one. \[ \begin{array}{l}4 \sin ^{2} \theta \cos ^{2} \theta=\text { Select... } v \\ \qquad \frac{1}{2}-\frac{1}{2}-\frac{1}{2} \cos 4 \theta \\ \frac{1}{2}+\frac{1}{2} \cos 2 \theta \\ 1\end{array}+\frac{1}{2} \cos 2 \theta \]

To rewrite \( 4 \sin^{2} \theta \cos^{2} \theta \) in terms with no power greater than one, you can use the double angle identity for sine. Recall that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \), therefore: \[ 4 \sin^{2} \theta \cos^{2} \theta = 4 \left( \frac{1}{2} \sin(2\theta) \right)^{2} = \sin^{2}(2\theta). \] Using the identity \( \sin^{2}(x) = \frac{1 - \cos(2x)}{2} \): \[ \sin^{2}(2\theta) = \frac{1 - \cos(4\theta)}{2}. \] Thus, multiplying by 2 gives: \[ 4 \sin^{2} \theta \cos^{2} \theta = \frac{1 - \cos(4\theta)}{2}. \] By rearranging this expression, it turns out that the correct option would be: \[ 4 \sin^{2} \theta \cos^{2} \theta = \frac{1}{2} - \frac{1}{2} \cos(4\theta). \]