Select the correct option. Rewrite \( 4 \sin ^{3} \theta \) in terms with no power greater than one. \( \begin{array}{l}2 \sin \theta-\sin \theta \cos 2 \theta \\ \text { (in } \theta-\sin \theta \cos 2 \theta \\ 2\end{array} 0 \frac{\sin \theta+\sin \theta-2 \sin \theta \cos 2 \theta}{2} \)

To rewrite \( 4 \sin^{3} \theta \) in terms with no power greater than one, we can use the identity for \( \sin^2 \theta \): \[ \sin^2 \theta = 1 - \cos^2 \theta \] Thus, we can express \( \sin^3 \theta \) as: \[ \sin^3 \theta = \sin \theta \cdot \sin^2 \theta = \sin \theta (1 - \cos^2 \theta) \] Now, substituting this into \( 4 \sin^3 \theta \): \[ 4 \sin^3 \theta = 4 \sin \theta (1 - \cos^2 \theta) = 4 \sin \theta - 4 \sin \theta \cos^2 \theta \] Next, we can use the double angle identity for cosine, which states: \[ \cos 2\theta = 2\cos^2 \theta - 1 \quad \Rightarrow \quad \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \] Substituting this into our expression gives: \[ 4 \sin^3 \theta = 4 \sin \theta - 4 \sin \theta \left(\frac{1 + \cos 2\theta}{2}\right) \] Simplifying this: \[ = 4 \sin \theta - 2 \sin \theta (1 + \cos 2\theta) = 4 \sin \theta - 2 \sin \theta - 2 \sin \theta \cos 2\theta \] Combining like terms: \[ = 2 \sin \theta - 2 \sin \theta \cos 2\theta \] Thus, we can rewrite \( 4 \sin^3 \theta \) as: \[ 2 \sin \theta (1 - \cos 2\theta) \] Now, we can express this in the form of the options provided. The correct option is: \[ 2 \sin \theta - 2 \sin \theta \cos 2\theta \] So, the final answer is: \[ 2 \sin \theta - \sin \theta \cos 2\theta \]