Precalculus (6th Edition) Blitzer

The required value is $\frac{1}{2}\left[ \cos 4x-\cos 8x \right]$.
One of the product-to-sum formulas is $\sin \alpha sin\beta =\frac{1}{2}\left[ \cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right) \right]$. So, in this question, according to the above-mentioned formula, the value of $\alpha$ is $6x$ and the value of $\beta$ is $2x$. Thus, the expression can be evaluated as pprovided below. \begin{align} & \sin 6xsin2x=\frac{1}{2}\left[ \cos \left( 6x-2x \right)-\cos \left( 6x+2x \right) \right] \\ & =\frac{1}{2}\left[ \cos 4x-\cos 8x \right] \end{align}