## Precalculus (6th Edition) Blitzer

The required solution is $-2\sin \frac{x}{2}\cos \frac{3x}{2}$.
One of the sum-to-product formula is $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$. So, in this question, according to the above-mentioned formula, the value of $\alpha$ is $x$ and the value of $\beta$ is $2x$. Now, the expression can be evaluated as provided below: \begin{align} & \sin x-\sin 2x=2\sin \frac{x-2x}{2}\cos \frac{x+2x}{2} \\ & =2\sin \frac{-x}{2}\cos \frac{3x}{2} \end{align} Then, applying the even-odd identity, which is $sin(-x)=-\sin x$, the expression can be further evaluated as given below: $2\sin \frac{-x}{2}\cos \frac{3x}{2}==-2\sin \frac{x}{2}\cos \frac{3x}{2}$ Hence, the provided expression can be written as $-2\sin \frac{x}{2}\cos \frac{3x}{2}$. So, it is not possible to find the exact value.