Precalculus (6th Edition) Blitzer

The exact value of the expression is $\frac{1-\sqrt{3}}{2}$.
Consider the following equation. $x=\sin \frac{\pi }{3}\cos \pi -\cos \frac{\pi }{3}\sin \frac{3\pi }{2}$ The values of the trigonometric functions are: \begin{align} & \sin \frac{\pi }{3}=\frac{\sqrt{3}}{2} \\ & \cos \pi =1 \\ & \cos \frac{\pi }{3}=\frac{1}{2} \\ & \sin \frac{3\pi }{2}=-1 \end{align} Now, substitute $\frac{\sqrt{3}}{2}$ for $\sin \frac{\pi }{3}$ , $-1$ for $\cos \pi$ , $\frac{1}{2}$ for $\cos \frac{\pi }{3}$ and $-1$ for $\sin \frac{3\pi }{2}$. \begin{align} & x=\frac{\sqrt{3}}{2}\left( -1 \right)-\left( \frac{1}{2} \right)\left( -1 \right) \\ & =-\frac{\sqrt{3}}{2}+\frac{1}{2} \\ & =\frac{-\sqrt{3}+1}{2} \\ & =\frac{1-\sqrt{3}}{2} \end{align} The exact value of the expression is $\frac{1-\sqrt{3}}{2}$.