## Precalculus (6th Edition) Blitzer

$\dfrac{2\sqrt 3-1}{2}$
Since, $\sin \dfrac{\pi}{3}=\dfrac{\sqrt 3}{2}$ Now, we have $f(\dfrac{\pi}{3})= 2 \sin \dfrac{\pi}{3}-\sin ( \dfrac{\dfrac{\pi}{3}}{2})$ or, $=\dfrac{2\sqrt 3}{2}-\dfrac{1}{2}$ After simplification, we have or, $=\dfrac{2\sqrt 3-1}{2}$