## Trigonometry 7th Edition

$-\frac{1}{2}$
$\sin(\frac{\pi}{6}-\frac{\pi}{3})$ = $\sin(\frac{-3\pi}{18}) = \sin(-\frac{\pi}{6})$ Using the identity $\sin(-x) = -\sin x$ we obtain, $\sin(-\frac{\pi}{6})$ = $-\sin\frac{\pi}{6}$ = $-\frac{1}{2}$