Find the exact value of the expression. \[ \sin \frac{\pi}{6} \cos \frac{2 \pi}{3}-\cos \frac{\pi}{6} \sin \frac{2 \pi}{3} \] \( \sin \frac{\pi}{6} \cos \frac{2 \pi}{3}-\cos \frac{\pi}{6} \sin \frac{2 \pi}{3}=\square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

To evaluate the expression \[ \sin \frac{\pi}{6} \cos \frac{2 \pi}{3} - \cos \frac{\pi}{6} \sin \frac{2 \pi}{3}, \] we can use the sine subtraction formula: \[ \sin(a - b) = \sin a \cos b - \cos a \sin b. \] Letting \( a = \frac{\pi}{6} \) and \( b = \frac{2\pi}{3} \), we find that \[ \sin \left( \frac{\pi}{6} - \frac{2\pi}{3} \right) = \sin \left( \frac{\pi}{6} - \frac{4\pi}{6} \right) = \sin \left( -\frac{3\pi}{6} \right) = \sin \left( -\frac{\pi}{2} \right). \] The sine of \(-\frac{\pi}{2}\) is \[ -1. \] Thus, \[ \sin \frac{\pi}{6} \cos \frac{2 \pi}{3} - \cos \frac{\pi}{6} \sin \frac{2 \pi}{3} = -1. \] In conclusion, the value of the expression is \[ \boxed{-1}. \]