## Trigonometry (11th Edition) Clone

$-\frac{1}{2}$
RECALL: In a unit circle, $\cos{s} = x$ (Refer to Figure 13 on page 111) The terminal side of the angle $-\frac{5\pi}{6}$ can be located by moving $\frac{5\pi}{6}$ clockwise from the positive x-axis. This angle has the same terminal side as $\frac{7\pi}{6}$ and intersects intersects the unit circle at the point $(-\frac{\sqrt3}{2}, -\frac{1}{2})$. This point has $x=-\frac{\sqrt3}{2}$ and $y=-\frac{1}{2}$. Since $\sin{s}=y$, then $\sin{(-\frac{5\pi}{6})} =\sin{\frac{7\pi}{6}} = -\frac{1}{2}$