## Trigonometry 7th Edition

$-\frac{\sqrt 3}{2}$
Since cosine is an even function, $\cos(−\frac{5\pi}{6})=\cos(\frac{5\pi}{6})$. By using the unit circle, $\frac{5\pi}{6}$ is in the second quadrant, therefore cosine is negative. By using the identity $\cos(\pi-x)=-cos(x)$, $\cos(\frac{5\pi}{6})=-\cos(\pi-\frac{5\pi}{6})=-\cos(\frac{\pi}{6})$ Since $\frac{\pi}{6}$ is a special angle where $\cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$, $\cos(−\frac{5\pi}{6})=-\frac{\sqrt 3}{2}$