## Precalculus (6th Edition) Blitzer

$-\sqrt 3$
Step 1. Let $u=cos^{-1}(-\frac{1}{2})$; we have $cos(u)=-\frac{1}{2}$ Step 2. As $u\in[0,\pi]$, we know $u$ is in quadrant II and $sin(u)\gt0$ Step 3. We have $sin(u)=\sqrt {1-cos^2(u)}=\sqrt {1-\frac{1}{4}}=\frac{\sqrt 3}{2}$ Step 4. Combining the above results, we have $tan(u)=\frac{sin(u)}{cos(u)}=-\sqrt 3$ Step 5. Thus we have $tan(cos^{-1}(-\frac{1}{2}))= -\sqrt 3$