jmc

Let $P$ be the point on the unit circle that is $300^{\circ }$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis. Triangle $POD$ is a 30-60-90 triangle, so $DO=\frac{1}{2}$ and $DP=\frac{\sqrt{3}}{2}$. Therefore, the coordinates of $P$ are $\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$, so $\cos 300^{\circ }=\boxed{\frac{1}{2}}$.