jmc

Let $P$ be the point on the unit circle that is $315^{\circ }$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.



Triangle $POD$ is a 45-45-90 triangle, so $DO=DP=\frac{\sqrt{2}}{2}$. Therefore, the coordinates of $P$ are $\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$, so $\cos 315^{\circ }=\boxed{\frac{\sqrt{2}}{2}}$.