smc

Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $MC=MD=\sqrt{3}.$ Let $O$ be the center of $△ABD,$ so $\overline{CO}\perp \overline{MOD}.$ Note that $MO=\frac{1}{3}MD=\frac{\sqrt{3}}{3}.$ In right $△CMO,$ we have $$\cos (\angle CMD)=\frac{MO}{MC}=\boxed{\frac{1}{3}}.$$