7. (8th "Hope Cup" Invitational Competition Question) In the cube $A C_{1}$, $M$ and $N$ are the midpoints of edges $C_{1} D_{1}$ and $A B$ respectively. Points $A_{1}$, $M$, $C$, and $N$ lie in the same plane. The sine value of the angle formed by $C D$ and the plane $A_{1} M C N$ is ( ).
A. $\frac{1}{2}$
B. $\frac{\sqrt{3}}{2}$
C. $\frac{\sqrt{6}}{3}$
D. $\frac{\sqrt{2}}{4}$
7.C. Reason: From the given conditions, we know that quadrilateral $A_{1} M C N$ is a rhombus, so $A_{1} C$ bisects $\angle M A_{1} N$ and $\angle M C N$. Also, since $\angle D C N=\angle M C D$, that is, $C D$ forms equal angles with the two sides of $\angle M C N$, the projection of $C D$ onto the plane $M C N$ is $A_{1} C$. Therefore, $\angle D C A_{1}$ is the angle formed by $C D$ and the plane $A_{1} M C N$. It is easy to know that $\tan \angle D C A_{1}=\sqrt{2}$, $\sin \angle D C A_{1}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$.