4. A semicircle with diameter $A B=2$, a perpendicular line to the plane of the circle is drawn through point $A$, and a point $S$ is taken on this perpendicular line such that $A S=A B$. $C$ is a moving point on the semicircle, and $M, N$ are the projections of point $A$ on $S B, S C$ respectively. When the volume of the pyramid $S-A M N$ is maximized, the sine of the angle formed by $S C$ and the plane $A B C$ is $\qquad$
4. $\frac{\sqrt{3}}{2}$.

It is easy to know that $B C \perp$ plane $S A C$
$\Rightarrow B C \perp A N \Rightarrow A N \perp$ plane $S B C$
$\Rightarrow A N \perp S M \Rightarrow S M \perp$ plane $A M N$.
Thus, $V_{S-A M N}=\frac{1}{3} S M \cdot S_{\triangle M N M}$.
Given $S A=A B=2$, we get $A M=S M=\sqrt{2}$.
And $A N \perp N M$, $\triangle A M N$ is a right-angled triangle with the hypotenuse length of $\sqrt{2}$, and its maximum area is achieved when $A N=M N=1$.

Therefore, when the volume of the tetrahedron $S-A M N$ is maximized, $A N=M N=1$. At this time, $\angle S C A=60^{\circ}$, and the sine value of the angle formed by $S C$ and plane $A B C$ is $\frac{\sqrt{3}}{2}$.