(4) A diameter $AB$ of a semicircle equals 2, and a perpendicular line to the plane of the circle is drawn through $A$. A point $S$ is taken on this perpendicular line such that $AS = AB$. $C$ is a moving point on the semicircle, and $M$ and $N$ are the projections of $A$ on $SB$ and $SC$, respectively. When the volume of the tetrahedron $S-AMN$ is maximized, the sine of the angle between $SC$ and the plane $ABC$ is $\qquad$.
(4) $\frac{\sqrt{3}}{2}$ Hint: It is easy to know that $B C \perp$ plane $S A C$, so $B C \perp A N$, thus $A N \perp$ plane $S B C$. Therefore, $A N \perp S M$, so $S M \perp$ plane $A M N$,
$$
V_{S-A M N}=\frac{1}{3} \cdot S M \cdot S_{\triangle A N M} .
$$

From $S A=A B=2$, we get $A M=S M=\sqrt{2}$, and $A N \perp N M, \triangle A M N$ is a right triangle with hypotenuse length $\sqrt{2}$, the maximum area is obtained 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, $\angle S C A=60^{\circ},$ the sine value of the angle formed by $S C$ and plane $A B C$ is $\frac{\sqrt{3}}{2}$.