2.3.1 * Given that $\triangle A B C$ is inscribed in a unit circle. Then the three segments of lengths $\sin A, \sin B, \sin C$ ( ).
(A) can form a triangle, whose area is greater than $\frac{1}{2}$ of the area of $\triangle A B C$
(B) can form a triangle, whose area is equal to $\frac{1}{2}$ of the area of $\triangle A B C$
(C) can form a triangle, whose area is less than $\frac{1}{2}$ of the area of $\triangle A B C$
(D) may not necessarily form a triangle
From the Law of Sines, we get $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2$, so the triangle formed by $\sin A, \sin B, \sin C$ as sides is similar to $\triangle ABC$, and its area is $\frac{1}{4}$ of the area of $\triangle ABC$. Therefore, the answer is $C$.