Example 11. In $\triangle \mathrm{A} B \mathrm{C}$, find the extremum of $\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}$.

 untranslated text remains the same as it is a mathematical expression.
Solve $\because \sin A+\sin B+\sin C=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$,
and $\cos \frac{A}{2} \cos \frac{B}{2}=\cos \left(\frac{A+B}{4}+\frac{A-B}{4}\right)$
$$
\begin{array}{l}
\cos \left(\frac{A+B}{4}-\frac{A-B}{4}\right) \\
=\cos ^{2}-\frac{A+B}{2}-\sin ^{2} \frac{A-B}{2},
\end{array}
$$

Since $\sin ^{2} \frac{A-B}{2} \geqslant 0$, it follows that when $A=B$.

Similarly, when $B=C$ and $C=A$, $\cos \frac{B}{2} \cos \frac{C}{2}$ and $\cos \frac{C}{2} \cos \frac{A}{2}$ are maximized.

Therefore, when $A=B=C$, $\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ is maximized.
$$
\begin{array}{l}
\text { Also, } \quad \cos \frac{A}{2}=\cos \frac{B}{2}=\cos \frac{C}{2}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \\
\therefore \quad \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \leqslant \frac{3 \sqrt{3}}{8}, \\
\therefore \quad(\sin A+\sin B+\sin C) \text { maximum value }=\frac{3 \sqrt{3}}{2} .
\end{array}
$$