Assuming $A \leqslant B \leqslant C$, then $A \leqslant 60^{\circ} \leqslant C$, so by the lemma, we have
$$\sin A+\sin B+\sin C \leqslant \sin 60^{\circ}+\sin \left(A+C-60^{\circ}\right)+\sin B$$

Assuming again that $A+C-60^{\circ} \leqslant B$, then because $\left(A+C-60^{\circ}\right)+B=A+B+C-$ $60^{\circ}=120^{\circ}$, we have $0^{\circ}<A+C-60^{\circ} \leqslant 60^{\circ} \leqslant B<180^{\circ}$. Using the lemma once more, we get
$$\begin{aligned}
\sin A+\sin B+\sin C & \leqslant \sin 60^{\circ}+\sin \left(A+C-60^{\circ}\right)+\sin B \\
& \leqslant \sin 60^{\circ}+\sin 60^{\circ}+\sin 60^{\circ} \\
& =3 \sin 60^{\circ}=\frac{3 \sqrt{3}}{2}
\end{aligned}$$

Therefore, the maximum value of $\sin A+\sin B+\sin C$ is $\frac{3 \sqrt{3}}{2}$.
Example 2 Let $A, B, C$ be the three interior angles of a triangle, find the maximum value of $\sin A+\sin B+\sin C$.

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