3. Calculate the maximum value of the expression $\sin \alpha+\sin \beta+\sin \gamma$, if $\alpha, \beta$ and $\gamma$ are angles in a triangle.
Solution. Let two of the angles $\alpha, \beta$ and $\gamma$ be acute. Let these angles be $\alpha$ and $\beta$. Since $\alpha$ and $\beta$ are acute angles, it follows that $0 \leq \cos \frac{\alpha-\beta}{2} \leq 1$, and since $0<\gamma<\pi$, it follows that $0 \leq \cos \left(\frac{\gamma}{2}-\frac{\pi}{6}\right) \leq 1$.

We obtain:

$$
\begin{aligned}
\sin \alpha+\sin \beta+\sin \gamma & =2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}+\sin \gamma \leq 2 \sin \frac{\alpha+\beta}{2}+\sin \gamma+\sin \frac{\pi}{3}-\sin \frac{\pi}{3} \\
& =2 \sin \frac{\alpha+\beta}{3}+2 \sin \left(\frac{\gamma}{2}+\frac{\pi}{6}\right) \cos \left(\frac{\gamma}{2}-\frac{\pi}{6}\right)-\sin \frac{\pi}{3} \\
& \leq 2 \sin \frac{\alpha+\beta}{3}+2 \sin \left(\frac{\gamma}{2}+\frac{\pi}{6}\right)-\sin \frac{\pi}{3} \\
& =4 \sin \left(\frac{1}{2}\left(\frac{\alpha+\beta+\gamma}{2}+\frac{\pi}{6}\right)\right) \cos \left(\frac{1}{2}\left(\frac{\alpha+\beta+\gamma}{2}-\frac{\pi}{6}\right)\right)-\sin \frac{\pi}{3} \\
& =4 \sin \frac{\pi}{3} \cos \left(\frac{\gamma}{2}-\frac{\pi}{6}\right)-\sin \frac{\pi}{3} \leq 4 \sin \frac{\pi}{3}-\sin \frac{\pi}{3}=\frac{3 \sqrt{3}}{2}
\end{aligned}
$$

If $\alpha=\beta=\gamma=\frac{\pi}{3}$, then $\sin \alpha+\sin \beta+\sin \gamma=\frac{3 \sqrt{3}}{2}$. Therefore, the maximum value that the expression $\sin \alpha+\sin \beta+\sin \gamma$ can take is $\frac{3 \sqrt{3}}{2}$.