78 Given $x \geqslant 0, y \geqslant 0$, and $x+y \leqslant 2 \pi$, try to find the maximum and minimum values of the function $f(x, y)=\sin x+\sin y- \sin (x+y)$.
78
$$
\begin{aligned}
f(x, y) & =\sin x+\sin y-\sin (x+y) \\
& =2 \cdot \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}-2 \cdot \sin \frac{x+y}{2} \cos \frac{x+y}{2} \\
& =2 \cdot \sin \frac{x+y}{2} \cdot\left(\cos \frac{x-y}{2}-\cos \frac{x+y}{2}\right) \\
& =4 \cdot \sin \frac{x+y}{2} \cdot \sin \frac{x}{2} \cdot \sin \frac{y}{2} \geqslant 0 .
\end{aligned}
$$

The equality holds if and only if \(x+y=2\pi\) or \(x=0\) or \(y=0\), so the minimum value of \(f(x, y)\) is 0.
From the above, we have
$$
\begin{aligned}
f(x, y) & =2 \sin \frac{x+y}{2} \cdot\left(\cos \frac{x-y}{2}-\cos \frac{x+y}{2}\right) \\
& \leqslant 2 \cdot \sin \frac{x+y}{2} \cdot\left(1-\cos \frac{x+y}{2}\right)
\end{aligned}
$$

Let \(t=\frac{x+y}{4}\left(0 \leqslant t \leqslant \frac{\pi}{2}\right)\), then
$$
\begin{aligned}
f(x, y) & \leqslant 2 \sin 2 t \cdot(1-\cos 2 t)=4 \sin t \cdot \cos t \cdot 2 \cdot \sin ^{2} t \\
& =8 \cdot\left[\left(\sin ^{2} t\right)^{3} \cdot \cos ^{2} t\right]^{\frac{1}{2}} \\
& =8 \cdot\left[\frac{1}{3} \cdot \sin ^{2} t \cdot \sin ^{2} t \cdot \sin ^{2} t \cdot\left(3-3 \cdot \sin ^{2} t\right)\right]^{\frac{1}{2}} \\
& \leqslant 8 \cdot\left[\frac{1}{3} \cdot\left(\frac{3}{4}\right)^{4}\right]^{\frac{1}{2}}=\frac{3}{2} \sqrt{3}
\end{aligned}
$$
(using the arithmetic-geometric mean inequality), the equality holds if and only if \(x=y\) and \(\sin ^{2} t=3-\sin ^{2} t\), i.e., \(x=y=\frac{2 \pi}{3}\), so the maximum value of \(f(x, y)\) is \(\frac{3 \sqrt{3}}{2}\).