(10) The sum of the maximum and minimum values of the function $y=\sin x+\sqrt{2+\cos ^{2} x}$ is

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(10) $2 \sqrt{2}$ Hint: From $-1 \leqslant \sin x \leqslant 1,0 \leqslant \cos ^{2} x \leqslant 1$, we know that when $x=\frac{3 \pi}{2}$, the minimum value of $\sin x+\sqrt{2+\cos ^{2} x}$ is $\sqrt{2}-1$.
Let $\sin x=\sqrt{3} \cos \theta$,
$\theta \in\left[\arccos \frac{1}{\sqrt{3}}, \arccos \left(-\frac{1}{\sqrt{3}}\right)\right] \subset\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$,
$\sqrt{2+\cos ^{2} x}=\sqrt{3} \sin \theta$, then
$$
y=\sqrt{3} \cos \theta+\sqrt{3} \cos \theta=\sqrt{6} \sin \left(\theta+\frac{\pi}{4}\right) .
$$

Since $\theta+\frac{\pi}{4} \in\left(\frac{\pi}{2}, \pi\right), y=\sqrt{6} \sin \left(\theta+\frac{\pi}{4}\right)$ is monotonically decreasing,
$$
\begin{aligned}
y_{\max } & =y\left(\arccos \frac{1}{\sqrt{3}}\right)=\sqrt{6} \sin \left(\arccos \frac{1}{\sqrt{3}}+\frac{\pi}{4}\right) \\
& =\sqrt{6}\left(\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{2}}\right)=\sqrt{2}+1 .
\end{aligned}
$$

Therefore, the sum of the maximum and minimum values is $2 \sqrt{2}$.