1. In $\triangle A B C$, if
$$
\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=2,
$$

then the maximum value of $\cos A+\cos B+2 \cos C$ is ( ).
(A) $\frac{3 \sqrt{2}}{2}$
(B) $\sqrt{5}$
(C) $\frac{4 \sqrt{3}}{3}$
(D) $\sqrt{6}$
-,1. B.
From the problem, we know
$$
\begin{array}{l}
4=2 \sin ^{2} A+2 \sin ^{2} B+2 \sin ^{2} C \\
\Rightarrow \cos 2 A+\cos 2 B+2 \cos ^{2} C=0 \\
\Rightarrow 2 \cos (A+B) \cdot \cos (A-B)+2 \cos ^{2} C=0 \\
\Rightarrow-4 \cos A \cdot \cos B \cdot \cos C=0 \\
\Rightarrow \cos A, \cos B, \cos C \text { must have one that is } 0 .
\end{array}
$$
$\Rightarrow \cos A, \cos B, \cos C$ must have one that is 0.
If $\cos A=0$, then $\angle A=\frac{\pi}{2}, \angle B+\angle C=\frac{\pi}{2}$.
$$
\begin{array}{l}
\text { Hence } f(B)=\cos A+\cos B+2 \cos C \\
=\cos B+2 \sin B\left(0<\angle B<\frac{\pi}{2}\right)
\end{array}
$$

The maximum value outside is $\sqrt{5}$.
If $\cos B=0$, then the maximum value is $\sqrt{5}$.
If $\cos C=0$, then the maximum value is $\sqrt{2}$.
In summary, the maximum value of $\cos A+\cos B+2 \cos C$ is $\sqrt{5}$.