1. In an acute triangle $\triangle A B C$, let $x=\sin A+\sin B+$ $\sin C, y=\cos A+\cos B+\cos C$. Then the relationship between $x$ and $y$ is $(\quad)$.
(A) $x>y$
(B) $x=y$
(C) $x<y$
(D) The relationship between $x$ and $y$ is uncertain
$$
\begin{array}{l}
-1 .(\Lambda) . \\
\angle A+\angle B=180^{\circ}-\angle C>90^{\circ} \\
\Rightarrow \sin A>\sin \left(90^{\circ}-B\right)=\cos B .
\end{array}
$$

Similarly, $\sin B>\cos C, \sin C>\cos A$.
$$
\therefore \sin A+\sin B+\sin C>\cos A+\cos B+\cos C \text {. }
$$

That is, $x>y$.