1. Given $\sin 18^{\circ}=\frac{-1+\sqrt{5}}{4}$. Then
$$
\begin{array}{l}
a=\sin 2007^{\circ}, b=\cos 2007^{\circ}, \\
c=-1+\tan 2007^{\circ}, \\
d=\frac{\sin 2007^{\circ}+\cos 2007^{\circ}}{\sqrt{2}}
\end{array}
$$

the size relationship is ( ).
(A) $b<a<d<c$
(B) $b<a<c<d$
(C) $c<b<a<d$
(D) $d<b<c<a$
-1.D.
Since $2007^{\circ}=5 \times 360^{\circ}+180^{\circ}+27^{\circ}$, we have:
$$
\begin{array}{l}
a=\sin 2000^{\circ}=-\sin 27^{\circ}, b=\cos 2007^{\circ}=-\cos 27^{\circ}, \\
c=-1+\tan 2007^{\circ}=-1+\tan 27^{\circ}, \\
d=\frac{\sin 2007^{\circ}+\cos 2007^{\circ}}{\sqrt{2}}=-\frac{\sin 27^{\circ}+\cos 27^{\circ}}{\sqrt{2}} \\
=-\sin \left(45^{\circ}+27^{\circ}\right)=-\sin 72^{\circ}=-\cos 18^{\circ} .
\end{array}
$$

Thus, $db \Leftrightarrow-1+\tan 27^{\circ}>-\cos 27^{\circ} \\
\Leftrightarrow \tan 27^{\circ}+\cos 27^{\circ}>1 \Leftrightarrow \sin \left(45^{\circ}+27^{\circ}\right)>\sin 45^{\circ},
\end{array}
$$

which is clearly true. Therefore, $c>b$ holds.
$$
\begin{array}{l}
c3-2 \sqrt{2}=(\sqrt{2}-1)^{2} \\
\Leftrightarrow 2 \sin 18^{\circ}>\sqrt{2}-1 \Leftrightarrow \frac{-1+\sqrt{5}}{2}>\sqrt{2}-1 \\
\Leftrightarrow 1+\sqrt{5}>2 \sqrt{2} \Leftrightarrow 6+2 \sqrt{5}>8 \Leftrightarrow \sqrt{5}>1,
\end{array}
$$

which is clearly true. Therefore, $c<a$ holds.
From the above, we know that $d<b<c<a$.