151 Let $0<x<\frac{\pi}{6}$, then the order of $\sin (\sin x), \cos (\cos x), \sin (\cos x), \cos (\sin x)$ is
A. $\sin (\sin x)<\cos (\cos x)<\sin (\cos x)<\cos (\sin x)$
B. $\sin (\sin x)<\sin (\cos x)<\cos (\cos x)<\cos (\sin x)$
C. $\sin (\sin x)<\sin (\cos x)<\cos (\sin x)<\cos (\cos x)$
D. $\cos (\cos x)<\sin (\sin x)<\cos (\sin x)<\sin (\cos x)$
151 A. Since $0<x<\frac{\pi}{6}$, we have $0<\sin x<\frac{1}{2}<\frac{\pi}{4}<\frac{\sqrt{3}}{2}<\cos x<1<\frac{\pi}{2}$, thus
$$
\cos (\cos x)<\sin (\cos x) .
$$

Furthermore, since $\sin x+\cos x=\sqrt{2} \cdot \sin \left(x+\frac{\pi}{4}\right)<\sqrt{2}<\frac{\pi}{2}$, we have $0<\sin x<\frac{\pi}{2}-$ $\cos x, 0<\cos x<\frac{\pi}{2}-\sin x$, thus
$$
\begin{array}{l}
\sin (\sin x)<\sin \left(\frac{\pi}{2}-\cos x\right)=\cos (\cos x), \\
\sin (\cos x)<\sin \left(\frac{\pi}{2}-\sin x\right)=\cos (\sin x)
\end{array}
$$

Therefore, the inequality $\sin (\sin x)<\cos (\cos x)<\sin (\cos x)<\cos (\sin x)\left(0<x<\frac{\pi}{6}\right)$ holds.