2. D. Given $\theta \in\left(0, \frac{\pi}{2}\right)$, we know $\theta>\sin \theta$, which means $\cos \theta>\sin (\cos \theta)$. Also, $\frac{\pi}{2}>\theta>\sin \theta>0$, and within $\left(0, \frac{\pi}{2}\right)$, the cosine function is a decreasing function, so we have $\sin (\cos \theta)\sin (\sin \theta)$.
When $\theta=0$, $\sin (\cos \theta)=\sin 1, \cos (\sin \theta)=1$, thus we have
$\sin (\cos \theta)<\cos (\sin \theta)$.
When $\theta \in\left(\frac{\pi}{2}, \pi\right]$, $\cos \theta \in[-1,0], \sin \theta \in[0,1)$, and $\sin (\cos \theta)<\cos (\sin \theta)$. In summary, when $\theta \in[0, \pi]$, $\sin (\cos \theta)<\cos (\sin \theta)$.
2. Below are three propositions: (甲) When $\theta \in\left(0, \frac{\pi}{2}\right)$, $\sin (\cos \theta)\sin (\sin \theta)$; (丙) When $\theta \in[0, \pi]$, $\sin (\cos \theta)<\cos (\sin \theta)$. The number of correct propositions is ( ).
A. 0
B. 1
C. 2
D. 3