14. Let $x \in[-1,1], \arcsin x=\theta, \theta \in\left[-\frac{\pi}{2},-\frac{\pi}{4}\right]$, then the value of $\arcsin \left(2 x \sqrt{1-x^{2}}\right)$ is ( ).
A. $2 \theta$
B. $\pi-2 \theta$
C. $-(\pi+2 \theta)$
D. $-(\pi-2 \theta)$
14. C. Since $\theta \in\left[-\frac{\pi}{2},-\frac{\pi}{4}\right], \sin \theta=x, \therefore \quad \cos \theta=\sqrt{1-x^{2}}$,
$$
\begin{array}{l}
\therefore \quad \arcsin \left(2 x \sqrt{1-x^{2}}\right)=\arcsin (\sin 2 \theta) . \\
\because \quad 2 \theta \in\left[-\pi,-\frac{\pi}{2}\right], \therefore \quad-\frac{\pi}{2} \leqslant-(\pi+2 \theta) \leqslant 0, \therefore \quad \arcsin \left(2 x \sqrt{1-x^{2}}\right)=\arcsin [\sin (-\pi-2 \theta)]= \\
-(\pi+2 \theta), \text { so the answer is C. }
\end{array}
$$