(4) Given $\sin \left(\frac{\pi}{4}-x\right)=\frac{3}{5}$, then the value of $\sin 2 x$ is ( ).
(A) $\frac{19}{25}$
(B) $\frac{16}{25}$
(C) $\frac{14}{25}$
(D) $\frac{7}{25}$
(4) From the given, we have
$$
\frac{\sqrt{2}}{2}(\cos x-\sin x)=\frac{3}{5},
$$

squaring both sides, we get
$$
\frac{1}{2}(1-\sin 2 x)=\frac{9}{25},
$$

solving for $\sin 2 x$ yields $\sin 2 x=\frac{7}{25}$. Alternatively, let $\frac{\pi}{4}-x=\alpha$, then $\sin \alpha=\frac{3}{5}$, so
$$
\sin 2 x=\sin \left(\frac{\pi}{2}-2 \alpha\right)=\cos 2 \alpha=1-2 \sin ^{2} \alpha=\frac{7}{25} .
$$

Therefore, the answer is D.