詳解：\(\theta \) 為第三象限角，如圖

故 \(\sin \frac{\theta }{2} = \frac{{ - 2}}{{\sqrt 5 }}\) ，  \(\cos \frac{\theta }{2} = \frac{{ - 1}}{{\sqrt 5 }}\)
(A)  \(\sin \frac{\theta }{2} = \frac{{ - 2}}{{\sqrt 5 }}\)
(B)  \(\sin \theta  = 2\sin \frac{\theta }{2}\cos \frac{\theta }{2} = 2 \cdot \frac{{ - 1}}{{\sqrt 5 }} \cdot \frac{{ - 2}}{{\sqrt 5 }} = \frac{4}{5}\)
(C) \(\because \) \(\cos \theta  = \cos 2\frac{\theta }{2} = 1 - 2{\sin ^2}\frac{\theta }{2} = 1 - 2 \cdot \frac{4}{5} =  - \frac{3}{5}\)
\(\therefore \)  \(\sin 2\theta  = 2\sin \theta \cos \theta  = 2\left( {\frac{4}{5}} \right)\left( { - \frac{3}{5}} \right) = \frac{{ - 24}}{{25}}\)
(D)  \(\sin 3\theta  = 3\sin \theta  - 4{\sin ^3}\theta  = 3\left( {\frac{4}{5}} \right) - 4{\left( {\frac{4}{5}} \right)^3} = \frac{{44}}{{125}}\)
故選(B)