Cho \(\cos a = - \dfrac{5}{{13}}\) và \(\pi jwplayer.key="...

* Hướng dẫn giải

Ta có: \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\)

\(\begin{array}{l} \Leftrightarrow {\sin ^2}\alpha  + {\left( { - \dfrac{5}{{13}}} \right)^2} = 1\\ \Leftrightarrow {\sin ^2}\alpha  = \dfrac{{144}}{{169}}\end{array}\)

\( \Rightarrow \sin \alpha  =  - \dfrac{{12}}{{13}}\) (do \(\pi  < \alpha  < \dfrac{{3\pi }}{2} \Rightarrow \sin \alpha  < 0\))

Khi đó \(\sin 2\alpha  = 2\sin \alpha \cos \alpha \) \( = 2.\dfrac{{ - 12}}{{13}}.\dfrac{{ - 5}}{{13}} = \dfrac{{120}}{{169}}\)

Chọn D