Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\pi < \alpha < \dfrac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha < 0\\
\cos \alpha < 0
\end{array} \right.\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\cos \alpha < 0 \Rightarrow \cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \dfrac{5}{{13}}\\
\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{5}{{12}}\\
\cos 2\alpha = 2{\cos ^2}\alpha - 1 = 2.{\left( { - \dfrac{5}{{13}}} \right)^2} - 1 = - \dfrac{{119}}{{169}}\\
\sin 2\alpha = 2\sin \alpha .\cos \alpha = 2.\dfrac{{ - 12}}{{13}}.\dfrac{{ - 5}}{{13}} = \dfrac{{120}}{{169}}
\end{array}\)
