Cho \(\tan \alpha  =  - \sqrt 5 \,\,\,\,\left( {\frac{\pi }{2}...

* Hướng dẫn giải

Do \(\frac{\pi }{2} < \alpha  < \pi  \Rightarrow \cos \alpha  < 0\)

Ta có \(\frac{1}{{{\rm{co}}{{\rm{s}}^2}\alpha }} = 1 + {\tan ^2}\alpha  = 6 \Rightarrow \cos \alpha  = \frac{{ - \sqrt 6 }}{6}\)

\(\sin \alpha  = \cos \alpha .\tan \alpha  = \frac{{\sqrt {30} }}{6} \Rightarrow \sin 2\alpha  = 2\sin \alpha \cos \alpha  =  - \frac{{\sqrt 5 }}{3}\)