Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\dfrac{\pi }{2} < \alpha < \pi \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
\cos \alpha < 0
\end{array} \right.\\
\tan \alpha = - 2 \Rightarrow \dfrac{{\sin \alpha }}{{\cos \alpha }} = - 2 \Leftrightarrow \sin \alpha = - 2\cos \alpha \\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\left( { - 2\cos \alpha } \right)^2} + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\cos ^2}\alpha = \dfrac{1}{5}\\
\cos \alpha < 0 \Rightarrow \cos \alpha = - \dfrac{1}{{\sqrt 5 }} \Rightarrow \sin \alpha = \dfrac{2}{{\sqrt 5 }}\\
\cos 2\alpha = 2{\cos ^2}\alpha - 1 = - \dfrac{3}{5}\\
2,\\
\sin 2A + \sin 2B + \sin 2C\\
= \left( {\sin 2A + \sin 2B} \right) + \sin 2C\\
= 2\sin \dfrac{{2A + 2B}}{2}.\cos \dfrac{{2A - 2B}}{2} + 2\sin C.\cos C\\
= 2\sin \left( {A + B} \right).\cos \left( {A - B} \right) + 2\sin C.\cos C\\
= 2\sin \left( {180^\circ - A - B} \right).\cos \left( {A - B} \right) + 2\sin C.\cos C\\
= 2\sin C.\cos \left( {A - B} \right) + 2\sin C.\cos C\\
= 2\sin C.\left( {\cos \left( {A - B} \right) + \cos C} \right)\\
= 2\sin C.2.\cos \dfrac{{A - B + C}}{2}.\cos \dfrac{{A - B - C}}{2}\\
= 4\sin C.\sin \left( {90^\circ - \dfrac{{A - B + C}}{2}} \right).\sin \left( {90^\circ - \dfrac{{A - B - C}}{2}} \right)\\
= 4\sin C.\sin \dfrac{{\left( {A + B + C} \right) - \left( {A - B + C} \right)}}{2}.\sin \dfrac{{\left( {A + B + C} \right) - \left( {A - B - C} \right)}}{2}\\
= 4\sin C.\sin B.\sin \left( {B + C} \right)\\
= 4\sin C.\sin B.\sin \left( {180^\circ - B - C} \right)\\
= 4\sin C.\sin B.\sin A\\
= 4\sin A.\sin B.\sin C
\end{array}\)
