Ta có: \(\tan \alpha = 2\sqrt 2 \Rightarrow \dfrac{{\sin \alpha }}{{\cos \alpha }} = 2\sqrt 2 \) \( \Rightarrow \sin \alpha = 2\sqrt 2 \cos \alpha \)
Mà \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) \( \Rightarrow 8{\cos ^2}\alpha + {\cos ^2}\alpha = 1\) \( \Leftrightarrow {\cos ^2}\alpha = \dfrac{1}{9} \Leftrightarrow \cos \alpha = \dfrac{1}{3}\) (vì trong khoảng \(\left( {{0^0};{{90}^0}} \right)\) thì \(\cos \alpha > 0\))
Suy ra \(\sin \alpha = \tan \alpha .\cos \alpha \)\( = 2\sqrt 2 .\dfrac{1}{3} = \dfrac{{2\sqrt 2 }}{3}\)
