Cho \(\sin \alpha = \frac{4}{5}\) \(\left( {0^\circ < \alpha < 90^\circ } \right)\). Tính \(\cos \alpha \).

Do \(0^\circ < \alpha < 90^\circ \Rightarrow \cos \alpha > 0\).>

Từ \({\sin ^2}\alpha + {\cos ^2}\alpha = 1 \Rightarrow {\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - {\left( {\frac{4}{5}} \right)^2} = \frac{9}{{25}} \Rightarrow \cos \alpha = \frac{3}{5}\). Chọn B.