Biết: \(\cos 2\alpha  = \frac{5}{9},{0^^\circ } < \alpha  < {90^^\circ }\). Khi đó: a) \(\sin \alpha  = \frac{{\sqrt {28} }}{9}\)

a) Đúng

b) Đúng

c) Sai

d) Đúng

 

Ta có: cos2α=59,0°<α<90°

\({\sin ^2}\alpha = \frac{{1 - \cos 2\alpha }}{2} = \frac{{1 - {{\left( {\frac{5}{9}} \right)}^2}}}{2} = \frac{{28}}{{81}} \Rightarrow \sin \alpha = \pm \frac{{\sqrt {28} }}{9}\)

Vì 0°<α<90° nên \(\sin \alpha = \frac{{\sqrt {28} }}{9}\)

\({\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - {\left( {\frac{{\sqrt {28} }}{9}} \right)^2} = \frac{{53}}{{81}} \Rightarrow \cos \alpha = \pm \frac{{\sqrt {53} }}{{81}}\)

Vì 0°<α<90° nên \(\cos \alpha = \frac{{\sqrt {53} }}{9}\)

\(\begin{array}{l}\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{2\sqrt {371} }}{{53}}\\\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{{\sqrt {371} }}{{14}}.\end{array}\)