Biết \(\cos 2\alpha  = \frac{5}{9},0^\circ  < \alpha  < 90^\circ \).

a) \(\sin \alpha  = \frac{{\sqrt {28} }}{9}\).

b) \(\cos \alpha  = \frac{{\sqrt {53} }}{9}\).

c) \(\tan \alpha  = \frac{{\sqrt {371} }}{{53}}\).

d) \(\cot \alpha  = \frac{{\sqrt {371} }}{{14}}\).

Ta có \(\cos 2\alpha = \frac{5}{9},0^\circ < \alpha < 90^\circ \)

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

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

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

Đáp án:           a) Đúng,          b) Đúng,         c) Sai,              d) Đúng.