Biết: $\cos 2\alpha =\frac{5}{9},0^{°}<\alpha <{90}^{°}$. Khi đó:

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: $\cos 2\alpha =\frac{5}{9},0^{°}<\alpha <{90}^{°}$

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

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

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