Biết: \(\sin \alpha = \frac{1}{3}\) và \(0 < \alpha < \frac{\pi }{2}\). Khi đó:

a) \(\sin 2\alpha = \frac{{\sqrt 2 }}{9}\)

b) \(\cos 2\alpha = \frac{7}{9}\)

c) \(\tan 2\alpha = \frac{{\sqrt 2 }}{7}\)

d) \(\cot 2\alpha = \frac{{7\sqrt 2 }}{2}\)

Ta có: \(\sin \alpha = \frac{1}{3}\) và \(0 < \alpha < \frac{\pi }{2}\).

\({\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - {\left( {\frac{1}{3}} \right)^2} = \frac{8}{9} \Rightarrow \cos \alpha = \pm \frac{{2\sqrt 2 }}{3}\)

Vì \(0 < \alpha < \frac{\pi }{2}\) nên \(\cos \alpha = \frac{{2\sqrt 2 }}{3}\)

\(\begin{array}{l}\sin 2\alpha = 2\sin \alpha \cdot \cos \alpha = 2 \cdot \frac{1}{3} \cdot \frac{{2\sqrt 2 }}{3} = \frac{{4\sqrt 2 }}{9}\\\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha = {\left( {\frac{{2\sqrt 2 }}{3}} \right)^2} - {\left( {\frac{1}{3}} \right)^2} = \frac{7}{9}\\\tan 2\alpha = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} = \frac{{4\sqrt 2 }}{2}:\frac{7}{9} = \frac{{4\sqrt 2 }}{7}\\\cot 2\alpha = \frac{{\cos 2\alpha }}{{\sin 2\alpha }} = \frac{{4\sqrt 2 }}{2}:\frac{7}{9} = \frac{{7\sqrt 2 }}{8}.\end{array}\)