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

a) \(\cos \alpha = - \frac{{2\sqrt 2 }}{3}\).

b) \(\sin 2\alpha = - \frac{{4\sqrt 2 }}{9}\).

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

d) \(\cot 2\alpha = \frac{{7\sqrt 2 }}{8}\).

a) \({\rm{ V\`i  }}\frac{\pi }{2} < \alpha  < \pi {\rm{ n\^e n }}\cos \alpha  < 0.{\rm{ }}\)

Ta có: \({\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = 1 - \frac{1}{9} = \frac{8}{9} \Rightarrow \cos \alpha  =  - \frac{{2\sqrt 2 }}{3}\).

b) \(\sin 2\alpha  = 2\sin \alpha \cos \alpha  = 2 \cdot \frac{1}{3} \cdot \left( { - \frac{{2\sqrt 2 }}{3}} \right) =  - \frac{{4\sqrt 2 }}{9}\).

c) \(\cos 2\alpha  = 1 - 2{\sin ^2}\alpha  = 1 - 2{\left( {\frac{1}{3}} \right)^2} = \frac{7}{9}\).

d) \(\tan 2\alpha  = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} =  - \frac{{4\sqrt 2 }}{7}\); \(\cot 2\alpha  = \frac{1}{{\tan 2\alpha }} =  - \frac{{7\sqrt 2 }}{8}\).

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