Cho biết sin alpha = 1/3 và pi/2 < alpha < pi. Khi đó:

a) Đúng. Vì \(\frac{\pi }{2} < \alpha < \pi \) nên \({\rm{cos}}\alpha < 0\). Ta có \({\rm{cos}}\alpha = - \sqrt {1 - {\rm{si}}{{\rm{n}}^2}\alpha } = - \sqrt {1 - \frac{1}{9}} = - \frac{{2\sqrt 2 }}{3}\).

b) Đúng. \({\rm{cos}}2\alpha = 1 - 2{\rm{si}}{{\rm{n}}^2}\alpha = 1 - 2 \cdot \frac{1}{9} = \frac{7}{9}\).

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

d) Sai. \({\rm{cot}}2\alpha = \frac{{{\rm{cos}}2\alpha }}{{{\rm{sin}}2\alpha }} = \frac{{7/9}}{{ - 4\sqrt 2 /9}} = - \frac{7}{{4\sqrt 2 }} = - \frac{{7\sqrt 2 }}{8}\).