Cho \(\sin \alpha = \frac{1}{3}\) với \(\alpha \in \left( {\frac{\pi }{2};\pi } \right)\). Tính cos α, tanα, cot α.

Vì \(\alpha \in \left( {\frac{\pi }{2};\pi } \right)\) nên cos α < 0.

Do đó từ sin² α + cos² α = 1, suy ra

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

Khi đó, \[\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} = - \frac{1}{{2\sqrt 2 }} = - \frac{{\sqrt 2 }}{4}\];

\(\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{{ - \frac{{\sqrt 2 }}{4}}} = - 2\sqrt 2 \).