Cho sin alpha = 2/3. Tính cos alpha, tan alpha biết 0 < alpha < 90 độ

Ta có: \[0 < \alpha < 90 \Rightarrow \left\{ {\begin{array}{*{20}{c}}{cos\,\alpha \,\, > 0}\\{\tan \alpha > 0}\end{array}} \right.\]

\[\sin \alpha = \frac{2}{3}\]

sin² α + cos² α = 1

\[ \Leftrightarrow {\left( {\frac{2}{3}} \right)^2} + co{s^2}\alpha = 1\]

\[\cot \,\,135^\circ = - 1\]

\[\tan \alpha = \frac{{\sin \alpha }}{{cos\,\alpha }} = \frac{2}{3}:\frac{{\sqrt 5 }}{3} = \frac{{2\sqrt 5 }}{5}\]

Vậy \[cos\,\alpha = \frac{{\sqrt 5 }}{3};\,\,\,\tan \alpha = \frac{{2\sqrt 5 }}{5}\].