\(\sin \alpha = \frac{2}{3}\)
A \(\cos \alpha = \pm \frac{{\sqrt 5 }}{3}\,\,;\,\,\,\tan \alpha = \pm \frac{{2\sqrt 5 }}{5}\,\,;\,\,\,\cot \alpha = \pm \frac{{\sqrt 5 }}{2}\)
B \(\cos \alpha = – \frac{{\sqrt 5 }}{3}\,\,;\,\,\,\tan \alpha = – \frac{{2\sqrt 5 }}{5}\,\,;\,\,\,\cot \alpha = – \frac{{\sqrt 5 }}{2}\)
C \(\cos \alpha = \frac{{\sqrt 5 }}{3}\,\,;\,\,\,\tan \alpha = \frac{{2\sqrt 5 }}{5}\,\,;\,\,\,\cot \alpha = \frac{{\sqrt 5 }}{2}\)
D \(\cos \alpha = \pm \frac{{\sqrt 5 }}{3}\,\,;\,\,\,\tan \alpha = \frac{{2\sqrt 5 }}{5}\,\,;\,\,\,\cot \alpha = \frac{{\sqrt 5 }}{2}\)
Hướng dẫn Chọn đáp án là: C
Phương pháp giải:
Sử dụng công thức lượng giác: \(\left\{ \begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\\tan \alpha .\cot \alpha = 1\\1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\end{array} \right..\)
Lời giải chi tiết:
Ta có: \(0 < \alpha < {90^0}\) \( \Rightarrow \left\{ \begin{array}{l}\sin \alpha > 0\\\cos \alpha > 0\\\tan \alpha > 0\\\cot \alpha > 0\end{array} \right..\)
\(\sin \alpha = \frac{2}{3}\)
*\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\left( {\frac{2}{3}} \right)^2} + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\cos ^2}\alpha = 1 – \frac{4}{9} = \frac{5}{9}\)\( \Rightarrow \cos \alpha = \frac{{\sqrt 5 }}{3}\)
*\(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{2}{3}:\frac{{\sqrt 5 }}{3} = \frac{{2\sqrt 5 }}{5}\)
*\(\cot \alpha = \frac{1}{{\tan \alpha }} = 1:\frac{{2\sqrt 5 }}{5} = \frac{{\sqrt 5 }}{2}\)
Chọn C.