by \({\rm{cos}}\alpha = \frac{3}{4}\)
A \(\sin \alpha = \pm \frac{4}{5}\,\,;\,\,\tan \alpha = \pm \frac{{16}}{{15}}\,\,;\,\,\cot \alpha = \pm \frac{{15}}{{16}}\)
B \(\sin \alpha = \frac{4}{5}\,\,;\,\,\tan \alpha = \frac{{16}}{{15}}\,\,;\,\,\cot \alpha = \frac{{15}}{{16}}\)
C \(\sin \alpha = \frac{4}{5}\,\,;\,\,\tan \alpha = \frac{{15}}{{16}}\,\,;\,\,\cot \alpha = \frac{{16}}{{15}}\)
D \(\sin \alpha = \pm \frac{4}{5}\,\,;\,\,\tan \alpha = \pm \frac{{15}}{{16}}\,\,;\,\,\cot \alpha = \pm \frac{{16}}{{15}}\)
Hướng dẫn Chọn đáp án là: A
Phương pháp giải:
Sử dụng công thức lượng giác: \(\left\{ \begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\\tan \alpha .\cot \alpha = 1\\1 + {\cot ^2}\alpha = \frac{1}{{si{n^2}\alpha }}\end{array} \right..\)
Lời giải chi tiết:
\({\rm{cos}}\alpha = \frac{3}{4}\)
*\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\sin ^2}\alpha + {\left( {\frac{3}{4}} \right)^2} = 1\)\( \Leftrightarrow {\sin ^2}\alpha = 1 – \frac{9}{{25}} = \frac{{16}}{{25}}\)\( \Rightarrow \sin \alpha = \pm \frac{4}{5}\)
*\(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \pm \frac{4}{5}:\frac{3}{4} = \pm \frac{{16}}{{15}}\)
*\(\cot \alpha = \frac{1}{{\tan \alpha }} = 1:\left( { \pm \frac{{16}}{{15}}} \right) = \pm \frac{{15}}{{16}}\)
Chọn A.
