Câu hỏi:

Tính các tỷ số lượng giác còn lại của \(\alpha \) biết:

Câu 1:

\({\rm{cos}}\alpha = \frac{3}{4}\)

  • 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 \(\sin \alpha = \pm \frac{4}{5}\,\,;\,\,\tan \alpha = \pm \frac{{15}}{{16}}\,\,;\,\,\cot \alpha = \pm \frac{{16}}{{15}}\)

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.


Câu 2:

\(cot\alpha = \frac{8}{{15}}\)

  • A \(\tan \alpha = \frac{{15}}{8}\,\,;\,\,\sin \alpha = \frac{{15}}{{17}}\,\,;\,\,\cos \alpha = \frac{8}{{17}}\)
  • B \(\tan \alpha = \pm \frac{{15}}{8}\,\,;\,\,\cos \alpha = \pm \frac{{15}}{{17}}\,\,;\,\,\sin \alpha = \pm \frac{8}{{17}}\)
  • C \(\tan \alpha = \frac{{15}}{8}\,\,;\,\,\cos \alpha = \frac{{15}}{{17}}\,\,;\,\,\sin \alpha = \frac{8}{{17}}\)
  • D \(\tan \alpha = \frac{{15}}{8}\,\,;\,\,\sin \alpha = \pm \frac{{15}}{{17}}\,\,;\,\,\cos \alpha = \pm \frac{8}{{17}}\)

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:

\(cot\alpha = \frac{8}{{15}}\)

* \(\tan \alpha .\cot \alpha = 1 \Leftrightarrow tan\alpha = \frac{1}{{\cot \alpha }} = \frac{1}{{\frac{8}{{15}}}} = \frac{{15}}{8}\)

* \(1 + {\cot ^2}\alpha = \frac{1}{{si{n^2}\alpha }}\)\( \Leftrightarrow 1 + {\left( {\frac{8}{{15}}} \right)^2} = \frac{1}{{si{n^2}\alpha }}\)\( \Leftrightarrow \frac{1}{{si{n^2}\alpha }} = \frac{{289}}{{225}}\)\( \Rightarrow si{n^2}\alpha = \frac{{225}}{{289}}\)\( \Rightarrow sin\alpha = \pm \frac{{15}}{{17}}\)

*\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\left( {\frac{{15}}{{17}}} \right)^2} + {\cos ^2}\alpha = 1\)\( \Leftrightarrow {\cos ^2}\alpha = 1 - \frac{{225}}{{289}} = \frac{{64}}{{289}}\)\( \Rightarrow \cos \alpha = \pm \frac{8}{{17}}\)

Chọn D.




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