Cho Cos α = 5/2 tính Sin α ,tan α,cot α 17/07/2021 Bởi Aubrey Cho Cos α = 5/2 tính Sin α ,tan α,cot α
Đáp án: \(\left[ \begin{array}{l}\sin a = \dfrac{{\sqrt {21} }}{5}\\\sin a = – \dfrac{{\sqrt {21} }}{5}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}Do:\cos a = \dfrac{2}{5}\\{\sin ^2}a + {\cos ^2}a = 1\\ \to \dfrac{4}{{25}} + {\sin ^2}a = 1\\ \to {\sin ^2}a = \dfrac{{21}}{{25}}\\ \to \left[ \begin{array}{l}\sin a = \dfrac{{\sqrt {21} }}{5}\\\sin a = – \dfrac{{\sqrt {21} }}{5}\end{array} \right.\\ \to \left[ \begin{array}{l}\tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{{\sqrt {21} }}{5}:\dfrac{2}{5} = \dfrac{{\sqrt {21} }}{2}\\\tan a = \dfrac{{\sin a}}{{\cos a}} = – \dfrac{{\sqrt {21} }}{5}:\dfrac{2}{5} = – \dfrac{{\sqrt {21} }}{2}\end{array} \right.\\ \to \left[ \begin{array}{l}\cot a = \dfrac{1}{{\tan a}} = \dfrac{2}{{\sqrt {21} }}\\\cot a = \dfrac{1}{{\tan a}} = \dfrac{2}{{ – \sqrt {21} }}\end{array} \right.\end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
\sin a = \dfrac{{\sqrt {21} }}{5}\\
\sin a = – \dfrac{{\sqrt {21} }}{5}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
Do:\cos a = \dfrac{2}{5}\\
{\sin ^2}a + {\cos ^2}a = 1\\
\to \dfrac{4}{{25}} + {\sin ^2}a = 1\\
\to {\sin ^2}a = \dfrac{{21}}{{25}}\\
\to \left[ \begin{array}{l}
\sin a = \dfrac{{\sqrt {21} }}{5}\\
\sin a = – \dfrac{{\sqrt {21} }}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{{\sqrt {21} }}{5}:\dfrac{2}{5} = \dfrac{{\sqrt {21} }}{2}\\
\tan a = \dfrac{{\sin a}}{{\cos a}} = – \dfrac{{\sqrt {21} }}{5}:\dfrac{2}{5} = – \dfrac{{\sqrt {21} }}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\cot a = \dfrac{1}{{\tan a}} = \dfrac{2}{{\sqrt {21} }}\\
\cot a = \dfrac{1}{{\tan a}} = \dfrac{2}{{ – \sqrt {21} }}
\end{array} \right.
\end{array}\)