Tính cos a, tan a, cot a biết sin a = 1/5 05/07/2021 Bởi Kaylee Tính cos a, tan a, cot a biết sin a = 1/5
$\sin a=\dfrac{1}{5}$ Mà $\sin^2a+\cos^2a=1$ $\to \cos a=\sqrt{1-\sin^2a}=\sqrt{1-\dfrac{1}{25}}=\dfrac{2\sqrt6}{5}$ $\tan a=\dfrac{\sin a}{\cos a}=\dfrac{\sqrt6}{12}$ $\cot a=\dfrac{1}{\tan a}=2\sqrt6$ Bình luận
Giải thích các bước giải: \({\sin ^2}x + {\cos ^2}x = 1\) Ta có: \(\begin{array}{l}{\sin ^2}a + {\cos ^2}a = 1\\ \Leftrightarrow {\left( {\dfrac{1}{5}} \right)^2} + {\cos ^2}a = 1\\ \Leftrightarrow \dfrac{1}{{25}} + {\cos ^2}a = 1\\ \Leftrightarrow {\cos ^2}a = \dfrac{{24}}{{25}}\\ \Leftrightarrow \left[ \begin{array}{l}\cos a = \dfrac{{2\sqrt 6 }}{5} \Rightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{1}{{2\sqrt 6 }} \Rightarrow \cot a = \dfrac{{\cos a}}{{\sin a}} = 2\sqrt 6 \\\cos a = – \dfrac{{2\sqrt 6 }}{5} \Rightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{{ – 1}}{{2\sqrt 6 }} \Rightarrow \cot a = \dfrac{{\cos a}}{{\sin a}} = – 2\sqrt 6 \end{array} \right.\end{array}\) Bình luận
$\sin a=\dfrac{1}{5}$
Mà $\sin^2a+\cos^2a=1$
$\to \cos a=\sqrt{1-\sin^2a}=\sqrt{1-\dfrac{1}{25}}=\dfrac{2\sqrt6}{5}$
$\tan a=\dfrac{\sin a}{\cos a}=\dfrac{\sqrt6}{12}$
$\cot a=\dfrac{1}{\tan a}=2\sqrt6$
Giải thích các bước giải:
\({\sin ^2}x + {\cos ^2}x = 1\)
Ta có:
\(\begin{array}{l}
{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\left( {\dfrac{1}{5}} \right)^2} + {\cos ^2}a = 1\\
\Leftrightarrow \dfrac{1}{{25}} + {\cos ^2}a = 1\\
\Leftrightarrow {\cos ^2}a = \dfrac{{24}}{{25}}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos a = \dfrac{{2\sqrt 6 }}{5} \Rightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{1}{{2\sqrt 6 }} \Rightarrow \cot a = \dfrac{{\cos a}}{{\sin a}} = 2\sqrt 6 \\
\cos a = – \dfrac{{2\sqrt 6 }}{5} \Rightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{{ – 1}}{{2\sqrt 6 }} \Rightarrow \cot a = \dfrac{{\cos a}}{{\sin a}} = – 2\sqrt 6
\end{array} \right.
\end{array}\)