1)cho COS=0,4 hãy tìm SIN;TAN 2biết COS=8/15;tìm SIN và COS 06/08/2021 Bởi Eloise 1)cho COS=0,4 hãy tìm SIN;TAN 2biết COS=8/15;tìm SIN và COS
1. $\sin x=\sqrt{1-\cos^2x}=\dfrac{\sqrt{21}}{5}$ $\to \tan x=\dfrac{\sin x}{\cos x}=\dfrac{\sqrt{21}}{2}$ 2. $\sin x=\sqrt{1-\cos^2x}=\dfrac{\sqrt{161}}{15}$ $\to \cot x=\dfrac{\cos x}{\sin x}=\dfrac{8}{\sqrt{161}}$ Bình luận
Đáp án: $\begin{array}{l}1)Do:{\sin ^2}x + {\cos ^2}x = 1\\ \Rightarrow {\sin ^2}x = 1 – {\cos ^2}x = 1 – 0,{4^2} = 0,84\\ \Rightarrow \sin x = \dfrac{{\sqrt {21} }}{5}\\ + {\tan ^2}x + 1 = \dfrac{1}{{{{\cos }^2}x}}\\ \Rightarrow {\tan ^2}x = \dfrac{1}{{0,{4^2}}} – 1 = \dfrac{{21}}{4}\\ \Rightarrow \tan x = \dfrac{{\sqrt {21} }}{2}\\2)\cos x = \dfrac{8}{{15}}\\ \Rightarrow {\sin ^2}x = 1 – {\left( {\dfrac{8}{{15}}} \right)^2} = \dfrac{{161}}{{225}}\\ \Rightarrow \sin x = \dfrac{{\sqrt {161} }}{{15}}\end{array}$ Bình luận
1.
$\sin x=\sqrt{1-\cos^2x}=\dfrac{\sqrt{21}}{5}$
$\to \tan x=\dfrac{\sin x}{\cos x}=\dfrac{\sqrt{21}}{2}$
2.
$\sin x=\sqrt{1-\cos^2x}=\dfrac{\sqrt{161}}{15}$
$\to \cot x=\dfrac{\cos x}{\sin x}=\dfrac{8}{\sqrt{161}}$
Đáp án:
$\begin{array}{l}
1)Do:{\sin ^2}x + {\cos ^2}x = 1\\
\Rightarrow {\sin ^2}x = 1 – {\cos ^2}x = 1 – 0,{4^2} = 0,84\\
\Rightarrow \sin x = \dfrac{{\sqrt {21} }}{5}\\
+ {\tan ^2}x + 1 = \dfrac{1}{{{{\cos }^2}x}}\\
\Rightarrow {\tan ^2}x = \dfrac{1}{{0,{4^2}}} – 1 = \dfrac{{21}}{4}\\
\Rightarrow \tan x = \dfrac{{\sqrt {21} }}{2}\\
2)\cos x = \dfrac{8}{{15}}\\
\Rightarrow {\sin ^2}x = 1 – {\left( {\dfrac{8}{{15}}} \right)^2} = \dfrac{{161}}{{225}}\\
\Rightarrow \sin x = \dfrac{{\sqrt {161} }}{{15}}
\end{array}$