Cho biết Tan α = √2. Tính cos α, sin α, cot α 14/08/2021 Bởi Josie Cho biết Tan α = √2. Tính cos α, sin α, cot α
$\cot\alpha=\dfrac{1}{\sqrt2}$ $\dfrac{1}{\cos^2\alpha}=1+\tan^2\alpha$ $\Rightarrow \cos\alpha=\dfrac{1}{\sqrt3}$ $\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{\sqrt6}{3}$ Bình luận
Đáp án: \(\left[ \begin{array}{l}\cos a = \dfrac{1}{{\sqrt 3 }}\\\cos a = – \dfrac{1}{{\sqrt 3 }}\end{array} \right. \to \left[ \begin{array}{l}\sin a = \dfrac{{\sqrt 2 }}{{\sqrt 3 }}\\\sin a = – \dfrac{{\sqrt 2 }}{{\sqrt 3 }}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}\tan a = \dfrac{{\sin a}}{{\cos a}} = \sqrt 2 \to \sin a = \sqrt 2 .\cos a\\ \to \cot a = \dfrac{1}{{\tan a}} = \dfrac{1}{{\sqrt 2 }}\\{\sin ^2}a + {\cos ^2}a = 1\\ \to 2.{\cos ^2}a + {\cos ^2}a = 1\\ \to {\cos ^2}a = \dfrac{1}{3}\\ \to \left[ \begin{array}{l}\cos a = \dfrac{1}{{\sqrt 3 }}\\\cos a = – \dfrac{1}{{\sqrt 3 }}\end{array} \right. \to \left[ \begin{array}{l}\sin a = \dfrac{{\sqrt 2 }}{{\sqrt 3 }}\\\sin a = – \dfrac{{\sqrt 2 }}{{\sqrt 3 }}\end{array} \right.\end{array}\) Bình luận
$\cot\alpha=\dfrac{1}{\sqrt2}$
$\dfrac{1}{\cos^2\alpha}=1+\tan^2\alpha$
$\Rightarrow \cos\alpha=\dfrac{1}{\sqrt3}$
$\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{\sqrt6}{3}$
Đáp án:
\(\left[ \begin{array}{l}
\cos a = \dfrac{1}{{\sqrt 3 }}\\
\cos a = – \dfrac{1}{{\sqrt 3 }}
\end{array} \right. \to \left[ \begin{array}{l}
\sin a = \dfrac{{\sqrt 2 }}{{\sqrt 3 }}\\
\sin a = – \dfrac{{\sqrt 2 }}{{\sqrt 3 }}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\tan a = \dfrac{{\sin a}}{{\cos a}} = \sqrt 2 \to \sin a = \sqrt 2 .\cos a\\
\to \cot a = \dfrac{1}{{\tan a}} = \dfrac{1}{{\sqrt 2 }}\\
{\sin ^2}a + {\cos ^2}a = 1\\
\to 2.{\cos ^2}a + {\cos ^2}a = 1\\
\to {\cos ^2}a = \dfrac{1}{3}\\
\to \left[ \begin{array}{l}
\cos a = \dfrac{1}{{\sqrt 3 }}\\
\cos a = – \dfrac{1}{{\sqrt 3 }}
\end{array} \right. \to \left[ \begin{array}{l}
\sin a = \dfrac{{\sqrt 2 }}{{\sqrt 3 }}\\
\sin a = – \dfrac{{\sqrt 2 }}{{\sqrt 3 }}
\end{array} \right.
\end{array}\)