Cho cos alpha = 1/3 với 0 < alpha < pi/2. Tính sin alpha, tan alpha, cot alpha 03/10/2021 Bởi Clara Cho cos alpha = 1/3 với 0 < alpha < pi/2. Tính sin alpha, tan alpha, cot alpha
$0<\alpha<\dfrac{\pi}{2}$ $\Rightarrow \sin\alpha>0$ $\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{2\sqrt2}{3}$ $\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=2\sqrt2$ $\cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{1}{2\sqrt2}$ Bình luận
Giải thích các bước giải: Ta có: \(\begin{array}{l}0 < \alpha < \dfrac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}\sin \alpha > 0\\\cos \alpha > 0\end{array} \right.\\{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\\sin \alpha > 0 \Rightarrow \sin \alpha = \sqrt {1 – {{\cos }^2}\alpha } = \sqrt {1 – {{\left( {\dfrac{1}{3}} \right)}^2}} = \dfrac{{2\sqrt 2 }}{3}\\\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = 2\sqrt 2 \\\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{1}{{2\sqrt 2 }}\end{array}\) Bình luận
$0<\alpha<\dfrac{\pi}{2}$
$\Rightarrow \sin\alpha>0$
$\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{2\sqrt2}{3}$
$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=2\sqrt2$
$\cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{1}{2\sqrt2}$
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
0 < \alpha < \dfrac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
\cos \alpha > 0
\end{array} \right.\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\sin \alpha > 0 \Rightarrow \sin \alpha = \sqrt {1 – {{\cos }^2}\alpha } = \sqrt {1 – {{\left( {\dfrac{1}{3}} \right)}^2}} = \dfrac{{2\sqrt 2 }}{3}\\
\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = 2\sqrt 2 \\
\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{1}{{2\sqrt 2 }}
\end{array}\)