Bài tập 18 trang 200 SGK Toán 10 NC

a) Ta có

\(\begin{array}{l}
\sin \alpha  =  - \sqrt {1 - c{\rm{o}}{{\rm{s}}^2}\alpha } \\
 =  - \sqrt {1 - \frac{1}{{16}}}  =  - \frac{{\sqrt {15} }}{4}\left( {do{\mkern 1mu} \sin \alpha  < 0} \right)
\end{array}\)

\(\begin{array}{l}
\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} =  - \sqrt {15} \\
\cot \alpha  = \frac{1}{{\tan \alpha }} =  - \frac{{\sqrt {15} }}{{15}}
\end{array}\)

b) Vì \(\frac{\pi }{2} < \alpha  < \frac{{3\pi }}{2} \)

\(\Rightarrow \cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha }  =  - \frac{{2\sqrt 2 }}{3}\)

\(\begin{array}{l}
\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{1}{{2\sqrt 2 }} = \frac{{\sqrt 2 }}{4}\\
\cot \alpha  = \frac{1}{{\tan \alpha }} = 2\sqrt 2 
\end{array}\)

c) 

Vì \(\left\{ \begin{array}{l}
 - \pi  < \alpha  < 0\\
\tan \alpha  = \frac{1}{2}
\end{array} \right. \Rightarrow \cos \alpha  < 0\)

\(\begin{array}{l}
 \Rightarrow \cos \alpha  =  - \frac{1}{{\sqrt {1 + {{\tan }^2}\alpha } }} =  - \frac{{2\sqrt 2 }}{5}\\
\sin \alpha  = \tan \alpha .\cot \alpha  =  - \frac{{\sqrt 5 }}{5}\\
\cot \alpha  = \frac{1}{{\tan \alpha }} = 2
\end{array}\)

-- Mod Toán 10 HỌC247