Bài tập 32 trang 206 SGK Toán 10 NC

a) Vì \(\cos \alpha  < 0\) nên ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha } \\
 =  - \sqrt {1 - \frac{{16}}{{25}}}  =  - \frac{3}{5}
\end{array}\\
{\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} =  - \frac{4}{3}}\\
{\cot \alpha  = \frac{1}{{\tan \alpha }} =  - \frac{3}{4}}
\end{array}\)

b) Vì \(\frac{\pi }{2} < \alpha  < \pi  \Rightarrow \sin \alpha  > 0\) nên:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
 \Rightarrow \sin \alpha  = \sqrt {1 - {{\cos }^2}\alpha } \\
 = \sqrt {1 - {{\left( {\frac{8}{{17}}} \right)}^2}}  = \frac{{15}}{{17}}
\end{array}\\
{\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} =  - \frac{{15}}{8}}\\
{\cot \alpha  = \frac{1}{{\tan \alpha }} =  - \frac{8}{{15}}}
\end{array}\)

c) Vì \(\pi  < \alpha  < \frac{{3\pi }}{2} \Rightarrow \cos \alpha  < 0\) nên:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
 \Rightarrow \cos \alpha  = \frac{{ - 1}}{{\sqrt {1 + {{\tan }^2}\alpha } }}\\
 =  - \frac{1}{{\sqrt {1 + {{\left( {\sqrt 3 } \right)}^2}} }} =  - \frac{1}{2}
\end{array}\\
{\sin \alpha  = \tan \alpha .\cos \alpha  - \frac{{\sqrt 3 }}{2}}\\
{\cot \alpha  = \frac{1}{{\tan \alpha }} = \frac{{\sqrt 3 }}{3}}
\end{array}\)

-- Mod Toán 10 HỌC247