Bài tập 41 trang 214 SGK Toán 10 NC

a) Ta có: 

\(\begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{\sin \alpha  = \frac{1}{3}}\\
{\frac{\pi }{2} < \alpha  < \pi }
\end{array}} \right.\\
 \Rightarrow \cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha } \\
 =  - \sqrt {1 - \frac{1}{9}}  =  - \frac{{2\sqrt 2 }}{3}
\end{array}\)

Khi đó:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sin 2\alpha  = 2\sin \alpha \cos \alpha \\
 = 2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) =  - \frac{{4\sqrt 2 }}{9}
\end{array}\\
{\cos 2\alpha  = 1 - 2{{\sin }^2}\alpha  = \frac{7}{9}}\\
{\tan 2\alpha  = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} =  - \frac{{4\sqrt 2 }}{7}}\\
{\cot 2\alpha  =  - \frac{{7\sqrt 2 }}{8}}
\end{array}\)

Ta có: 

\(\frac{\pi }{4} < \frac{\alpha }{2} < \frac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\cos \frac{\alpha }{2} > 0\\
\sin \frac{\alpha }{2} > 0
\end{array} \right.\)

\(\begin{array}{*{20}{r}}
\begin{array}{l}
\cos \alpha  = 2{\cos ^2}\frac{\alpha }{2} - 1\\
 \Rightarrow \cos \frac{\alpha }{2} = \sqrt {\frac{{1 + \cos \alpha }}{2}} \\
 = \sqrt {\frac{{3 - 2\sqrt 2 }}{6}} 
\end{array}\\
\begin{array}{l}
\cos \alpha  = 1 - 2{\sin ^2}\frac{\alpha }{2}\\
 \Rightarrow \sin \frac{\alpha }{2} = \sqrt {\frac{{1 - \cos \alpha }}{2}} \\
 = \sqrt {\frac{{3 + 2\sqrt 2 }}{6}} 
\end{array}
\end{array}\)

\(\begin{array}{l}
\tan \frac{\alpha }{2} = \frac{{\sin \frac{\alpha }{2}}}{{2\cos \frac{\alpha }{2}}} = 3 + 2\sqrt 2 ;\\
\cot \frac{\alpha }{2} = 3 - 2\sqrt 2 
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
2{\cos ^2}{15^0} = 1 + \cos {30^0}\\
 = 1 + \frac{{\sqrt 3 }}{2} \Rightarrow \cos {15^0} = \sqrt {\frac{{2 + \sqrt 3 }}{2}} 
\end{array}\\
\begin{array}{l}
2{\sin ^2}{15^0} = 1 - \cos {30^0}\\
 = 1 - \frac{{\sqrt 3 }}{2} \Rightarrow \sin {15^0} = \sqrt {\frac{{2 - \sqrt 3 }}{2}} 
\end{array}\\
\begin{array}{l}
\tan {15^0} = \sqrt {\frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }}}  = 2 - \sqrt 3 ,\\
\cot {15^0} + 2 + \sqrt 3 
\end{array}
\end{array}\)

-- Mod Toán 10 HỌC247