a) Biết \(\sin \alpha = \frac{1}{3};\alpha \in \left( {\frac{\pi }{2};\pi } \right)\). Hãy tính giá trị lượng của góc \(2\alpha \) và góc \(\frac{\alpha }{2}\)
b) Sử dụng \({15^0} = \frac{{{{30}^0}}}{2}\), hãy kiểm nghiệm lại kết quả của bài tập 39.
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
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