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Bài tập 42 trang 214 SGK Toán 10 NC

Bài tập 42 trang 214 SGK Toán 10 NC

Chứng minh rằng:

a) \(\sin \frac{{11\pi }}{{12}}{\rm{cos}}\frac{{5\pi }}{{12}} = \frac{1}{4}\left( {2 - \sqrt 3 } \right)\)

b) \({\rm{cos}}\frac{\pi }{7}{\rm{cos}}\frac{{3\pi }}{7}{\rm{cos}}\frac{{5\pi }}{7} =  - \frac{1}{8}\)

c) \(\sin {6^0}\sin {42^0}\sin {66^2}\sin {78^0} = \frac{1}{16}\)

ADSENSE

Hướng dẫn giải chi tiết

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sin \frac{{11\pi }}{{12}}{\rm{cos}}\frac{{5\pi }}{{12}}\\
 = \sin \left( {\pi  - \frac{\pi }{{12}}} \right){\rm{cos}}\left( {\frac{\pi }{2} - \frac{\pi }{{12}}} \right)
\end{array}\\
\begin{array}{l}
 = {\sin ^2}\frac{\pi }{{12}} = \frac{1}{2}\left( {1 - {\rm{cos}}\frac{\pi }{6}} \right)\\
 = \frac{1}{2} = \left( {1 - \frac{{\sqrt 3 }}{2}} \right) = \frac{1}{4}\left( {2 - \sqrt 3 } \right)
\end{array}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
{{\rm{cos}}\frac{{3\pi }}{7} = {\rm{cos}}\left( {\pi  - \frac{{4\pi }}{7}} \right) =  - {\rm{cos}}\frac{{4\pi }}{7}}\\
{{\rm{cos}}\frac{{5\pi }}{7} = {\rm{cos}}\left( {\pi  - \frac{{2\pi }}{7}} \right) =  - {\rm{cos}}\frac{{2\pi }}{7}}\\
\begin{array}{l}
 \Rightarrow {\rm{cos}}\frac{\pi }{7}{\rm{cos}}\frac{{3\pi }}{7}{\rm{cos}}\frac{{5\pi }}{7}\\
 = {\rm{cos}}\frac{\pi }{7}{\rm{cos}}\frac{{2\pi }}{7}{\rm{cos}}\frac{{4\pi }}{7}
\end{array}\\
{ = \frac{1}{{\sin \frac{\pi }{7}}}\left( {{\rm{sin}}\frac{\pi }{7}{\rm{cos}}\frac{\pi }{7}} \right){\rm{cos}}\frac{{2\pi }}{7}{\rm{cos}}\frac{{4\pi }}{7}}\\
{ = \frac{1}{{\sin \frac{\pi }{7}}}.\frac{1}{2}\left( {\sin \frac{{2\pi }}{7}{\rm{cos}}\frac{{2\pi }}{7}} \right).{\rm{cos}}\frac{{4\pi }}{7}}\\
{ = \frac{1}{{\sin \frac{\pi }{7}}}.\frac{1}{4}\sin \frac{{4\pi }}{7}{\rm{cos}}\frac{{4\pi }}{7}}\\
{ = \frac{1}{{8\sin \frac{\pi }{7}}}.\sin \frac{{8\pi }}{7} = \frac{{ - \sin \frac{\pi }{7}}}{{8\sin \frac{\pi }{7}}} =  - \frac{1}{8}}
\end{array}\)

c)

\(\begin{array}{l}
\sin {6^0}\sin {42^0}\sin {66^2}\sin {78^0}\\
 = \sin {6^0}{48^0}\cos {24^0}\cos {12^0}\\
 = \frac{1}{{{6^0}}}\left( {\sin {6^0}\cos {6^0}} \right)\cos {12^0}\cos {24^0}\cos {48^0}\\
 = \frac{1}{{\cos {6^0}}}\left( {\frac{1}{2}\sin {{12}^0}\cos {{12}^0}} \right)\cos {24^0}\cos {48^0}\\
 = \frac{1}{{\cos {6^0}}}.\frac{1}{4}\sin {24^0}\cos {24^0}\cos {48^0}\\
 = \frac{1}{{\cos {6^0}}}\left( {\frac{1}{8}\sin {{48}^0}\cos {{48}^0}} \right)\\
 = \frac{1}{{\cos {6^0}}}.\frac{1}{{16}}\sin {96^0} = \frac{{\cos {6^0}}}{{16\cos {6^0}}} = \frac{1}{{16}}
\end{array}\)

-- Mod Toán 10 HỌC247

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