Bài 42 trang 214 SGK Đại số 10 Nâng cao

Chứng minh rằng:

LG a

\(\sin {{11\pi } \over {12}}\cos {{5\pi } \over {12}} = {1 \over 4}(2 - \sqrt 3 )\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \sin {{11\pi } \over {12}}\cos {{5\pi } \over {12}} \cr&= \sin (\pi - {\pi \over {12}})cos({\pi \over 2} - {\pi \over {12}}) \cr 
& = {\sin ^2}{\pi \over {12}} = {1 \over 2}(1 - \cos {\pi \over 6}) \cr&= {1 \over 2}(1 - {{\sqrt 3 } \over 2})\cr& = {1 \over 4}(2 - \sqrt 3 ) \cr} \)

Cách khác:

\[\begin{array}{l}
\sin \frac{{11\pi }}{{12}}\cos \frac{{5\pi }}{{12}}\\
= \frac{1}{2}\left[ {\sin \left( {\frac{{11\pi }}{{12}} + \frac{{5\pi }}{{12}}} \right) + \sin \left( {\frac{{11\pi }}{{12}} - \frac{{5\pi }}{{12}}} \right)} \right]\\
= \frac{1}{2}\left( {\sin \frac{{4\pi }}{3} + \sin \frac{\pi }{2}} \right)\\
= \frac{1}{2}\left( { - \sin \frac{\pi }{3} + 1} \right)\\
= \frac{1}{2}\left( {1 - \frac{{\sqrt 3 }}{2}} \right) = \frac{{2 - \sqrt 3 }}{4}
\end{array}\]

LG b

\(\cos {\pi  \over 7}\cos {{3\pi } \over 7}\cos {{5\pi } \over 7} =  - {1 \over 8}\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \cos {{3\pi } \over 7} = \cos (\pi - {{4\pi } \over 7}) = - \cos {{4\pi } \over 7} \cr 
& \cos {{5\pi } \over 7} = \cos (\pi - {{2\pi } \over 7}) = - \cos {{2\pi } \over 7} \cr} \) 

Đặt

\(\eqalign{
& A=\cos {\pi \over 7}\cos {{3\pi } \over 7}\cos {{5\pi } \over 7} \cr& = \cos \frac{\pi }{7}.\left( { - \cos \frac{{4\pi }}{7}} \right).\left( { - \cos \frac{{2\pi }}{7}} \right)\cr&= \cos {\pi \over 7}\cos {{2\pi } \over 7}\cos {{4\pi } \over 7} \cr } \)

\(\begin{array}{l}
\Rightarrow 8A\sin \frac{\pi }{7}\\
= 8\sin \frac{\pi }{7}\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7}\\
= 4.\left( {2\sin \frac{\pi }{7}\cos \frac{\pi }{7}} \right)\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7}\\
= 4.\sin \frac{{2\pi }}{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7}\\
= 2.\left( {2\sin \frac{{2\pi }}{7}\cos \frac{{2\pi }}{7}} \right)\cos \frac{{4\pi }}{7}\\
= 2.\sin \frac{{4\pi }}{7}\cos \frac{{4\pi }}{7}\\
= \sin \frac{{8\pi }}{7} = \sin \left( {\pi + \frac{\pi }{7}} \right)\\
= - \sin \frac{\pi }{7}\\
\Rightarrow 8A\sin \frac{\pi }{7} = - \sin \frac{\pi }{7}\\
\Rightarrow 8A = - 1\\
\Leftrightarrow A = - \frac{1}{8}\left( {dpcm} \right)
\end{array}\)

LG c

\(\sin {6^0}\sin {42^0}\sin {66^0}\sin {78^0} = {1 \over {16}}\) (Hướng dẫn: Nhân hai vế với cos 6⁰)

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}
\sin {42^0} = \sin \left( {{{90}^0} - {{48}^0}} \right) = \cos {48^0}\\
\sin {66^0} = \sin \left( {{{90}^0} - {{24}^0}} \right) = \cos {24^0}\\
\sin {78^0} = \sin \left( {{{90}^0} - {{12}^0}} \right) = \cos {12^0}
\end{array}\)

Do đó,

\(\eqalign{
& A=\sin {6^0}\sin {42^0}\sin {66^0}\sin {78^0} \cr&= \sin {6^0}\cos {48^0}\cos {24^0}\cos {12^0} 
\cr} \)

\(\begin{array}{l}
\Rightarrow A\cos {6^0}\\
= \sin {6^0}\cos {6^0}\cos {12^0}\cos {24^0}\cos {48^0}\\
= \frac{1}{2}.\left( {2\sin {6^0}\cos {6^0}} \right)\cos {12^0}\cos {24^0}\cos {48^0}\\
= \frac{1}{2}\sin {12^0}\cos {12^0}\cos {24^0}\cos {48^0}\\
= \frac{1}{2}.\frac{1}{2}.\left( {2\sin {{12}^0}\cos {{12}^0}} \right)\cos {24^0}\cos {48^0}\\
= \frac{1}{4}\sin {24^0}\cos {24^0}\cos {48^0}\\
= \frac{1}{4}.\frac{1}{2}.\left( {2\sin {{24}^0}\cos {{24}^0}} \right)\cos {48^0}\\
= \frac{1}{8}\sin {48^0}\cos {48^0}\\
= \frac{1}{{16}}.2\sin {48^0}\cos {48^0} = \frac{1}{{16}}\sin {96^0}\\
= \frac{1}{{16}}\sin \left( {{{90}^0} + {6^0}} \right)\\
= \frac{1}{{16}}\left( {\sin {{90}^0}\cos {6^0} + \cos {{90}^0}\sin {6^0}} \right)\\
= \frac{1}{{16}}\cos {6^0}\\
\Rightarrow A\cos {6^0} = \frac{1}{{16}}\cos {6^0}\\
\Leftrightarrow A = \frac{1}{{16}}
\end{array}\)