- LG a
- LG b
- LG c
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,
\[\eqalign{
& A=\sin {6^0}\sin {42^0}\sin {66^0}\sin {78^0} \cr&= \sin {6^0}\cos {48^0}\cos {24^0}\cos {12^0}
\cr} \]
\[\begin{array}{l}
\Rightarrow A\cos {6^0}\\
= \sin {6^0}\cos {6^0}\cos {12^0}\cos {24^0}\cos {48^0}\\
= \frac{1}{2}.\left[ {2\sin {6^0}\cos {6^0}} \right]\cos {12^0}\cos {24^0}\cos {48^0}\\
= \frac{1}{2}\sin {12^0}\cos {12^0}\cos {24^0}\cos {48^0}\\
= \frac{1}{2}.\frac{1}{2}.\left[ {2\sin {{12}^0}\cos {{12}^0}} \right]\cos {24^0}\cos {48^0}\\
= \frac{1}{4}\sin {24^0}\cos {24^0}\cos {48^0}\\
= \frac{1}{4}.\frac{1}{2}.\left[ {2\sin {{24}^0}\cos {{24}^0}} \right]\cos {48^0}\\
= \frac{1}{8}\sin {48^0}\cos {48^0}\\
= \frac{1}{{16}}.2\sin {48^0}\cos {48^0} = \frac{1}{{16}}\sin {96^0}\\
= \frac{1}{{16}}\sin \left[ {{{90}^0} + {6^0}} \right]\\
= \frac{1}{{16}}\left[ {\sin {{90}^0}\cos {6^0} + \cos {{90}^0}\sin {6^0}} \right]\\
= \frac{1}{{16}}\cos {6^0}\\
\Rightarrow A\cos {6^0} = \frac{1}{{16}}\cos {6^0}\\
\Leftrightarrow A = \frac{1}{{16}}
\end{array}\]