Bài 47 trang 215 SGK Đại số 10 Nâng cao

LG a

\(\cos {10^0}\cos {50^0}\cos {70^0} \) \(= \sin {20^0}\sin {40^0}\sin {80^0} = {{\sqrt 3 } \over 8}\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \cos {10^0}\cos {50^0}\cos {70^0}\cr& = \cos {10^0}{\rm{[}}{1 \over 2}(cos{120^0} + \cos {20^0}){\rm{]}} \cr 
& = - {1 \over 4}\cos {10^0} + {1 \over 2}\cos {10^0}\cos {20^0} \cr 
& = - {1 \over 4}\cos {10^0} + {1 \over 4}(cos{30^0} + \cos {10^0})\cr& = {1 \over 4}\cos {30^0} = {{\sqrt 3 } \over 8} \cr} \)

\(\begin{array}{l}
\sin {20^0}\sin {40^0}\sin {80^0}\\= \sin {20^0}.\frac{1}{2}\left( {\cos {{40}^0} - \cos {{120}^0}} \right)\\
= \frac{1}{2}\sin {20^0}\left( {\cos {{40}^0} + \frac{1}{2}} \right)\\
= \frac{1}{2}\sin {20^0}\cos {40^0} + \frac{1}{4}\sin {20^0}\\
= \frac{1}{2}.\frac{1}{2}\left( {\sin {{60}^0} - \sin {{20}^0}} \right) + \frac{1}{4}\sin {20^0}\\
= \frac{1}{4}\sin {60^0} = \frac{1}{4}.\frac{{\sqrt 3 }}{2} = \frac{{\sqrt 3 }}{8}
\end{array}\)

LG b

\(\sin {10^0}\sin {50^0}\sin {70^0}\) \( = \cos {20^0}\cos {40^0}\cos {80^0} = {1 \over 8}\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \sin {10^0}\sin {50^0}\sin {70^0}\cr& = {1 \over 2}(cos{20^0} - \cos {120^0})\sin {10^0} \cr 
& = {1 \over 4}\sin {10^0} + {1 \over 2}\sin {10^0}\cos {20^0} \cr 
& = {1 \over 4}\sin {10^0} + {1 \over 4}(\sin {30^0} - \sin {10^0}) \cr&= {1 \over 4}\sin {30^0} = {1 \over 8} \cr 
& \cos {20^0}\cos {40^0}\cos {80^0} \cr} \)

\(\begin{array}{l}
= \cos {20^0}.\frac{1}{2}\left( {\cos {{120}^0} + \cos {{40}^0}} \right)\\
= \frac{1}{2}\cos {20^0}\left( { - \frac{1}{2} + \cos {{40}^0}} \right)\\
= - \frac{1}{4}\cos {20^0} + \frac{1}{2}\cos {20^0}\cos {40^0}\\
= - \frac{1}{4}\cos {20^0} + \frac{1}{2}.\frac{1}{2}\left( {\cos {{60}^0} + \cos {{20}^0}} \right)\\
= - \frac{1}{4}\cos {20^0} + \frac{1}{4}\cos {60^0} + \frac{1}{4}\cos {20^0}\\
= \frac{1}{4}\cos {60^0} = \frac{1}{4}.\frac{1}{2} = \frac{1}{8}
\end{array}\)

Loigiaihay.com