Đáp ána) Ta có:\(\eqalign{
& \cos {10^0}\cos {50^0}\cos {70^0}\cr& = \cos {10^0}{\rm{[}}{1 \over 2}(cos{120^0} + \cos {20^0}){\rm{]}} \cr
& = - {1 \over 4}\cos {10^0} + {1 \over 2}\cos {10^0}\cos {20^0} \cr
& = - {1 \over 4}\cos {10^0} + {1 \over 4}(cos{30^0} + \cos {10^0})\cr& = {1 \over 4}\cos {30^0} = {{\sqrt 3 } \over 8} \cr
& \sin {20^0}\sin {40^0}\sin {80^0} = \cos {70^0}\cos {50^0}\cos {10^0} \cr&= {{\sqrt 3 } \over 8} \cr} \)b) Ta có:\(\eqalign{
& \sin {10^0}\sin {50^0}\sin {70^0}\cr& = {1 \over 2}(cos{20^0} - \cos {120^0})\sin {10^0} \cr
& = {1 \over 4}\sin {10^0} + {1 \over 2}\sin {10^0}\cos {20^0} \cr
& = {1 \over 4}\sin {10^0} + {1 \over 4}(\sin {30^0} - \sin {10^0}) \cr&= {1 \over 4}\sin {30^0} = {1 \over 8} \cr
& \cos {20^0}\cos {40^0}\cos {80^0} = \sin {10^0}\sin {50^0}\sin {70^0} = {1 \over 8} \cr} \)
