求值:log2cosπ/9+log2cos2π/9+log2cos4π/9

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/19 11:40:48
求值:log2cosπ/9+log2cos2π/9+log2cos4π/9求值:log2cosπ/9+log2cos2π/9+log2cos4π/9求值:log2cosπ/9+log2cos2π/9+

求值:log2cosπ/9+log2cos2π/9+log2cos4π/9
求值:log2cosπ/9+log2cos2π/9+log2cos4π/9

求值:log2cosπ/9+log2cos2π/9+log2cos4π/9
利用2倍角公式
cosπ/9*cos2π/9*cos4π/9
=[sinπ/9*cosπ/9*cos2π/9*cos4π/9]/sinπ/9
=[1/2sin2π/9*cos2π/9*cos4π/9]/sinπ/9
=[1/4sin4π/9*cos4π/9]/sinπ/9
=[1/8sin8π/9]/sinπ/9
=[1/8sinπ/9]/[sinπ/9]
=1/8
log2cosπ/9+log2cos2π/9+log2cos4π/9
=log2(cosπ/9*cos2π/9*cos4π/9)
=log2(1/8)
=-3

'

误解