cos 2π/7 +cos 4π/7 +cos 6π/7=?

来源:学生作业帮助网 编辑:六六作业网 时间:2024/07/01 13:45:03
cos2π/7+cos4π/7+cos6π/7=?cos2π/7+cos4π/7+cos6π/7=?cos2π/7+cos4π/7+cos6π/7=?原式=2sin(π/7)[cos(2π/7)+co

cos 2π/7 +cos 4π/7 +cos 6π/7=?
cos 2π/7 +cos 4π/7 +cos 6π/7=?

cos 2π/7 +cos 4π/7 +cos 6π/7=?
原式
=2sin(π/7)[cos(2π/7)+cos(4π/7)+cos(6π/7)]/2sin(π/7)
=[2sin(π/7)cos(2π/7)+2sin(π/7)cos(4π/7)+2sin(π/7)cos(6π/7)]/2sin(π/7)
=[sin(3π/7)-sin(π/7)+sin(5π/7)-sin(3/7)+sin(7π/7)-sin(5π/7)]/2sin(π/7)
=[sin(7π/7)-sin(π/7)]/2sin(π/7)
=-sin(π/7)/2sin(π/7)
=-1/2.