cosπ/5cos2π/5
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/27 13:01:50
cosπ/5cos2π/5cosπ/5cos2π/5cosπ/5cos2π/5cos(π/5)cos(2π/5)=4sin(π/5)cos(π/5)cos(2π/5)/[4sin(π/5)]=2sin
cosπ/5cos2π/5
cosπ/5cos2π/5
cosπ/5cos2π/5
cos(π/5)cos(2π/5)
=4sin(π/5) cos(π/5)cos(2π/5)/ [4sin(π/5)]
=2sin(2π/5) cos(2π/5)/ [4sin(π/5)]
=sin(4π/5)/ [4sin(π/5)]
=sin(π/5)/ [4sin(π/5)]
=1/4.
求COSπ/5*COS2π/5
cosπ/5×cos2π/5=
COSπ/5乘以COS2π/5
cosπ/5×cos2π/5
cosπ/5cos2π/5
cosπ/5cos2π/5=?
cosπ/5 - cos2π/5 怎样变成 cosπ/5 + cos2π/5rt
（cosπ/5）（cos2π/5)的值等于
cosπ/5·cos2π/5等于多少?
cosπ/5*cos2π/5的值等于
cosπ/5 乘以cos2π/5等于多少
化简cosπ/5cos2π/5RT.
若cos(π-θ)=3/5,cos2θ=
cos(π-θ)=3/5 求cos2θ
cos(π- θcos2θ)=3/5 求
cosπ/5*cos2π/5 =(2sinπ/5*cosπ/5*cos2π/5)/(2sinπ/5) =(sin2π/5*cos2π/5)/(2sinπ/5) =(2sin2π/低2部咋算的?
cosπ/5 +cos2π/5 +cos3π/5 +cos4π/5 =可以简写.
求cosπ/5乘cos2π/5乘cos3π/5乘cos4π/5