cosπ/5cos2π/5=?
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cosπ/5cos2π/5=?cosπ/5cos2π/5=?cosπ/5cos2π/5=?cosπ/5cos2π/5=sinπ/5cosπ/5cos2π/5/sinπ/5=4sinπ/5cosπ/5c
cosπ/5cos2π/5=?
cosπ/5cos2π/5=?
cosπ/5cos2π/5=?
cosπ/5cos2π/5
=sinπ/5cosπ/5cos2π/5/sinπ/5
=4sinπ/5cosπ/5cos2π/5/4sinπ/5
=2sin2π/5cos2π/5/4sinπ/5
=sin4π/5/4sinπ/5
=sin(π-π/5)/4sinπ/5
=(1/4)sinπ/5/sinπ/5
=1/4
cosπ/5×cos2π/5=
cosπ/5cos2π/5=?
求COSπ/5*COS2π/5
COSπ/5乘以COS2π/5
cosπ/5×cos2π/5
cosπ/5cos2π/5
若cos(π-θ)=3/5,cos2θ=
cos(π-θ)=3/5 求cos2θ
cos(π- θcos2θ)=3/5 求
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.
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.
设sin(π/4+α)+cos(π/4+α)=-1/5,则cos2α=