cosπ/5cos2π/5在每一步的后面写上运用了哪条公式
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cosπ/5cos2π/5在每一步的后面写上运用了哪条公式cosπ/5cos2π/5在每一步的后面写上运用了哪条公式cosπ/5cos2π/5在每一步的后面写上运用了哪条公式sin2α＝2sinαco
cosπ/5cos2π/5在每一步的后面写上运用了哪条公式
cosπ/5cos2π/5
在每一步的后面写上运用了哪条公式
cosπ/5cos2π/5在每一步的后面写上运用了哪条公式
sin2α＝2sinαcosα
利用二倍角公式
原式=[(2sin∏/5cos∏/5)cos2∏/5]/(2sin∏/5)
=[sin2∏/5cos2∏/5]/(2sin∏/5)
=[2sin2∏/5cos2∏/5]/(4sin∏/5)
=[sin4∏/5]/[4sin∏/5]
此时利用诱导公式sin（π－α）＝sinα
原式=[sin(∏-4∏/5)]/(4sin∏/5)
=[sin∏/5]/(4sin∏/5)
=1/4
cosπ/5cos2π/5
=2sinπ/5cosπ/5cos2π/5/(2sinπ/5)
=sin2π/5cos2π/5/(2sinπ/5)
=2sin2π/5cos2π/5/(4sinπ/5)
=sin4π/5/(4sinπ/5)
=sin(π-π/5)/(4sinπ/5)
=sinπ/5/(4sinπ/5)
=1/4
cosπ/5cos2π/5
=2sinπ/5cosπ/5cos2π/5/(2sinπ/5)
=sin2π/5cos2π/5/(2sinπ/5)
=2sin2π/5cos2π/5/(4sinπ/5)
=sin4π/5/(4sinπ/5)
=sin(π-π/5)/(4sinπ/5)
=sinπ/5/(4sinπ/5)
=1/4
没有什么特殊公式
cosπ/5cos2π/5在每一步的后面写上运用了哪条公式
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)的值等于
cosπ/5*cos2π/5的值等于
cosπ/5 - cos2π/5 怎样变成 cosπ/5 + cos2π/5rt
cosπ/5·cos2π/5等于多少?
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的值等于 A1/4 B1/2 C2 D4