Question

cos2π/7cos4π/7cos6π/7=?详细过程,谢谢!

Answer 1

cos6π/7=-cosπ/7
故
cos2π/7cos4π/7cos6π/7=
-cosπ/7cos2π/7cos4π/7
然后乘上sinπ/7，有
-sinπ/7*cosπ/7*cos2π/7*cos4π/7
=-1/2sin2π/7*cos2π/7*cos4π/7
=-1/4sin4π/7*cos4π/7
=-1/8sin8π/7=1/8sinπ/7
然后两边除以sinπ/7
得
cos2π/7cos4π/7cos6π/7=1/8
希望我的回答可以帮到你~不懂的可以再问我哈！

追问

cos6π/7=-cosπ/7这个是如何算出的?

追答

这是因为
cos(π-A)=-cosA
理解这个式子你可以用画图的办法，你可以在单位圆内画图，也可以画出cos(x)的函数图象（如果你学过的话）

Answer 2

cos2π/7cos4π/7cos6π/7
=cos6π/7cos2π/7cos4π/7
=-cosπ/7cos2π/7cos4π/7
=-[2sinπ/7cosπ/7cos2π/7cos4π/7]/(2sinπ/7)
=-[sin2π/7cos2π/7cos4π/7]/(2sinπ/7)
=-[2sin2π/7cos2π/7cos4π/7]/(4sinπ/7)
=-[sin4π/7cos4π/7]/(4sinπ/7)
=-[2sin4π/7cos4π/7]/(8sinπ/7)
=-(sin8π/7)/(8sinπ/7)
=sinπ/7)/(8sinπ/7)
=1/8

Answer 3

cos2π/7cos4π/7cos6π/7
=(8sin2π/7cos2π/7cos4π/7cos6π/7)/(8sin2π/7)
=(4sin4π/7cos4π/7cos6π/7)/(8sin2π/7)
=(2sin8π/7cos6π/7)/(8sin2π/7)
=(2sinπ/7cosπ/7)/(8sin2π/7)
=(sin2π/7)/(8sin2π/7)
=1/8