cos(2π/7)cos(4π/7)cos(6π/7)怎么解·要过程
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/06/30 15:11:25
cos(2π/7)cos(4π/7)cos(6π/7)怎么解·要过程
xiexie````
xiexie````
cos(2π/7)cos(4π/7)cos(6π/7)
=cos(2π/7)cos(4π/7)[-cos(π-6π/7)]
=-cos(π/7)cos(2π/7)cos(4π/7)
=-2sin(π/7)cos(π/7)cos(2π/7)cos(4π/7)/[2sin(π/7)]
=-sin(2π/7)cos(2π/7)cos(4π/7)/[2sin(π/7)]
=-2sin(2π/7)cos(2π/7)cos(4π/7)/[4sin(π/7)]
=-sin(4π/7)cos(4π/7)/[4sin(π/7)]
=-2sin(4π/7)cos(4π/7)/[8sin(π/7)]
=-sin(8π/7)/[8sin(π/7)]
=-sin(π+π/7)/[8sin(π/7)]
=sin(π/7)/[8sin(π/7)]
=1/8
=cos(2π/7)cos(4π/7)[-cos(π-6π/7)]
=-cos(π/7)cos(2π/7)cos(4π/7)
=-2sin(π/7)cos(π/7)cos(2π/7)cos(4π/7)/[2sin(π/7)]
=-sin(2π/7)cos(2π/7)cos(4π/7)/[2sin(π/7)]
=-2sin(2π/7)cos(2π/7)cos(4π/7)/[4sin(π/7)]
=-sin(4π/7)cos(4π/7)/[4sin(π/7)]
=-2sin(4π/7)cos(4π/7)/[8sin(π/7)]
=-sin(8π/7)/[8sin(π/7)]
=-sin(π+π/7)/[8sin(π/7)]
=sin(π/7)/[8sin(π/7)]
=1/8
cos(2派/7)+cos(4派/7)+cos(6派/7)=?
化简sin(2π+a)cos(π-a)cos(π/2-a)cos(7π/2-a)/cos(π-a)sin(3π-a)si
不用计算器,求(cos(π/8))^4+(cos(3π/8))^4+(cos(5π/8))^4+(cos(7π/8))^
cos^2(π/3-a)+cos^2(π/6+a)
cos^2(π/6)
cos(-19/6π)
已知cos(π4
【紧急求助】计算:cosα+cos(2π/3 +α)+cos(4π/3 +α)
怎么比较cos(3π /5),cos(π /4)的大小?是要根据图像吗?
1:COS(-7π/12) =
∫(0~π) x根号(cos^2x-cos^4x) dx 怎么算
一个定积分的计算过程∫(上π/2下0)(cos^5 x-cos^7 x)dx=2*4/(1*3*5)-2*4*6/(1*