分别计算：cosπ/3,cosπ/5cos2π/5,cosπ/7cos2π/7cos3π/7

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/09/21 17:39:12

分别计算：cosπ/3,cosπ/5cos2π/5,cosπ/7cos2π/7cos3π/7
根据上述结果猜想一个一般的结论,并给予证明

1.cosπ/3=1/2
2.cosπ/5cos2π/5
=2sin(π/5)*cos(π/5)*cos(2π/5) / 2sin(π/5)
=sin(2π/5)*cos(2π/5) / 2sin(π/5)
=sin(4π/5) / 4sin(π/5)
=sin(π/5) / 4sin(π/5)
=1/4
3.cos(π/7)cos(2π/7)cos(3π/7)
=2sin(π/7)cos(π/7)cos(2π/7)cos(3π/7) / 2sin(π/7)
=sin(2π/7)cos(2π/7)(cos3π/7) / 2sin(π/7)
=sin(4π/7)cos(3π/7) / 4sin(π/7)
=sin(8π/7) / 8sin(π/7)
=sin(π/7) / 8sin(π/7)
=1/8

一般结论：
cos[π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)]=1/(2^n)
证明：
cos[π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)]
=2sin[π/(2n-1)]cos[π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)] / 2sin[π/(2n-1)]
【分子分母补2sin[π/(2n-1)]】
=sin[2π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)] / 2sin[π/(2n-1)]
=2sin[2π/(2n-1)]cos[2π/(2n-1)]……cos[nπ/(2n-1)] / 4sin[π/(2n-1)]
=sin[4π/(2n-1)]……cos[nπ/(2n-1)] / 4sin[π/(2n-1)]
=……
=2sin[nπ/(2n-1)]cos[nπ/(2n-1)] / (2^n)sin[π/(2n-1)]
=cos[2nπ/(2n-1)] / (2^n)sin[π/(2n-1)]
=sin[π/(2n-1)] / (2^n)sin[π/(2n-1)] 【cos[2nπ/(2n-1)] =sin[π/(2n-1)] 】
=1/(2^n)
重点：不断地利用二倍角公式：sin2θ=2sinθcosθ
希望采纳~~~~~~~~
再问： cos[2nπ/(2n-1)] =sin[π/(2n-1)] 这个为什么
再答：哦，这里有个笔误。那段应该改成下面这个： =…… =2sin[nπ/(2n-1)]cos[nπ/(2n-1)] / (2^n)sin[π/(2n-1)] =sin[2nπ/(2n-1)] / (2^n)sin[π/(2n-1)]

求证cos2π/7+cosπ/3=cosπ/7+cos3π/7 COSπ/5乘以COS2π/5 求COSπ/5*COS2π/5 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 sinπ.cosπ.sin2π.cos2π分别是多少 cosπ/5*cos2π/5 =(2sinπ/5*cosπ/5*cos2π/5)/(2sinπ/5) =(sin2π/5 （二倍角的三角函数）cosπ/5cos2/5π的值是______cosπ/5cos2π/5 =sinπ/5cosπ/5c 已知(sinα+sin2α+sin3α)/(cosα+cos2α+cos3α)=﹣4/tanα,且π＜2α＜3π/2,求已知cos2α=-7/25,2α属于(π,3π/2).求sinα,cosα 求值：cos2π7