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

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
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

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~~~~~~~~

buhui

cosπ/3=1/2
cosπ/5cos2π/5
利用二倍角公式：sin2θ=2sinθcosθ
cos(π/5)*cos(2π/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)/4s...

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cosπ/3=1/2
cosπ/5cos2π/5
利用二倍角公式：sin2θ=2sinθcosθ
cos(π/5)*cos(2π/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
cosπ/7cos2π/7cos3π/7
=sinπ/7*cosπ/7*cos2π/7*cos3π/7/sinπ/7
=2sinπ/7*cosπ/7*cos2π/7*cos(π-4π/7)/2sinπ/7
=sin2π/7*cos2π/7*(-cos4π/7)/2sinπ/7
=-2sin2π/7*cos2π/7*cos4π/7/4sinπ/7
=-sin4π/7*cos4π/7/4sinπ/7
=-2sin4π/7*cos4π/7/8sinπ/7
=-sin8π/7/8sinπ/7
=sinπ/7/8sinπ/7
=1/8

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