Question

高一数学必修四题目

观察以下各式①cosπ/3=1/2②cosπ/5cos2π/5=1/4③cosπ/7cos2π/7cos3π/7=1/8④cosπ/9cos2π/9cos3π/9cos4π/9=1/16 分析上述各式的特征，写出能反映一般规律的等式 请对③式给予证明 请对一般规律的等式给予证明

Answer 1

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)]
=sin[2nπ/(2n-1)] / (2^n)sin[π/(2n-1)]
=sin[π/(2n-1)] / (2^n)sin[π/(2n-1)] 【sin[2nπ/(2n-1)] =sin[π/(2n-1)] 】
=1/(2^n)

重点：不断地利用二倍角公式：sin2θ=2sinθcosθ