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求值Cos[π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)...
求值Cos[π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]
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Cos[π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]
=sin[π/(2^(n+1)+1)]*Cos[π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
=sin[2π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/2sin[π/(2^(n+1)+1)]
=sin[4π/(2^(n+1)+1)]Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/2^2sin[π/(2^(n+1)+1)]
=……
=sin[2^nπ/(2^(n+1)+1)]cos[2^nπ/(2^(n+1)+1)]/2^nsin[π/(2^(n+1)+1)]
=sin[2^(n+1)π/(2^(n+1)+1)]/2^(n+1)sin[π/(2^(n+1)+1)]
=sin[π-π/(2^(n+1)+1)]/2^(n+1)sin[π/(2^(n+1)+1)]
=sin[π/(2^(n+1)+1)]/2^(n+1)sin[π/(2^(n+1)+1)]
=1/2^(n+1).
=sin[π/(2^(n+1)+1)]*Cos[π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
=sin[2π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/2sin[π/(2^(n+1)+1)]
=sin[4π/(2^(n+1)+1)]Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/2^2sin[π/(2^(n+1)+1)]
=……
=sin[2^nπ/(2^(n+1)+1)]cos[2^nπ/(2^(n+1)+1)]/2^nsin[π/(2^(n+1)+1)]
=sin[2^(n+1)π/(2^(n+1)+1)]/2^(n+1)sin[π/(2^(n+1)+1)]
=sin[π-π/(2^(n+1)+1)]/2^(n+1)sin[π/(2^(n+1)+1)]
=sin[π/(2^(n+1)+1)]/2^(n+1)sin[π/(2^(n+1)+1)]
=1/2^(n+1).
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Cos[π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]
乘以sin[π/(2^(n+1)+1)],再除以sin[π/(2^(n+1)+1)]
=sin[π/(2^(n+1)+1)]*Cos[π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
=1/2sin[2π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
一直使用而被角公式结合下去
=1/2^nsin[2^（n+1）π/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
=[1/2^nsin[π-(2^(n+1)π/(2^(n+1)π+1)]/sin[π/(2^(n+1)+1)]
=[1/2^n*sin[π/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
=1/2^n
乘以sin[π/(2^(n+1)+1)],再除以sin[π/(2^(n+1)+1)]
=sin[π/(2^(n+1)+1)]*Cos[π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
=1/2sin[2π/(2^(n+1)+1)]*Cos[2π/(2^(n+1)+1)]*Cos[4π/(2^(n+1)+1)].......*cos[2^nπ/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
一直使用而被角公式结合下去
=1/2^nsin[2^（n+1）π/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
=[1/2^nsin[π-(2^(n+1)π/(2^(n+1)π+1)]/sin[π/(2^(n+1)+1)]
=[1/2^n*sin[π/(2^(n+1)+1)]/sin[π/(2^(n+1)+1)]
=1/2^n
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