cos(x)cos(x/2)cos(x/4)...cos(x/2^n) = cos(x)cos(x/2)cos(x/4)...cos(x/2^n)sin(x/2^n)/sin(x/2^n)
=sin2x/[2^(n-1) sin(x/2^n)]
n趋于无穷大时，上式=sin2x/[2^(n-1) sin(x/2^n)] ~ sin2x /[2^(n-1) *x/2^n] = sin2x/2x
lim sin2x/2x =1得证
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