用数学归纳法证明：cos(x/2)*cos[x/(2^2)]*…*cos[x/(2^n)]=sinx/[(2^n)*sinx/(2^n)]我证不出来,
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证明：（1）当n=1时,
右边=sinx/[2*sin(x/2)]=2sin(x/2)*cos(x/2)/[2*sin(x/2)]=cos(x/2)=左边；
（2）假设当n=k时,等式成立,
即cos(x/2)*cos[x/(2^2)]*…*cos[x/(2^k)]=sinx/[(2^k)*sinx/(2^k)] ；
当n=k+1时,
右边=sinx/{[2^(k+1)]*sinx/[2^(k+1)]}
=sinx/[(2^k)*sinx/(2^k)]}* {[sinx/(2^k)]/[2sinx/2^(k+1)]}
=sinx/[(2^k)*sinx/(2^k)]}* {[2sinx/2^（k+1)*cosx/2^（k+1)]/[2sinx/2^(k+1)]}
=sinx/[(2^k)*sinx/(2^k)]}* cosx/2^（k+1)
=cos(x/2)*cos[x/(2^2)]*…*cos[x/(2^k)]*cos[x/2^（k+1)]
=左边（3）综上所述,原等式成立.
这题的关键是用到公式：sinx=2[sin(x/2)]*[cos(x/2)];而且是从右边证到左边,并不是常规的从左边证到右边.