sinχcosχcos2χcos4χcos8χcos16χ……cos2²﹣¹χ
问题描述：
sinχcosχcos2χcos4χcos8χcos16χ……cos2²﹣¹χ
答
原式=0.5sin2xcos2χcos4χcos8χcos16χ……cos2²﹣¹χ=1/4*sin4xcos4χcos8χcos16χ……cos2^(n-1)χ
=...=1/2^n*cos(2^n*x)