由|f(x)|<=|sinx|
则在0附近且x>0时
-sinx<=f(x)<=sinx（在0附近）
-sinx/x<=[f(x)-f(0)]/(x-0)<=sinx/x
x→0+得-1<=lim(x→0+)[f(x)-f(0)]/(x-0)=f'(0+)<=1
同理：在0附近且x<0时-1<=f'(0-)<=1
从而|f'(0)|<=1
又f'(x)=a1*cosx+2*a2*cos(2x)+...+n*an*cos(nx)
f'(0)=a1+2*a2+...+n*an
故：|a1+2*a2+...+n*an|<=1