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1-cosx=1-{[cos(x/2)]^2-[sin(x/2)]^2}
=1-[cos(x/2)]^2+[sin(x/2)]^2
=[sin(x/2)]^2+[sin(x/2)]^2
=2[sin(x/2)]^2
又 sin(x/2)<x/2
所以 2[sin(x/2)]^2<2(x/2)^2
=1-[cos(x/2)]^2+[sin(x/2)]^2
=[sin(x/2)]^2+[sin(x/2)]^2
=2[sin(x/2)]^2
又 sin(x/2)<x/2
所以 2[sin(x/2)]^2<2(x/2)^2
2017-09-27
展开全部
Let f(x)=x-sinx
f'(x)=1-cosx>=0, f(x)是增函数
f(0)=0
if x>0, then f(x)>0
x>sinx
(x/2)^2>(sin (x/2) )^2
f'(x)=1-cosx>=0, f(x)是增函数
f(0)=0
if x>0, then f(x)>0
x>sinx
(x/2)^2>(sin (x/2) )^2
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