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提供另外一种证法.1-cosx=1-cos(2x/2)=2[sin(x/2)]^2；函数y=sin(x/2)在x=0处的切线方程为y=x/2,显然当x＞0时,y=sin(x/2)在直线y=x/2下方,即sin(x/2)＜x/2①,当0＜x＜π/2时,0＜sin(x/2)＜x/2,所以[sin(x/2)]^2＜(x^2)/4,即2[sin(x/2)]^2＜(x^2)/2②；当x=π/2时,sin(x/2)=1/√2,过点(0,0)和(π/2,1/√2)的直线方程为y=√2x/π,显然当0＜x＜π/2时,函数y=sin(x/2)在直线y=√2x/π的上方,所以√2x/π＜sin(x/2),所以2(x^2)/π^2＜[sin(x/2)]^2,即4(x^2)/π^2＜2[sin(x/2)]^2,因(x^2)/π＜4(x^2)/π^2,所以(x^2)/π＜2[sin(x/2)]^2③；综合②、③得(x^2)/π＜2[sin(x/2)]^2＜(x^2)/2,即(x^2)/π＜1-cosx＜(x^2)/2(证毕).