f(x)=e^x-(1+x)-(1-cosx),f'(x)=e^x-1-sinx,f''(x)=e^x-cosx>0,当x>0时,因此f'(x)递增函数,故f'(x)>f'(0)=0,于是f(x)是递增函数,f(x)>f(0)=0,即e^x-(1+x)>1-cosx