已知函数f（x）＝f′（½p）sinx+cosx，则f（¼p）＝ 求这道题做法
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f（x）＝f′（p/2）sinx+cosx,①
则f'(x)=f'(p/2)cosx-sinx,
f'(p/2)=f'(p/2)cos(p/2)-sin(p/2),
f'(p/2)[1-cos(p/2)]=-sin(p/2),
f'(p/2)=-sin(p/2)/[1-cos(p/2)]②
由①，f(p/4)=f'(p/2)sin(p/4)+cos(p/4)
=-2{sin(p/4)]^2cos(p/4)/[1-cos(p/2)]+cos(p/4)
=0.
则f'(x)=f'(p/2)cosx-sinx,
f'(p/2)=f'(p/2)cos(p/2)-sin(p/2),
f'(p/2)[1-cos(p/2)]=-sin(p/2),
f'(p/2)=-sin(p/2)/[1-cos(p/2)]②
由①，f(p/4)=f'(p/2)sin(p/4)+cos(p/4)
=-2{sin(p/4)]^2cos(p/4)/[1-cos(p/2)]+cos(p/4)
=0.
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