已知函数f(x)满足f(π/4-x)=f(x) f(-x)=-f(x) f(π/3)=1 则f(π/2+x)-f(x)=

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已知函数f(x)满足f(π/4-x)=f(x)f(-x)=-f(x)f(π/3)=1则f(π/2+x)-f(x)=已知函数f(x)满足f(π/4-x)=f(x)f(-x)=-f(x)f(π/3)=1则

已知函数f(x)满足f(π/4-x)=f(x) f(-x)=-f(x) f(π/3)=1 则f(π/2+x)-f(x)=
已知函数f(x)满足f(π/4-x)=f(x) f(-x)=-f(x) f(π/3)=1 则f(π/2+x)-f(x)=

已知函数f(x)满足f(π/4-x)=f(x) f(-x)=-f(x) f(π/3)=1 则f(π/2+x)-f(x)=
f(π/2+x) - f(x) = f[π/4 -(π/2+x)] - f(x)
= f(-π/4 - x) - f(x)
= - f(π/4 + x) - f(x)
= - f[π/4 - (π/4 + x)] - f(x)
= -f(-x) - f(x)
=f(x) - f(x)
= 0