已知函数f(x)=cos(2π-x) cos(π/2-x)-sin^2x (1)求函数f(x)的最小正周期

1、f(x)=cos(2π-x) cos(π/2-x)-sin^2x =-cosxsinx-sin^2x=-½sin2x-(1-cos2x)/2=-1/2sin2x+1/2cos2x-1/2= - √2/2sin(2x-π/4)-1/2所以函数f(x)的最小正周期为π2、当x∈[-π/8,3π/8]时,-π/2≤2x-π/4≤π...