已知函数f(x)=sin²x-sin²(x-π/6) 求f(x)最小正周期 求f在负三分之派 四分之派

解： f(x)=sin²x-sin²(x-π/6) =½[1-cos(2x)]-½[1-cos(2x-π/3)] =-½[cos(2x)-cos(2x-π/3)] =-½[cos(2x)-cos(2x)cos(π/3)-sin(2x)sin(π/3)] =-½[cos(2x)-½cos(2x)-sin(2x)sin(π/3)] =-½[½cos(2