已知函数f(x)=sin(2x-π/6)-2cos(x-π/4)cos(x+π/4）+1,求f(x)的最小周期及在区间【0,π/2】上的值域
f(x) = √3/2*sin(2x) - 1/2*cos(2x) - (cosx + sinx)(cosx - sinx) + 1
= √3/2*sin(2x) - 1/2*cos(2x) - cos(2x) + 1
= √3[1/2*sin(2x) - √3/2*cos(2x)] + 1
= √3sin(2x - π/3) + 1
所以最小正周期为π
x∈[0,π/2],则(2x - π/3)∈[-π/3,2π/3]
所以值域为[-1/2,√3+1]