1.已知函数f(x)=(sinx-cosx)sinx x∈R,求f（x）的值域或及最小正周期

原函数可化为f(x)=sin^2x-cosxsinx=1/2(1-cos2x)-1/2sin2x=1/2[1-(cos2x+sin2x)]=-√2/2sin（2x+π/4）+1/2因为x∈R,所以sin（2x+π/4）∈[-1,1]-√2/2sin(2x+π/4）∈[-√2/2,√2/2]-√2/2sin(2x+π/4+1/2∈[-√2+1...