求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域.

y=sin^4x+cos^4x+4sin^2xcos^2x-1=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=1/2*sin^2(2x)=(1-cos4x)/4=1/4-1/4*cos4x周期T=2π/4=π/2值域是：[0,1/2]如果你认可我的回答,请点击左下角...