求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域.
问题描述:
求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域.
答
y=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1
=1+2sin^2xcos^2x-1
=2sin^2xcos^2x
=sin^2(2x)/2
=(1-cos4x)/4
周期显然是pi/2而不是pi/4
值域是:[0,1/2]