Question

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

Answer 1

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]