（sin^4 x+cos^4 x+sin^2 x * cos^2 x)/2-sin2x的最小正周期,最大值和最小值..

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sin^4 x+cos^4 x+sin^2 x*cos^2 x
=sin^4 x+cos^4 x+2sin^2 x*cos^2 x-sin^2 x*cos^2 x
=(sin^2 x+cos^2 x)^2-sin^2 x*cos^2 x
=1-sin^2 x*cos^2 x
=(1+sinxcosx)(1-sinxcosx)
2-sin2x=2-2sinxcosx=2(1-sinxcosx)
所以f(x)=(1+sinxcosx)/2-(sinxcosx)/2+(cos2x)/4
=1/2+(cos2x)/4
所以T=2π/2=π
-1所以cosx=-1,f(x)最小=1/2-1/4=1/4
cosx=1,f(x)最大=1/2+1/4=3/4

sin^4 x+cos^4 x+sin^2 x*cos^2 x
=sin^4 x+cos^4 x+2sin^2 x*cos^2 x-sin^2 x*cos^2 x
=(sin^2 x+cos^2 x)^2-sin^2 x*cos^2 x
=1-sin^2 x*cos^2 x
=(1+sinxcosx)(1-sinxcosx)
2-sin2x=2-2sinxcosx=2(1-sinxcosx)
所以
（sin^4 x+cos^4 x+sin^2 x * cos^2 x)/2-sin2x
=(1+sinxcosx)/2
=1/2+1/4sin2x
所以T=2π/2=π
-1所以
sin2x=-1,（sin^4 x+cos^4 x+sin^2 x * cos^2 x)/2-sin2x最小=1/2-1/4=1/4
sin2x=1,（sin^4 x+cos^4 x+sin^2 x * cos^2 x)/2-sin2x最大=1/2+1/4=3/4
-1所以cosx=-1,f(x)最小=1/2-1/4=1/4
cosx=1,f(x)最大=1/2+1/4=3/4