（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。。

有详细过程。。谢谢

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<=sin2x<=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<=cos2x<=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f(x)=(1+sinxcosx)/2-(sinxcosx)/2+(cos2x)/4 =1/2+(cos2x)/4 所以T=2π/2=π -1<=cos2x<=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

f(x)=(sin^4x+cos^4x+sin^2cos^2)/(2-sin2x) ={[(sinx)^2+(cosx)^2]^2-(sinxcosx)^2}/(2-2sinxcosx) =(1-(sinxcosx)^2)/(2-2sinxcosx) =(1+sinxcosx)(1-sinxcosx)/2(1-sinxcosx) =1/2+1/4sin2x 最小正周期t=π 最大值1/2+1/4=0.75 最小值1/2-1/4=0.25 单调递增区间 (-π/4+kπ,π/4+kπ)k为任意整数