已知函数f (x)=(sin^4x+cos^4x+sin^2cos^2)/(2-sin2x)最小正周期.最大值.最小值.单调递增区间
问题描述：
已知函数f (x)=(sin^4x+cos^4x+sin^2cos^2)/(2-sin2x)最小正周期.最大值.最小值.单调递增区间
答
sin^4x+cos^4x
=(sin^2x+cos^2x)^2-2sin^2cos^2
=1-2sin^2cos^2
sin^4x+cos^4x+sin^2cos^2
=1-sin^2cos^2=1-(sin2x)^2/4
=(4-(sin2x)^2)/4
f (x)=(4-(sin2x)^2)/4 /(2-sin2x)
=(2+sin2x)/4
最小正周期为 T=π
最大值为3/4
最小值为1/4
当 2kπ-π/2此时kπ-π/4所以单调区间为：[kπ-π/4,kπ+π/4] k属于整数集
答
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