f(x)=2sinx(sinx+cosx)
=2sin²x+2sinxcosx
=1-cos2x+sin2x
=1+√2 (√2/2sin2x-√2/2cos2x)
=1+√2(sin2xcosπ/4-cos2xsinπ/4)
=1+√2sin(2x-π/4)
所以
最小正周期=π
最大值=1+√2
原始=2sinx^2+sin2x=1-cos2x+sin2x=1+genhao2sin(2x-pai/4)所以最小正周期为π,最大值为1+genhao2
f(x)=2sin(^2)x+2sinxcosx
=1-cos2x+sin2x
=√(2)(sin2x-(π/4))+1
所以f(x)的最小正周期为π,最大值为 √(2+1
最小正周期π
最大值2*2^0.5