求f(x)=2sin(x+π/4)sin(x-π/4)+sin2X的最大值.(答案是根号2,

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求f(x)=2sin(x+π/4)sin(x-π/4)+sin2X的最大值.(答案是根号2,
求f(x)=2sin(x+π/4)sin(x-π/4)+sin2X的最大值.(答案是根号2,

求f(x)=2sin(x+π/4)sin(x-π/4)+sin2X的最大值.(答案是根号2,
f(x)=2sin(x+π/4)sin(x-π/4)+sin2X
=2(根号2/2*sinx+根号2/2*cosx)(根号2/2*sinx-根号2/2*cosx)+sin2x
=sinx^2-cosx^2+sin2x
=-cos2x+sin2x
=根号2sin(2x-π/4)
因为sin(2x-π/4)最大值是1
所以原式最大值为根号2