f(x)=2sin(x+π/4)sin(x-π/4)+sin2x
=2sin(x+π/4)sin(-π/2+x+π/4)+sin2x
=-2sin(x+π/4)cos(x+π/4)+sin2x
=-sin(2x+π/2)+sin2x
=-cos2x+sin2x
=√2sin(2x-π/4)
所以由正弦型函数的性质易知：
函数f(x)的值域为[-√2,√2]
函数f(x)=2sin(x+π/4)sin(x-π/4)+sin2x的值域
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