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由于 sin(π/4 +x)=sin[π/2 -(π/4 -x)]=cos(π/4 -x)，所以
f(x)=sin(π/3 +2x) -sin(π/3 -2x) +2√3sin(π/4 -x)sin(π/4 +x)
=sin(π/3)cos2x+cos(π/3)sin2x -sin(π/3)cos2x+cos(π/3)sin2x +2√3sin(π/4 -x)cos(π/4 -x)
=2cos(π/3)sin2x +√3sin2(π/4 -x)
=sin2x +√3sin(π/2 -2x)
=sin2x +√3cos2x
=2[(1/2)sin2x +(√3/2)cos2x]
=2sin(2x+π/3)
f(x)=sin(π/3 +2x) -sin(π/3 -2x) +2√3sin(π/4 -x)sin(π/4 +x)
=sin(π/3)cos2x+cos(π/3)sin2x -sin(π/3)cos2x+cos(π/3)sin2x +2√3sin(π/4 -x)cos(π/4 -x)
=2cos(π/3)sin2x +√3sin2(π/4 -x)
=sin2x +√3sin(π/2 -2x)
=sin2x +√3cos2x
=2[(1/2)sin2x +(√3/2)cos2x]
=2sin(2x+π/3)
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