所用公式：
cos(A+B)=cosAcosB+sinAsinB.......①
cos(A-B)=cosAcosB-sinAsinB.......②
①-②：cos(A+B)-cos(A-B)=2sinAsinB........③
①+②：cos(A+B)+cos(A-B)=2cosAcosB........④
令C=A+B，D=A-B，则④可改为cosC+cosD=2cos(C/2+D/2)cos(C/2-D/2)......④改
解：
f(x)=cos(2x-π/3)+2sin(x-π/4)×sin(x+π/4)
=cos(2x-π/3)+cos(2x)-cos(-π/2) ............利用公式③
=cos(2x-π/3)+cos(2x)-0
=2cos(x-π/6+x)cos(x-π/6-x) ........利用公式④改
=2cos(2x-π/6)cos(-π/6)
=√3cos(2x-π/6)
故最小正周期为π
f(x) = cos(2x-π/3)+2*sin(x-π/4)*sin(x+π/4)
最小正周期等于π
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