高手来了,等等哈
cosxcos(x+π/3)+cos^2(π/3+x)
=cos(x+π/3)[cosx+cos(x+π/3)]
=cos(x+π/3)(cosx+cosxcosπ/3-sinxsinπ/3)
=(cosxcosπ/3-sinxsinπ/3)(3cosx/2-√3sinx/2)
=(cosx/2-√3sinx/2)(3cosx/2-√3/2sinx)
=3cos^2x/4-√3sinxcosx+3sin^2x/4
=3/4-√3sinxcosx
故原式=cos^2-√3sinxcosx+3/4
=(1+cos2x)/2-√3sin2x/2+3/4
=1/2(cos2x+√3sin2x)+5/4
=1/2sin(2x+π/6)+5/4
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