Ответ:(-3cosx+√3sinx)/2

Объяснение:

cos∝-cosβ=-(2sin((∝-β)/2))*(2sin((∝+β)/2))

(sinπ/3)=√3/2

cos(2π/3-x)-cosx=-2(sin((2π/3-x-x)/2))*(sin(2π/3-x-+x)/2))=

(-2√3/2)sin(π/3-x)=-√3sin(π/3-x)=

=-√3(sin(π/3)*cosx-cos(π/3)*sinx)=

-√3((√3/2)*cosx-(1/2)sinx)=(-3cosx+√3sinx)/2

2 cпособ

cos(∝-β)=cos∝*cosβ+sin∝*sinβ

cos(2π/3-x)-cosx=cos(2π/3)*cosx+sin(2π/3)*sinx-cosx=

-(cosx)/2+(√3sinx)/2-cosx=(-3cosx+√3sinx)/2