Question

数f(x)=2cos(x-π/3)+2sin(3π/2-x),怎么化简求具体过程

Answer 1

f(x)=2cos(x-π/3)+2sin(3π/2-x)
=2[cosxcos(π/3)+sinxsin(π/3)]+2[cosxsin(3π/2)-sinxcos(3π/2)]
=2(1/2*cosx+√3/2*sinx)+2(-cosx*1-sinx*0)
=cosx+√3sinx-2cosx
=-cosx+√3sinx
=2(√3/2*sinx+1/2*cosx)
=2sin(x-π/6)

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Answer 2

f(x)= 2cos(x-π/3)+2sin(3π/2-x)
=2cos(x-π/3)-2cosx
=2(cosxcosπ/3+sinxsinπ/3)-2cosx
=√3sinx+cosx-2cosx
=√3sinx-cosx
=2（sinxcosπ/6-cosxsinπ/6）
=2sin(x-π/6）

追问

第二部2cosx哪来的

追答

你不会换算吗，2sin(3π/2-x)=2sin(π+π/2-x)=-2cosx。同学。