cosx-1/2(cosxcosπ/3+sinxsinπ/3)如何得到-根号3除以2(sinxcosπ/3-cosxsinπ/3)?

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cosx-1/2(cosxcosπ/3+sinxsinπ/3)如何得到-根号3除以2(sinxcosπ/3-cosxsinπ/3)?

cosx-1/2(cosxcosπ/3+sinxsinπ/3)如何得到-根号3除以2(sinxcosπ/3-cosxsinπ/3)?
cosx-1/2(cosxcosπ/3+sinxsinπ/3)如何得到-根号3除以2(sinxcosπ/3-cosxsinπ/3)?

cosx-1/2(cosxcosπ/3+sinxsinπ/3)如何得到-根号3除以2(sinxcosπ/3-cosxsinπ/3)?
原式=cosx-1/2(cosx(1/2)+sinx(√3/2))
=3/4cosx-√3/4sinx
=-√3/2(1/2sinx-√3/2cosx )
=-√3/2(sinxcosπ/3-cosxsinπ/3)

cosx-(1/2)[cosxcos(π/3)+sinxsin(π/3)]如何得到-(√3/2)[sinxcos(π/3)-cosxsin9π/3)]?
解:cosx-(1/2)[cosxcos(π/3)+sinxsin(π/3)]=cosx-(1/2)[(1/2)cosx+(√3/2)sinx]
=cosx-(1/4)cosx-(√3/4)sinx=(3/4)cosx-(√3/4)sinx
=(-√3/2)[(1/2)sinx-(√3/2)cosx]
=(-√3/2)[sinxcos(π/3)-cosxsin(π/3)]

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