Question

f(x)=cos(2x-2兀/3)-cosx=√3/2sinx-3/2cos2x

f(x)=cos(2x-2兀/3)-cosx=√3/2sinx-3/2cos2x=√3sin(2x-兀/3)能一步步算吗，我看不懂

Answer 1

• f(x)=cos(2x-⅔π)-cos2x吧 展开：

  =cos2xcos⅔π+sin2xsin⅔π-cos2x

  =-½cos2x+√3/2sin2x-cos2x

  =√3/2sin2x-3/2cos2x

  =√3(½sin2x-√3/2cos2x)

  =√3[sin2xcos(⅓π)-cos2xsin(⅓π)] 

  =√3sin(2x-⅓π)

• 或者用和差化积公式

  cos(2x-⅔π)-cos2x=-2sin[(2x-⅔π+2x)/2]·sin[(2x-⅔π-2x)/2]

  =-2sin(2x-⅓π)·sin(-⅓π)

  =√3sin(2x-⅓π) 更简单