解:
1-cosxcos(2x)cos(3x)
=1-½[cos(2x+x)+cos(2x-x)]cos(3x)
=1-½[cos(3x)+cosx]cos(3x)
=1-½cos²(3x)-½cosxcos(3x)
=1-¼[1+cos(6x)]-¼[cos(3x+x)+cos(3x-x)]
=¾-¼cos(6x)-¼[cos(4x)+cos(2x)]
=¼[1-cos(6x)+1-cos(4x)+1-cos(2x)]
=¼[½·(6x)²+½·(4x)²+½·(2x)²]
=7x²
7x²与axⁿ是等价无穷小
a=7,n=2
用到的等价无穷小:1-cosx~½x²