高等数学,跪求详解。第1-103题怎么做,谢谢

2025-05-10 05:52:30
推荐回答(1个)
回答(1):


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