有关三角函数的问题,最好化成为同角三角函数解答

1个回答

  • y

    =sin(x+π/3)sin(x+π/2)=-[sinxcos(π/3)+cosxsin(π/3)]cosx

    =-(1/2)sinxcosx-(√3/2)(cosx)^2=-(1/4)sin2x-(√3/4)(1+cos2x)

    =-1-(1/4)sin2x-(√3/4)cos2x=-1-(1/2)[sin2xcos(π/3)+cos2xsin(π/3)]

    =-1-(1/2)sin(2x+π/3).

    于是:

    1、函数的值域是[-3/2,-1/2].

    2、函数的定义域是R.

    3、当2kπ-π/2<2x+π/3<2kπ+π/2时,函数递减,

    当2kπ+π/2<2x+π/3<2kπ+3π/2时,函数递增.

    由2kπ-π/2<2x+π/3<2kπ+π/2,得:kπ-5π/12<x<kπ-π/12.

    由2kπ+π/2<2x+π/3<2kπ+3π/2,得:kπ-π/12<x<kπ+7π/12.

    ∴函数在区间(kπ-5π/12,kπ-π/12)上递减,在区间(kπ-π/12,kπ+7π/12)递增.

    4、函数为非奇非偶函数.

    5、函数的最小正周期是π.