已知向量a=(cosx-3,sinx)向量b=(cosx,sinx-3),f(x)=向量a乘以向量b,后附小题

5个回答

  • (1)∵a=(cosx-3,sinx)b=(cosx,sinx-3),f(x)=a·b,

    ∴f(x)=(cosx-3)cosx+sinx(sinx-3)

    =cos²x-3cosx+sin²x-3sinx

    =-3(sinx+cosx)+1

    =-3√2sin(x+π/4)+1

    ∴函数f(x)的单调递增区间为:

    π/2+2kπ≤x+π/4≤3π/2+2kπ,k∈z(sin(x+π/4)前面系数<0)

    即π/4+2kπ≤x≤5π/4+2kπ,k∈z

    ∵x∈[2π,3π]

    ∴函数f(x)的单调递增区间为:[9π/4,3π](当k=1时)

    (2)∵f(x)=-1,

    ∴ -3√2sin(x+π/4)+1=-1

    sin(x+π/4)=√2/3,

    ∵x∈(π/2,3π/4),∴3π/4<x+π/4<π

    ∴cos(x+π/4)=-√7/3

    ∴cos2x=sin(2x+π/2)=2sin(x+π/4)cos(x+π/4)=-2√14/9

    ∵x∈(π/2,3π/4),∴2x∈(π,3π/2)

    ∴sin2x=-5/9

    ∴tan2x=sin2x/cos2x=5√14/28