设0≤θ≤π/2,函数y=(1/sinθ-1)(1/cosθ-1),求函数的值域

1个回答

  • ∵0≦θ≦π/2,∴函数的定义域是(0,π/2),∴0≦θ/2≦π/4,∴sin(θ/2)、cos(θ/2)不等.

    ∴y

    =(1-sinθ)(1-cosθ)/(sinθcosθ)

    =[cos(θ/2)-sin(θ/2)]^2×2[sin(θ/2)]^2/[2sin(θ/2)cos(θ/2)cosθ]

    =sin(θ/2)[cos(θ/2)-sin(θ/2)]/{cos(θ/2)[cos(θ/2)+sin(θ/2)]

    =tan(θ/2)[1-tan(θ/2)]/[1+tan(θ/2)].

    令tan(θ/2)=x,则:

    y=x(1-x)/(1+x),∴y+yx=x-x^2,∴x^2+(y-1)x+y=0.

    显然,x是实数,∴需要:(y-1)^2-4y≧0,∴y^2-6y≧-1,∴(y-3)^2≧8,

    ∴y-3≦-2√2,或y-3≧2√2,∴y≦3-2√2,或y≧3+2√2.

    ∴函数的值域是(-∞,3-2√2]∪[3+2√2,+∞).