已知x,y≠kπ+[π/2](k∈Z),sinx是sinθ,cosθ的等差中项,siny是sinθ,cosθ的等比中项.

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  • 证明:(1)∵sinθ与cosθ的等差中项是sinx,等比中项是siny,

    ∴sinθ+cosθ=2sinx①,sinθcosθ=sin2y②,

    2-②×2,可得(sinθ+cosθ)2-2sinθcosθ=4sin2x-2sin2y,即4sin2x-2sin2y=1.

    ∴4×[1−cos2x/2]-2×[1−cos2y/2]=1,即2-2cos2x-(1-cos2y)=1.

    故证得cos2x=[1/2]cos2y;

    (2)要证

    2(1−tan2x)

    1+tan2x=

    1−tan2y

    1+tan2y,只需证

    1−

    sin2x

    cos2x

    1+

    sin2x

    cos2x=

    1−

    sin2y

    cos2y

    2(1+

    sin2y

    cos2y),

    即证

    cos2x−sin2x

    cos2x+sin2x=

    cos2y−sin2y

    2(cos2y+sin2y),即证cos2x-sin2x=[1/2](cos2y-sin2y),只需证cos2x=[1/2]cos2y.

    由(1)的结论,cos2x=[1/2]cos2y显然成立.

    所以

    2(1−tan2x)

    1+tan2x=

    1−tan2y

    1+tan2y.