证明:(1)∵sinθ与cosθ的等差中项是sinx,等比中项是siny,
∴sinθ+cosθ=2sinx①,sinθcosθ=sin2y②,
①2-②×2,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.