试题:
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.
求证:(1)cos2x=
1
2
cos2y;(2)
2(1-tan2x)
1+tan2x
=
1-tan2y
1+tan2y

答案:

我来补答
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
 
 
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