sinθ,sin2x,cosθ是等差数列,sinθ,sinx,cosθ为等比数列,求cos2x的值~(+﹏+)~
人气:274 ℃ 时间:2020-04-05 02:24:48
解答
sinθ,sin2x,cosθ是等差数列,
那么2sin2x=sinθ+cosθ①
sinθ,sinx,cosθ为等比数列,
那么sin²x=sinθcosθ②
①²-②×2:
4sin²2x-2sin²x=(sinθ+cosθ)²-2sinθcosθ
∵(sinθ+cosθ)²-2sinθcosθ
=sin²θ+cos²θ+2sinθcosθ-2sinθcosθ
=1
∴4(1-cos²2x)-(1-cos2x)=1
∴4cos²2x-cos2x-2=0
 cos2x=(1±√33)/8
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