若sin2x、sinx分别是sinθ与cosθ的等差中项和等比中项,则cos2x的值为:(  ) A.1+338 B.1−338 C.1±338 D.1−24

问题描述:

若sin2x、sinx分别是sinθ与cosθ的等差中项和等比中项,则cos2x的值为:(  )
A.

1+
33
8

B.
1−
33
8

C.
33
8

D.
1−
2
4

依题意可知2sin2x=sinθ+cosθsin2x=sinθcosθ∵sin2θ+cos2θ=(sinθ+cosθ)2-2sinθcosθ=4sin22x-2sin2x=1∴4(1-cos22x)+cos2x-2=0,即4cos22x-cos2x-2=0,求得cos2x=1±338∵sin2x=sinθcosθ∴cos2x=1-2si...