摘要:15.t x 已知cos2θ=.θ∈(.π). (I)求sinθ的值, (II)求sin (θ+)-sin2θ的值. 解:(I)∵ cos2θ=.∴ 1-2sin2θ=. ∴ sin2θ=. ∵θ∈(.π).∴ sinθ=. (II)∵ sinθ=且θ∈(.π). ∴ cosθ=.∴ sin2θ= 2sinθcosθ= 2××() =. ∴ sin (θ+)-sin2θ= sinθ·cos+ cosθ·sin-sin2θ =×+ ()×-() =.

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