tanβ=-3,求2sin^2β+sinβcosβ-cos^2β+2的值.

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tanβ=-3,求2sin^2β+sinβcosβ-cos^2β+2的值.

tanβ=-3,求2sin^2β+sinβcosβ-cos^2β+2的值.
tanβ=-3,求2sin^2β+sinβcosβ-cos^2β+2的值.

tanβ=-3,求2sin^2β+sinβcosβ-cos^2β+2的值.
2sin^2β+sinβcosβ-cos^2β+2=2+(2sin^2β+sinβcosβ-cos^2β)/(sin^2β+cos^2β)
(2sin^2β+sinβcosβ-cos^2β)/(sin^2β+cos^2β)分式上下同除cos^2β
(2sin^2β+sinβcosβ-cos^2β)/(sin^2β+cos^2β)
=(2tan^2β+tanβ-1)/(tan^2β+1)=-7
所以原式2sin^2β+sinβcosβ-cos^2β+2=-7+2=-5