已知tanA=2,则sinAcosA一2sin^2A=?

2025-04-11 11:10:57
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回答1:

解:
tanA=2
sinAcosA-2sin²A
=(sinAcosA-2sin²A)/(sin²A+cos²A)
=(tanA-2tan²A)/(tan²A+1)
=(2-2·2²)/(2²+1)
=-6/5