若tanα=3,则sin2α-2sinαcosα+3cos2α=

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若tanα=3,则sin2α-2sinαcosα+3cos2α=
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若tanα=3,则sin2α-2sinαcosα+3cos2α=
若tanα=3,则sin2α-2sinαcosα+3cos2α=

若tanα=3,则sin2α-2sinαcosα+3cos2α=
∵tanα=3
∴sin²α-2sinαcosα+3cos²α
=(sin²α-2sinαcosα+3cos²α)/(sin²α+cos²α)
=(sin²α/cos²α-2sinα/cosα+3)/(sin²α+1) (分子分母同时除以cos²α)
=(tan²α-2tanα+3)/(tan²α+1)
=(9-6+3)/(9+1)
=3/5

解1: tan(π/4+α)=3 [tan(π/4)+tanα]/[1-tan(π/4)tanα][2sinαcosα+3cos2α)/(5cos2α-3sin2α) =[2sinαcosα+3(cos