若tan(π+α)=3,α∈(π,3/2π),求sin^2α+3sinαcosα+2cos^2α

问题描述:

若tan(π+α)=3,α∈(π,3/2π),求sin^2α+3sinαcosα+2cos^2α

tan(π+α)=3,α∈(π,3/2π),
所以
tanα=3
所以
sin^2α+3sinαcosα+2cos^2α
=(sin^2α+3sinαcosα+2cos^2α)/(sin²α+cos²α)
=(tan²α+3tanα+2)/(tan²α+1)
=(9+9+2)/(9+1)
=2