(1)已知tanα=2,求
sin^2α++sinαcosα-2cos^2α
=(sin^2α++sinαcosα-2cos^2α)/(sin^2α+cos^2α)
=(tan^2α+tanα-2)/(tan^2α+1)
=(4+2-2)/(4+1)
=4/5
(2)已知sinα=2sinβ,tanα=3tanβ,
cosα=sinα/tanα
=2sinβ/3tanβ
=2/3 cosβ