Question

已知tanθ＝2,则sin2θ＋sinθcosθ－2cos2θ＝?

Answer 1

sin2θ+sinθcosθ-2cos2θ

=3sinθcosθ-4(cosθ)^2+2

=[3sinθcosθ-4(cosθ)^2+2(sinθ)^2+2(cosθ)^2]/[(sinθ)^2+(cosθ)^2](分子分母同除(cosθ)^2)

=[3tanθ-4+2(tanθ)^2+2]/[(tanθ)^2+1]

=(6-4+8+2)/(4+1)

=12/5