Question

已知tanθ=2,则sin^θ+sinθcosθ-2cos^2θ=?

Answer 1

(sinθ)^2+sinθcosθ-2(cosθ)^2

=[(sinθ)^2+sinθcosθ-2(cosθ)^2]/[(sinθ)^2+(cosθ)^2]

={[(sinθ)^2/(cosθ)^2]+[(sinθcosθ)/(cosθ)^2]-[2(cosθ)^2/(cosθ)^2]}/{[(sinθ)^2/(cosθ)^2]+[(cosθ)^2/(cosθ)^2]}

=[(tanθ)^2+tanθ-2]/[(tanθ)^2+1]

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

=4/5