cotθ=2,则5cosθ2+3sinθcosθ-2sin2θ=?

cotθ=2,则1+(cotθ)^2=5;
(sinθ)^2=1/(1+(cotθ)^2)=1/5;
(cosθ)^2=1-(sinθ)^2=4/5;
sinθcosθ=(cosθ/sinθ)·(sinθ)^2=cotθ·(sinθ)^2=2·(1/5)=2/5;
∴5cosθ2+3sinθcosθ-2sin2θ=5×4/5+3×2/5-2×1/5=24/5

cotθ – tanθ =cosθ/sinθ-sinθ/cosθ
=( cos^2θ-sin^2θ)/(sinθcosθ)
=cos2θ/ (1/2 sin2θ)
=2cos2θ/ (sin2θ)
=2 cot 2θ
所以原式成立