①利用公式sin(π-θ)=sinθ和sin(∏+θ)=-sinθ证明：sin（-θ）=-sinθ

①sin(-θ)=sin[π-(π+θ)]
=sin(π+θ) (∵sin(π-θ)=sinθ)
=-sinθ (∵sin(π+θ)=-sinθ).
②tanθsinθ∕(tanθ-sinθ)=(sinθ)^2/(sinθ-sinθcosθ) (分子分母同乘cosθ)
=[1-(cosθ)^2]/[sinθ(1-cosθ)]
=(1+cosθ)(1-cosθ)/[sinθ(1-cosθ)]
=(1+cosθ)/sinθ.
③∵sinα-2cosα+1=0 ==>sinα=2cosα-1
==>[2cosα-1]^2+(cosα)^2=1
==>5(cosα)^2-4cosα=0
∴cosα(5cosα-4)=0
∵α≠kπ+π∕2,k∈z ==>cosα≠0
∴5cosα-4=0 ==>cosα=4/5
==>sinα=2cosα-1=3/5
故1∕(sin2α-sinαcosα+1)=1/(2sinαcosα-sinαcosα+1) (应用倍角公式)
=1/(sinαcosα+1)
=1/[(3/5)(4/5)+1]
=25/37.