来自刘洪军的问题
已知(sinαcosα)/(1-cos2α)=1,tan(α-β)=-2/3,则tan(β-2α)=
已知(sinαcosα)/(1-cos2α)=1,tan(α-β)=-2/3,则tan(β-2α)=
1回答
2020-06-25 01:41
已知(sinαcosα)/(1-cos2α)=1,tan(α-β)=-2/3,则tan(β-2α)=
已知(sinαcosα)/(1-cos2α)=1,tan(α-β)=-2/3,则tan(β-2α)=
∵(sinαcosα)/(1-cos2α)=(sinαcosα)/(2sin²α)=(cosα)/(2sinα)=1/(2tanα)=1
∴tanα=1/2
∴tan(β-2α)=﹣tan(2α-β)=﹣tan[α+(α-β)]=﹣[tanα+tan(α-β)]/[1-tanαtan(α-β)]
=﹣(1/2-2/3)/[1-1/2×(﹣2/3)]=1/6×3/4=1/8