Question

已知π/2<β<α<3π/4,且cos( α-β)=12/13, sin(α+β)=-3/5,求tanαtanβ

,求tanαtanβ

Answer 1

解：∵π/2<β<α<3π/4
==>π<α+β<3π/2
∴cos(α+β)<0
∵sin(α+β)=-3/5
∴cos(α+β)=-√[1-sin²(α+β)]=-4/5
∵cos( α-β)=12/13
∴tanαtanβ=(sinαsinβ)/(cosαcosβ)
=(2sinαsinβ)/(2cosαcosβ)
=[cos(α-β)-cos(α+β)]/[cos(α-β)+cos(α+β)] (应用余弦和差角公式)
=[(12/13)-(-4/5)]/[(12/13)+(-4/5)]
=14。