Question

已知角α、β均为锐角,且有sinα=√2/2,cos(α+β)=-3/5,则tanβ=

Answer 1

∵角α、β均为锐角

==>00

∵sinα=√2/2,cos(α+β)=-3/5

∴cosα=√[1-(sinα)^2]=√2/2

sin(α+β)=√[1-(cos(α+β))^2]=4/5

于是,得

tanα=sinα/cosα=1

tan(α+β)=sin(α+β)/cos(α+β)=-4/3

故 tanβ=tan[(α+β)-α]

=[tan(α+β)-tanα]/[1+tan(α+β)tanα] (应用差角公式)

=[(-4/3)-1]/[1+(-4/3)*1]

=7.