Question

已知tanα=1/3,tan（π-β）=-1/2,sin（α+β）-2sinαcosβ除以2sinαsinβ+cos（α

Answer 1

tanα = 1/3,tan(π - β) = - tanβ = 1/2

[sin(α + β) - 2sinαcosβ]/[2sinαsinβ + cos(α + β)]

= [sinαcosβ + cosαsinβ - 2sinαcosβ]/[2sinαsinβ + cosαcosβ - sinαsinβ]

= [cosαsinβ - sinαcosβ]/[cosαcosβ + sinβsinβ]

= - sin(α - β)/cos(α - β)

= - tan(α - β)

= - [tanα - tanβ]/[1 + tanαtanβ]

= - [(1/3) - (1/2)]/[1 + (1/3)(1/2)]

= 1/7