Question

已知tanα,tanβ是方程x^2+4x-3=0的两个根，求sin^2（α+β）+sin(α+β)cos(α+β)=

Answer 1

tanα+tanβ=-4
tanαtanβ=-3
tan(α+β)=-4/(1+3)=-1

sin^2（α+β）+sin(α+β)cos(α+β)
=[sin^2（α+β）+sin(α+β)cos(α+β)]/1
=[sin^2（α+β）+sin(α+β)cos(α+β)]/[sin^2（α+β）+cos^2(α+β)]
=[tan^2(α+β)+tan(α+β)]/(1+tan^2(α+β))
=(1-1)/(1+1)
=0

Answer 2

tanα+tanβ=-4
tanαtanβ=-3
tan(α+β)=-4/(1+3)=-1
sin²(α+β)+sin(α+β)cos(α+β)
=[sin²(α+β)+sin(α+β)cos(α+β)]/1
=[sin²(α+β)+sin(α+β)cos(α+β)]/[sin²(α+β)+cos²(α+β)]
=[tan²(α+β)+tan(α+β)]/(1+tan²(α+β))
=(1-1)/(1+1)
=0