Question

已知方程x^2+4ax+3a+1(a>1)的两根为tana,tanb,a,b∈(-π/2,π/2)求tan((a+b)/2)

Answer 1

方程x^2+4ax+3a+1(a>1)的两根为tana,tanb
tana+tanb=-4a, tanatanb=3a+1
tan(a+b)=(tana+tanb)/(1- tanatanb)
=(-4a)/(1-3a-1)=4/3>0
∵a>1∴3a+1>4>0 4a>4>0
∴tana+tanb=-4a<0, tanatanb>0
∴tana,tanb同号且同为负，
∵a,b∈(-π/2,π/2)
∴a,b∈(-π/2,0)
又tan(a+b)>0
∴a+b∈ (-π,-π/2)，即a+b为第三象限的角，于是
sin(a+b)=-4/5,cos(a+b)=-3/5
tan[(a+b)/2]=[1-cos(a+b)]/sin(a+b)=-2