Question

已知tanα，tanβ是方程x²+3x-5=0的两个根，

求sin²（α+β）+2sin（α+β）*cos（α+β）的值

Answer 1

tana，tanb是方程x²+3x-5=0的两个根
tana+tanb=-3 ;tana*tanb=-5
tan(a+b)=(tana+tanb)/(1-tana*tanb)
=-3/(1+5)
=-1/2
sin²(a+b)+2sin(a+b)cos(a+b) (同除以sin²a+cos²b)
=[sin²(a+b)+2sin(a+b)cos(a+b)]/[sin²(a+b)+cos²(a+b)] (分子分母同除以cos²(a+b) )
=[tan²(a+b)+2tan(a+b)]/[tan²(a+b)+1]
=[(-1/2)²+2*(-1/2)]/[(-1/2)²+1]
=(-3/4)/(5/4)
=-3/5