已知tanα,tanβ是方程x²-3x-3=0的两个根 求sin²（α+β）-3sin（α+β）-3

tana+tanb=3,tanatanb=-3,∴tan(a+b)=(tana+tanb)/(1-tanatanb)=3/4
又∵1+tan^2(a+b)=sec^2(a+b)=1/cos^2(a+b),∴cos^2(a+b)=16/25,
∴sin^2(a+b)=9/25,
又∵sin(2a+2b)=2sin(a+b)cos(a+b)=2cos^2(a+b)tan(a+b)=
2tan(a+b)/[1+tan^2(a+b)]=24/25.
∴3sin(a+b)cos(a+b)=3sin(2a+2b)/2[二倍角公式]=36/25
又∵tan(2a+2b)=2tan(a+b)/[1-tan^2(a+b)]=24/7.
÷得cos(2a+2b)=7/25,
∴原式=9/25-36/25+7/25=-20/25=-4/5