已知tanα，tanβ是方程x2-3x-3=0的两根，试求sin²(α+β)-3sin(α+β)cos(α+β)-3cos²(α+β)

解：（一）易知由题设及韦达定理可得：tana+tanb=3,tana*tanb=-3.∴tan(a+b)=(tana+tanb)/(1-tana*tanb)=3/4.(二）∵sin²(a+b)+cos²(a+b)=1.∴原式=[sin²(a+b)-3sin(a+b)cos(a+b)-3cos²(a+b)]/[sin²(a+b)+cos²(a+b)]=[tan²(a+b)-3tan(a+b)-3]/[tan²(a+b)+1]=[(3/4)²-(9/4)-3]/[(3/4)²+1]=-3.

由根与系数的关系
tanα+tanβ=3
tanα*tanβ=-3
tan(α+β)=[tanα+tanβ]/(1-tanα*tanβ)=3/(1-(-3))=3/4
1+[tan(α+β)]^2=[sec(α+β)]^2=1/[cos(α+β)]^2
25/16=1/[cos(α+β)]^2
[cos(α+β)]^2=16/25

[sin(α+β)]^2-3sin(α+β)cos(α+β)-[3cos(α+β)]^2
={[sin(α+β)]^2-3sin(α+β)cos(α+β)-[3cos(α+β)]^2}/{[cos(α+β)]^2}*[cos(α+β)]^2
={[tan(α+β)]^2-3tan(α+β)-3}*[cos(α+β)]^2
=(9/16-9/4-3)*16/25
=-3