Question

tanα tanβ 是方程 x2-3x-3=0的两根 求sin(α+β)2-3sin(α+β)cos(α+β)-3cos(

tanα tanβ 是方程 x2-3x-3=0的两根 求sin(α+β)2-3sin(α+β)cos(α+β)-3cos(α+β）2

Answer 1

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