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

由根与系数的关系tanα+tanβ=3tanα*tanβ=-3tan(α+β)=[tanα+tanβ]/(1-tanα*tanβ)=3/(1-(-3))=3/41+[tan(α+β)]^2=[sec(α+β)]^2=1/[cos(α+β)]^225/16=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