作业帮已知TanX,TanY是方程X^-3X-3=0的俩根,求sin^(x+y)-3sin(x+y)cos(x+y)-3cos
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已知TanX,TanY是方程X^-3X-3=0的俩根,求sin^(x+y)-3sin(x+y)cos(x+y)-3cos^(x+y)的值,
tanx+tany=3
(tanx)(tany)=-3
tan(x+y)=(tanx+tany)/(1-tanxtany)=3/4
[sin(x+y)]^2+[cos(x+y)]^2=1
[sin(x+y)]^2=9/25,[cos(x+y)]^2=16/25,sin(x+y)cos(x+y)={[cos(x+y)]^2}tan(x+y)=12/25
[sin(x+y)]^2-3sin(x+y)cos(x+y)-3[cos(x+y)]^2=-3
希望采纳
再问： 求公式，谢谢
(tanx)(tany)=-3
tan(x+y)=(tanx+tany)/(1-tanxtany)=3/4
[sin(x+y)]^2+[cos(x+y)]^2=1
[sin(x+y)]^2=9/25,[cos(x+y)]^2=16/25,sin(x+y)cos(x+y)={[cos(x+y)]^2}tan(x+y)=12/25
[sin(x+y)]^2-3sin(x+y)cos(x+y)-3[cos(x+y)]^2=-3
希望采纳
再问： 求公式，谢谢
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