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,

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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,
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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,
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,

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

tanx+tany=3，tanxtany=-3
tan(x+y)=(tanx+tany)/(1-tanxtany)=3/[1-(-3)]=3/4
故
sin^2 (x+y)-3sin(x+y)cos(x+y)-3cos^2 (x+y)
=[1-cos(2x+2y)]/2-3/2*sin(2x+2y)-3[cos(2x+2y)+1]/2
=-3/2*sin(2...

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tanx+tany=3，tanxtany=-3
tan(x+y)=(tanx+tany)/(1-tanxtany)=3/[1-(-3)]=3/4
故
sin^2 (x+y)-3sin(x+y)cos(x+y)-3cos^2 (x+y)
=[1-cos(2x+2y)]/2-3/2*sin(2x+2y)-3[cos(2x+2y)+1]/2
=-3/2*sin(2x+2y)-2cos(2x+2y)-1
=-3/2*2tan(x+y)/{1+[tan(x+y)]^2}-2*{1-[tan(x+y)]^2}/{1+[tan(x+y)]^2}-1
=-3/2*3/4*1/{1+[3/4]^2}-2*{1-[3/4]^2}/{1+[3/4]^2}-1
=-179/100=-1.79

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由根与系数关系有 tanx+tany=3 tanxtany=-3
sin(x+y)=sinxcosy+cosxsiny=(tanx+tany)cosxcosy=3cosxcosy
sinxsiny=-3cosxcosy
cos(x+y)=cosxcosy-sinxsiny=4cosxcosy
tan(x+y)=3/4 [cos(x+y)]^2=1/[...

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由根与系数关系有 tanx+tany=3 tanxtany=-3
sin(x+y)=sinxcosy+cosxsiny=(tanx+tany)cosxcosy=3cosxcosy
sinxsiny=-3cosxcosy
cos(x+y)=cosxcosy-sinxsiny=4cosxcosy
tan(x+y)=3/4 [cos(x+y)]^2=1/[1+(tan(x+y)^2]=1/(1+9/16)=16/25
(cosxcosy)^2=[cos(x+y)]^2/16=1/25
[sin(x+y)]^2-3sin(x+y)cos(x+y)-3[cos(x+y)]^2
=(3cosxcosy)^2-3(3cosxcosy)(4cosxcosy)-3[4cosxcosy]^2
=[9-36-48][cosxcosy]^2
= -75/25
=-3

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