Question

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

Answer 1

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

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