已知tanx,tany是方程x^2+6x+7=0的两个根,求证sin(x+y)=cos(x+y).

tanx+tany=-6;
tanx*tany=7;
将sin(x+y)=cos(x+y)展开：
即证
sinxcosy+cosxsiny=cosxcosy-sinxsiny;
(sinx-cosx)cosy+(cosx+sinx)siny=0;
cosx(tanx-1)cosy+cosx(1+tanx)tanycosy=0;
消去cosx、cosy：
tanx-1+tany+tanx*tany=0;
得证。

因为tanx,tany是方程x^2+6x+7=0的两个根，所以
tanx+tany=-6，tanx*tany=7，
即（sinx*cosy+cosx*siny)/(cosx*cosy)=-6,
(sinx*siny)/(cosx*cosy)=7.
所以sinx*siny=7cosx*cosy，sinx*cosy+cosx*siny=-6cosx*cosy，
sin(x+y)=sinx*cosy+cosx*siny=-6cosx*cosy，
cos(x+y)=cosx*cosy-sinx*siny=-6cosx*cosy,
所以sin(x+y)=cos(x+y).

tanx+tany=-6,tanx*tany=7
tan(x+y)=(tanx+tany)/(1-tanxtany)
=(-6)/(1-7)=1
tan(x+y)=sin(x+y)/cos(x+y)=1
sin(x+y)=cos(x+y).