已知tanx,tany是方程x平方+Px+Q=0的两个根求sin（x+y）平方+

因为tanx、tany是方程x²＋Px＋Q=0的两根,则：
tanx＋tany=－P
tanxtany=Q
所以,tan(x＋y)=[tanx＋tany]/[1－tanxtany]=(－P)/(1－Q)
sin²(x＋y)＋Psin(x＋y)cos(x＋y)＋Qcos²(x＋y) 【利用分母1=sin²w＋cos²w】
=[sin²(x＋y)＋Psin(x＋y)cos(x＋y)＋cos²(x＋y)]/[sin²(x＋y)＋cos²(x＋y)] 【分子分母同
=[tan²(x＋y)＋Ptan(x＋y)＋Q]/[1＋tan²(x＋y)] 除以cos²(x＋y)】
=[P²－P²(1－Q)＋Q(1－Q)²]/[P²＋(1－Q)²]
=Q

看不懂

tanx和tany是方程的两个根，
则tanx+tany=-P，tanx*tany=Q
将PQ代入所求式子中：
sin(x+y)^2-P*sin(x+y)*cos(x+y)+Q*cos(x+y)^2
=sin(x+y)^2-(tanx+tany)*sin(x+y)*cos(x+y)+tanx*tany*cos(x+y)^2
=[sin(x+y)-tanx*c...

tanx+tany=-p
tanxtany=Q
sin²（x+y）+psin(x+y)cos(x+y)+Qcos²(x+y)
=【sin²（x+y）+psin(x+y)cos(x+y)+Qcos²(x+y)】/[sin²(x+y)+cos²(x+y)]
同除以cos²(x+y)