已知tanx,tany是方程x平方+Px+Q=0的两个根求sin（x+y）平方+sinP（x+y）cos（x+y）+Qcos（x+Y）平方的值

已知tanx,tany是方程x平方+Px+Q=0的两个根求sin（x+y）平方+sinP（x+y）cos（x+y）+Qcos（x+Y）平方的值
已知tanx,tany是方程x平方+Px+Q=0的两个根求sin（x+y）平方+
sinP（x+y）cos（x+y）+Qcos（x+Y）平方的值

已知tanx,tany是方程x平方+Px+Q=0的两个根求sin（x+y）平方+sinP（x+y）cos（x+y）+Qcos（x+Y）平方的值
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

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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...

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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*cos(x+y)]*[sin(x+y)-tany*cos(x+y)]
=[sinx*cosy+cosx*siny-(sinx/cosx)*(cosx*cosy-sinx*siny)]*[sinx*cosy+cosx*siny-(siny/cosy)*(cosx*cosy-sinx*siny)]
=(siny/cosx)*(sinx/cosy)=tanx*tany=Q
故所求值为Q

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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)

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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)
=【tan²(x+y)+Ptan(x+y)+Q】/[tan²(x+y)+1]
tan²(x+y)+Ptan(x+y)+Q
=tan²(x+y)+tan(x+y)(-tanx-tany)+tanxtany
=tan²(x+y)-tan(x+y)[tan(x+y)(1-tanxtany)+tanxtany
=tan²(x+y)tanxtany+tanxtany
所以，
原式=【tan²(x+y)tanxtany+tanxtany】/[tan²(x+y)+1]
=tanxtany

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