在△ABC中，已知（a2+b2）sin（A-B)=（a2-b2）sin（A+B），△ABC为什么三角形？ 三角形ABC中，(a2+b2)sin(A-B)=(a2-b2...
\u5728\u25b3ABC\u4e2d\uff0c\u5df2\u77e5\uff08a2+b2\uff09sin\uff08A-B\uff09=\uff08a2-b2\uff09sin\uff08A+B\uff09\uff0c\u5224\u65ad\u25b3ABC\u7684\u5f62\u72b6\u5728\u25b3ABC\u4e2d\uff0c\u7531\u6b63\u5f26\u5b9a\u7406\u53ef\u77e5asinA=bsinB=k\uff0c\u5219a=ksinA\uff0cb=ksinB\uff0c\u4ee3\u5165\uff08a2+b2\uff09sin\uff08A-B\uff09=\uff08a2-b2\uff09sin\uff08A+B\uff09\uff0c\u5e76\u628ak\u7ea6\u5206\u53ef\u5f97\uff08sin2A+sin2B\uff09sin\uff08A-B\uff09=\uff08sin2A-sin2B\uff09sin\uff08A+B\uff09\uff0csin2Asin\uff08A-B\uff09+sin2Bsin\uff08A-B\uff09=sin2Asin\uff08A+B\uff09-sin2Bsin\uff08A+B\uff09\uff0csin2A[sin\uff08A+B\uff09-sin\uff08A-B\uff09]=sin2B[sin\uff08A-B\uff09+sin\uff08A+B\uff09]\uff0c\u5229\u7528\u548c\u89d2\u516c\u5f0f\uff0c\u6574\u7406\u6709 sin2A2cosAsinB=sin2B?2sinAcosB\uff0c\u5373sin2A2cosAsinB-sin2B2sinAcosB=0\uff0c\u5373 sinAsinB\uff082sinAcosA-2sinBcosB\uff09=0\uff0c\u5373 sinAsinB\uff08sin2A-sin2B\uff09=0\uff0e\u53c8 sinA\uff1e0\uff0csinB\uff1e0\uff0c\u6240\u4ee5sin2A=sin2B\uff0c2A=2B \u62162A+2B=180\u5ea6\uff0c\u6545 A=B\u6216A+B=90\u5ea6\uff0c\u6240\u4ee5\uff0c\u25b3ABC\u662f\u7b49\u8170\u4e09\u89d2\u5f62\u6216\u76f4\u89d2\u4e09\u89d2\u5f62\uff0e
\uff08a²+b²)sin(A-B)=(a²-b²)sin(A+B),
(sin²A+sin²B)sin(A-B)=(sin²A-sin²B)sin(A+B)
sin²A\u00d7[sin(A+B)-sin(A-B)]=sin²B\u00d7[sin(A-B)+sin(A+B)]
sin²A\u00d72cosAsinB=sin²B\u00d72sinAcosB
sin²A\u00d72cosAsinB-sin²B\u00d72sinAcosB=0
sinAsinB(sin2A-sin2B)=0
sin2A=sin2B
2A=2B \u62162A+2B=180\u00b0
A=B\u6216A+B=90\u00b0
\u6545\u25b3ABC\u662f\u7b49\u8170\u4e09\u89d2\u5f62\u6216\u76f4\u89d2\u4e09\u89d2\u5f62
a²sinAcosB-a²cosAsinB+b²sinAcosB-b²cosAsinB=a²sinAcosB+a²cosAsinB-b²sinAcosB-b²cosAsinB
a²cosAsinB=b²sinAcosB
∴a²cosA/sinA=b²cosB/sinB
∴ a²sinAcosA/sin²A=b²sinBcosB/sin²B
由正弦定理可以知道a/sinA=b/sinB ∴a²/sin²A=b²/sin²B
∴ sinAcosA=sinBcosB ∴ 2sinAcosA=2sinBcosB
∴ sin2A=sin2B
∴ 2A=2B 或者 2A=180°-2B
∴ A=B或者A+B=90°
∴ △ABC是等腰三角形或者直角三角形
解：由正弦定理，
a=2RsinA
b=2RsinB
代入上式得：
（sin²A+sin²B）sin（A-B)=（sin²A-sin²B）sin（A+B）
‘整理得：
sin²A[sin（A+B）-sin（A-B)]=sin²B[sin（A+B）+sin（A-B)]
故：
sin²AsinBcosA=sin²BsinAcosB
sinAsinB(sinAcosA-sinBcosB)=0
即sinAsinB(sin2A-sin2B)=0
由sinAsinB≠0
得sin2A=sin2B
故A=B或2A+2B=π
所以，△ABC是一个等腰三角形，或是一个直角三角形
展开化简会得到sinAcosA=sinBcosB，所以A=B
即为等腰三角形
^表示平方
(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),
利用正弦定理得
(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)
sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B))
sin^A*2cosAsinB=sin^B*2sinAcosB
sin^A*2cosAsinB-sin^B*2sinAcosB=0
sinAsinB(sin2A-sin2B)=0
sin2A=sin2B
2A=2B 或2A+2B=180度
A=B或A+B=90度
故△ABC是等腰三角形或直角三角形
绛旓細瑙ｏ細(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B)(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B))sin^A*2cosAsinB=sin^B*2sinAcosB sin^A*2cosAsinB-sin^B*2sinAcosB=0 sinAsinB(sin2A-sin2B)=0 sin2A...
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