你好!
(sinA+sinB)(cosA+cosB)=2sinC
sinAcosA+sinAcosB+sinBcosA+sinBcosB=2sinC
1/2*sin2A+sin(A+B)+1/2*sin2B=2sinC
又sin(A+B) = sin(π-C) = sinC
所以sin2A+sin2B=2sin(A+B)
由和差化积公式得
2sin(A+B)cos(A-B) = 2sin(A+B)
即 cos(A-B) = 1
∴A-B=0 即 A=B
故三角形ABC为等腰三角形