怀疑2asinA=(2a+c)sinB+(2c+b)sinC
应为2asinA=(2b+c)sinB+(2c+b)sinC,
若是,由正弦定理可得
2a^2=(2b+c)b+(2c+b)c,
整理得a^2=b^2+c^2+bc,
所以cosA=(b^2+c^2-a^2)/2bc=-1/2,
所以A=2π/3.B+C=2π/3
sinB+sinC =2sin(B+C)/2.cos(B-C)/2
=2sinπ/3.cos(B-C)/2
=√3cos(B-C)/2≤√3.