2AB*AC=2cbcosA=a²-(b+c)²
由余弦定理得a²=b²+c²-2bccosA
于是2bccosA=b²+c²-2bccosA-(b+c)²
得cosA=-1/2,A=2π/3
B+C=π/3,得B=π/3-C
2√3cos²(C/2)-sin(4π/3-B)
=√3(1+cosC)-sin(π+C)
=√3+√3cosC+sinC
=√3+2sin(C+π/3)
当C=π/6时,原式有最大值2+√3
此时B=C=π/6
2AB*AC=2cbcosA=a²-(b+c)²
由余弦定理得a²=b²+c²-2bccosA
于是2bccosA=b²+c²-2bccosA-(b+c)²
得cosA=-1/2,A=2π/3
B+C=π/3,得B=π/3-C
2√3cos²(C/2)-sin(4π/3-B)
=√3(1+cosC)-sin(π+C)
=√3+√3cosC+sinC
=√3+2sin(C+π/3)
当C=π/6时,原式有最大值2+√3
此时B=C=π/6