设三角形ABC的内角A,B,C的对边分别为a,b,c已知b2+c2=a2+根号3bc求2sinBcosC-sin(B-C)的值

由已知,余弦定理有：cosA=(b^2+c^2-a^2)/2bc=根号3bc/2bc=根号3/2
A=30度,
2sinBcosC-sin(B-C)
=2sinBcosC-sinBcosC+cosBsinC
=sinBcosC+cosBsinC
=sin(B+C)
=sin(180-A)
=sinA=1/2