sinA-sin[(B-C)]=sinA+sin[(C-B)]=2sin[(A+C-B)/2]cos[A+B-C)/2] =2sin[(π-2B)/2]cos[(π-2C)/2] =2sin(π/2-B)cos(π/2-C) =2cosBsinC 所以tanB=cos(C-B)/[sinA+sin(C-B)]=cos(C-B)/2cosBsinC 即2sinBsinC=cosBcos...