在三角形ABC中，求证：sinA+sinB+sinC=4cos(?
在三角形AB中，求证：sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2).
sinA+sinB+sinC =2sin(A/2)cos(A/2) + 2sin[(B+C)/2]cos[(B-C)/2] =2sin[(π-B-C)/2]cos(A/2) + 2sin[(π-A)/2]cos[(B-C)/2] =2cos[(B+C)/2]cos(A/2) + 2cos(A/2)cos[(B-C)/2] =2cos(A/2){cos[(B+C)/2]+cos[(B-C)/2]} =4cos(A/2)cos(B/2)cos(C/2)