1)sinA+sinB+sinC=4*cos A/2*cosB/2*cosC/2Ta có : sinA+sinB+sinC=2 sin[(A+B)/2].cos[(A-B)/2] + sinC=2 cosC/2cos[(A-B)/2] + 2cos[C/2]*sinC/2= 2 cosC/2 [cos[(A-B)/2]+ sinC/2]= 2 cosC/2[cos[(A-B)/2]+ cos(A+B)/2]= 4 cosA/2 * cosB/2 *cos C/2
1)sinA+sinB+sinC=4*cos A/2*cosB/2*cosC/2Ta có : sinA+sinB+sinC=2 sin[(A+B)/2].cos[(A-B)/2] + sinC=2 cosC/2cos[(A-B)/2] + 2cos[C/2]*sinC/2= 2 cosC/2 [cos[(A-B)/2]+ sinC/2]= 2 cosC/2[cos[(A-B)/2]+ cos(A+B)/2]= 4 cosA/2 * cosB/2 *cos C/2câu còn lại làm tương tự