Question

在三角形ABC中.求证:sinA+sinB+sinC=4*cosA/2*cosB/2*cosC/2

Answer 1

根据和差化积

sinA+sinB+sinC

=2sin[(A+B)/2]cos[(A-B)/2]+sinC

=2cos(C/2)cos[(A-B)/2]+2sin(C/2)cos(C/2)

=2cos(C/2){cos[(A-B)/2]+cos[(A+B)/2]}

=2cos(C/2)[2cos(A/2)cos(B/2)]

=4（cosA/2）（cosB/2)(cosC/2)