Question 1014707
Prove first that

sin2A + sin2B + sin2C = 4sinA*sinB*sinC.  

sin2A + sin2B + sin2C
={{{sin2A + sin2B + sin(2*pi-2(A+B))}}}
={{{sin2A + sin2B - sin2(A+B)}}}
={{{sin2A + sin2B - sin(2A+2B)}}}
={{{sin2A + sin2B - sin2A*cos2B - cos2A*sin2B}}}
= sin2A(1 - cos2B) + sin2B(1 - cos2A)
={{{2sin2A*(sinB)^2 + 2sin2B*(sinA)^2}}}
={{{4sinA*cosB*(sinB)^2 + 4sinB*cosB*(sinA)^2}}}
=4sinAsinB(cosAsinB+sinAcosB)
=4sinAsinBsin(A+B)
={{{4sinAsinBsin(pi-C)}}}
=4sinAsinBsinC

Since {{{(4sinA*sinB*sinC)/(4cos(A/2)*cos(B/2)*cos(C/2))}}}
={{{(32sin(A/2)cos(A/2)sin(B/2)cos(B/2)sin(C/2)cos(C/2))/(4cos(A/2)*cos(B/2)*cos(C/2))}}},
the main result follows.