First, test the validity of the proposition, let A=B=C=60°, then sin(A/2)=sin(30°)=½, so:
sin2(A/2)+sin2(B/2)+sin2(C/2)=¾.
1-2sin(A/2)sin(B/2)sin(C/2)=1-2(½)3=1-2/8=¾.
To conserve text space and aid readability, let sA=sin(A/2) and cA=cos(A/2) and similarly for other angles.
So we need to prove: sA2+sB2+sC2=1-2sAsBsC.
In a triangle A+B+C=180 so C=180-(A+B), C/2=90-½(A+B) and:
sC=cos(90-C/2)=cos(½(A+B))=cA+B=cAcB-sAsB.
Therefore, sC2=cA+B2=(cAcB-sAsB)2=cA2cB2-2sAsBcAcB+sA2sB2.
sA2+sB2+sC2=
sA2+sB2+cA2cB2-2sAsBcAcB+sA2sB2; but cA2cB2=(1-sA2)(1-sB2)=1-sB2-sA2+sA2sB2.
sA2+sB2+sC2=sA2+sB2+1-sB2-sA2+sA2sB2-2sAsBcAcB+sA2sB2,
sA2+sB2+sC2=1+2sA2sB2-2sAsBcAcB; but sC=cAcB-sAsB, so cAcB=sC+sAsB.
sA2+sB2+sC2=1+2sA2sB2-2sAsB(sC+sAsB)=1+2sA2sB2-2sAsBsC-2sA2sB2=1-2sAsBsC.
Therefore, sA2+sB2+sC2=1-2sAsBsC.
Hence, sin2(A/2)+sin2(B/2)+sin2(C/2)=1-2sin(A/2)sin(B/2)sin(C/2).