Prove that sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=1-2.sin(A/2)sin(B/2).sin(C/2)

Steps to prove sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=1-2.sin(A/2)sin(B/2).sin(C/2)

L.H.S

=sin^2(A/2)+sin^2(B/2)+sin^2(C/2)

=(1-cosA)/2+(1-cosB)/2+(1-cosC)/2

=(1-cosA+1-cosB+1-cosC)/2

={3-(cosA+cosB+cosC)}/2

=(1/2)[3-{2.cos(A+B)/2.cos(A-B)/2+1-2sin^2(c/2)}]

=(1/2)[3-2.cos(A+B)/2.cos(A-B)/2-1+2sin^2(c/2)}]

=(1/2)[2-2sin(C/2){cos(A-B)/2-cos(A+B)/2}]

=(1/2).2[1-sin(C/2){cos(A-B)/2-cos(A+B)/2}]

=1-sin(C/2){cos(A/2-B/2)-cos(A/2+B/2)}

=1-2.sin(A/2)sin(B/2).sin(C/2)

Hence sin^2(A/2)+sin^2(B/2)+sin^2(C/2)=1-2.sin(A/2)sin(B/2).sin(C/2) is proved.

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