Since the sum of three internal angles of a triangle is equal to \pi

A+B+C=\pi

\cos C=-\cos(A+B)

\dfrac{C}{2} =\dfrac{\pi}{2} -\dfrac{A+B}{2}

\sin\dfrac{C}{2}=\cos\dfrac{A+B}{2}

Using the double angle identity

\sin^2\dfrac{A}{2} +\sin^2\dfrac{B}{2} +\sin^2\dfrac{C}{2}

=\dfrac{1-\cos A}{2} +\dfrac{1-\cos B}{2} +\sin^2\dfrac{C}{2}

=1-\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2} +\sin^2\dfrac{C}{2}

=1-\sin\dfrac{C}{2} \cos \dfrac{A-B}{2} +\sin^2\dfrac{C}{2}

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

1-2\sin\dfrac{A}{2} \sin\dfrac{B}{2}\sin\dfrac{C}{2}