Using Law of cosine

\cos C = \dfrac{a^2+b^2-c^2}{2 a b}

and double angle angle identity

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

Then we have the equality

1-2 \sin^2 \dfrac{C}{2}= \dfrac{a^2+b^2-c^2}{2 a b}

\sin^2 \dfrac{C}{2} = \dfrac{1}{2} \Big( 1- \dfrac{a^2+b^2-c^2}{2 a b}\Big)

=\dfrac{2ab-a^2-b^2+c^2}{4ab}

=\dfrac{(c-a+b)(c+a-b)}{4ab}

=\dfrac{(s-a)(s-b)}{ab}

Therefore,

\sin\dfrac{C}{2} = \sqrt{\dfrac{(s-a)(s-b)}{ab} }

Using Law of cosine

\cos C = \dfrac{a^2+b^2-c^2}{2 a b}

and double angle angle identity

\cos C =2 \cos^2 \dfrac{C}{2} - 1

We get the equality

2 \cos^2 \dfrac{C}{2} - 1 = \dfrac{a^2+b^2-c^2}{2 a b}

Then,

\cos^2 \dfrac{C}{2} = \dfrac{1}{2} \Big( \dfrac{a^2+b^2-c^2}{2 a b}+1\Big)

=\dfrac{a^2+b^2-c^2+2ab}{4ab}

=\dfrac{(a+b-c)(a+b+c)}{4ab}

=\dfrac{(s-c)s}{ab}

Therefore,

\cos\dfrac{C}{2} = \sqrt{\dfrac{s(s-c)}{ab} }