How to prove $\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}};\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}$ I have read these formulae in my book but i could not understand how these are proved.
$\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}};\sin\frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{ca}};\sin\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}$ and 
$\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}};\cos\frac{B}{2}=\sqrt{\frac{s(s-b)}{ca}};\cos\frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}}$
where $A,B,C$ are the angles of a triangle $ABC$ and $a,b,c$ are the sides opposite to the angles $A,B,C$ respectively.

I know that area of triangle$\Delta=\frac{1}{2}bc\sin A$
But area of triangle by Heron's formula is $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$\sin A=\frac{2\Delta}{bc}$
$2\sin\frac{A}{2}\cos\frac{A}{2}=\frac{2\Delta}{bc}$
$\sin\frac{A}{2}\cos\frac{A}{2}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{bc}$
I do not know how to prove it further or there is some other method to prove it.

\begin{aligned}
      \cos A &= \frac{b^2+c^2-a^2}{2bc} \\
      \cos A &= 2\cos^2\frac{A}{2} - 1 \\
      2\cos^2\frac{A}{2} - 1 &= \frac{b^2+c^2-a^2}{2bc}\\
      \cos^2 \frac{A}{2} &= \frac{b^2+c^2-a^2+2bc}{4bc} = \frac{(b+c-a)(b+c+a)}{4bc}\\
      \cos \frac{A}{2} &= \sqrt{\frac{(2s-2a)(2s)}{4bc}} = \sqrt{\frac{s(s-a)}{bc}}
\end{aligned}
$\sin \frac{A}{2}$ can be solved in the same way
