By the half angle formula and Law of Cosines, we have \begin{align*} \sin^2\frac{A}{2}&=\frac{1-\cos A}{2}\\ &=\frac{1}{2}\big(1- \frac{b^2+c^2-a^2}{2bc}\big)\\ &=\frac{1}{2}\cdot\frac{1}{bc}\cdot\big(2bc-(b^2+c^2-a^2)\big)\\ &=\frac{1}{4bc}\cdot\big(a^2 -(b^2+c^2-2bc)\big)\\ &=\frac{1}{4bc}\cdot\big(a^2 -(b-c)^2\big)\\ &=\frac{1}{4bc}\cdot(a+b-c)(a-b+c)\\ &=\frac{1}{4bc}\cdot((a+b+c)-2c)((a+b+c)-2b)\\ &=\frac{1}{4bc}\cdot(2p-2c)(2p-2b)\\ &=\frac{(p-b)(p-c)}{bc} \end{align*} $$\therefore\quad \sin\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}}$$