Prove the formula :
$r_{1}$ = 4R sin (A/2) cos (B/2) cos (C/2) and similar expression for $r_{2}$ and $r_{3}$
$r_{2}$ = 4R cos (A/2) sin (B/2) cos (C/2)
$r_{3}$ = 4R cos (A/2) cos (B/2) sin (C/2)
where $r_{1}, r_{2}, r_{3}$ have their usual meanings.

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Solution

(i) $4 R sin (A / 2) cos (B / 2) cos (C / 2)$
$= 4 R \sqrt{\frac{(s - b) (s - c)}{b c}} \sqrt{\frac{s (s - b)}{a c}} \sqrt{\frac{s (s - c)}{a b}}$
$= \frac{4 R}{a b c} \sqrt{s^{2} (s - b)^{2} (s - c)^{2}}$
$= \frac{4 R}{a b c} S (s - b) (s - c)$
$= \frac{1}{S} \times \frac{S^{2}}{(s - a)} = \frac{S}{s - a} = r_{1}$
(ii) $4 R cos A / 2 sin B / 2 cos C / 2$
$= 4 R \sqrt{\frac{s (s - a)}{b c}} \sqrt{\frac{(s - a) (s - c)}{a c}} \sqrt{\frac{s (s - c)}{a b}}$
$= \frac{4 R}{a b c} \sqrt{s^{2} (s - a)^{2} (s - c)^{2}}$
$= \frac{1}{S} \frac{S^{2}}{(s - b)} = \frac{S}{s - b} = r_{2}$
(iii) $4 R cos A / 2 cos B / 2 sin C / 2$
$= 4 R \sqrt{\frac{s (s - a)}{b c}} \sqrt{\frac{s (s - b)}{a c}} \sqrt{\frac{(s - a) (s - b)}{a b}}$
$\Rightarrow \frac{4 R}{a b c} \sqrt{s^{2} (s - a)^{2} (s - b)^{2}}$
$\frac{1}{S} \frac{S^{2}}{s - c} = \frac{S}{s - c} = r_{3}$ proved.

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