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\[
\sin \frac{A}{2} = \sqrt {\frac{{\left( {p - b} \right)\left( {p - c} \right)}}{{bc}}}
\]
\[
\sin \frac{B}{2} = \sqrt {\frac{{\left( {p - c} \right)\left( {p - a} \right)}}{{ca}}}
\]
\[
\sin \frac{C}{2} = \sqrt {\frac{{\left( {p - a} \right)\left( {p - b} \right)}}{{ab}}}
\]
\[
\cos \frac{A}{2} = \sqrt {\frac{{p\left( {p - a} \right)}}{{bc}}} ,\cos \frac{B}{2} = \sqrt {\frac{{p\left( {p - b} \right)}}{{ca}}} ,\cos \frac{C}{2} = \sqrt {\frac{{p\left( {p - c} \right)}}{{ab}}}
\]
\[
\tan \frac{A}{2} = \frac{r}{{p - a}} = \sqrt {\frac{{\left( {p - b} \right)\left( {p - c} \right)}}{{p\left( {p - a} \right)}}}
\]
\[
\tan \frac{B}{2} = \frac{r}{{p - b}} = \sqrt {\frac{{\left( {p - c} \right)\left( {p - a} \right)}}{{p\left( {p - b} \right)}}}
\]
\[
\tan \frac{C}{2} = \frac{r}{{p - c}} = \sqrt {\frac{{\left( {p - a} \right)\left( {p - b} \right)}}{{p\left( {p - c} \right)}}}
\]
式中\[
p = \frac{1}{2}\left( {a + b + c} \right)
\]
r为△ABC的内切圆半径，且
\[
r = \sqrt {\frac{{\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)}}{p}} = \frac{S}{p} = p\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}
\]
式中S为△ABC的面积．
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