Semiperimeter and Half Angle Formulae

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Property - 3: Semi-perimeter and half-angle formulae

For a  \(\Delta ABC\) , with sides a, b, c, its semi perimeter is the quantity

\[\boxed{\;s = \frac{{a + b + c}}{2}\;}\]

One of the most important sets of properties followed by triangles is the set of half-angle formulae tabulated below.

\({\sin \;\;{\text{of}}\;\;{\text{half - angles}}}\)
\[{\sin \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{bc}}} }\] \[{\sin \frac{B}{2} = \sqrt {\frac{{(s - c)(s - a)}}{{ca}}} }\] \[{\sin \frac{A}{2} = \sqrt {\frac{{(s - a)(s - b)}}{{ab}}\;} }\]
\({\cos\;\; {\text{of}}\;\;{\text{half - angles}}}\)
\[{\cos \frac{A}{2} = \sqrt {\frac{{s(s - a)}}{{bc}}} }\] \[{\cos \frac{B}{2} = \sqrt {\frac{{s(s - b)}}{{ac}}} }\] \[{\cos \frac{C}{2} = \sqrt {\frac{{s(s - c)}}{{ab}}} }\]
\({\tan \;\;{\text{of}}\;\;{\text{half - angles}}}\)
\[{\tan \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{s(s - a)}}} }\] \[{\tan \frac{B}{2} = \sqrt {\frac{{(s - c)(s - a)}}{{s(s - b)}}} }\] \[{\tan \frac{C}{2} = \sqrt {\frac{{(s - a)(s - b)}}{{s(s - c)}}\;} }\]

These are widely used expressions and should be memorized. Let us justify these by proving the relations for  \(\begin{align}\sin \frac{A}{2}\;\;\text{ }\!\!~\!\!\text{ }and\text{ }\!\!~\!\!\text{  }\;\;\cos \frac{A}{2}\end{align}\) , from which the other relations automatically follow.

\[\begin{align}&{\sin ^2}\frac{A}{2} = \frac{{1 - \cos A}}{2} = \frac{1}{2}\left( {1 - \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right) \\  \,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\;\;= \frac{1}{2}\left( {\frac{{{a^2} - {{(b - c)}^2}}}{{2bc}}} \right)  \\ \,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\;\;= \frac{1}{2}\frac{{(a + b - c)(a + c - b)}}{{2bc}} \\ \,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\;\;= \frac{1}{2}\frac{{\left\{ {(a + b + c) - 2c} \right\}\left\{ {(a + b + c) - 2b} \right\}}}{{2bc}}  \\ \,\,\,\,\,\,\,&\qquad\qquad\qquad\qquad\;\;= \frac{1}{2}\frac{{(2s - 2c)(2s - 2b)}}{{2bc}}  \\ \,\,\,\,\,\,\,&\qquad\qquad\qquad\qquad\;\; = \frac{{(s - b)(s - c)}}{{bc}}  \\  \;\;\;\;\;\;\;\;\; &\qquad\quad\Rightarrow \;\,\,\,\,\,\,\,\sin \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{bc}}}  \\ \end{align} \]

\[\begin{align}&{\cos ^2}\frac{A}{2} = \frac{{1 + \cos A}}{2} = \frac{{{{(b + c)}^2} - {a^2}}}{{4bc}} \\   \,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\;\;\;= \frac{{(b + c + a)(b + c - a)}}{{4bc}}\\   \,\,\,\,\,\,\,\,  &\qquad\qquad\qquad\qquad\;\;\;= \frac{{s(s - a)}}{{bc}} \\ &\qquad\quad\;  \Rightarrow \,\,\,\,\cos \frac{A}{2}\,\,\, = \sqrt {\frac{{s(s - a)}}{{bc}}}  \\ \end{align} \]

\(\begin{align}\tan \frac{A}{2}\end{align}\) is simply  \(\begin{align}\frac{{\sin \frac{A}{2}}}{{\cos \frac{A}{2}}}\end{align}\) . The other relations follow similarly. A very important corollary of the half-angle formulae is the Heron’s formula for the area of a triangle:

\[\begin{align}&\Delta  = \frac{1}{2}bc\;\sin A \\ \,\,\,\,\, &\quad= bc\sin \frac{A}{2}\cos \frac{A}{2} \\ \,\,\,\,\, &\quad= bc \cdot \sqrt {\frac{{(s - b)(s - c)}}{{bc}}}  \cdot \sqrt {\frac{{s(s - a)}}{{bc}}}   \\   \Rightarrow \;\;\;\;\;\;&\boxed{\;\Delta  = \sqrt {s(s - a)(s - b)(s - c)} }\; \qquad\qquad {\text{Heron's}}\;{\text{formula}}{\text{.}} \\ \end{align} \]

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