# Examples on Semiperimeter and Half Angle Formulae

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Example - 50

Evaluate  \begin{align}S = (b - c)\cot \frac{A}{2} + (c - a)\cot \frac{B}{2} + (a - b)\cot \frac{C}{2}\end{align}  .

Solution: By the half-angle formula for  \begin{align}tan \frac{A}{2},\end{align}  we have

\begin{align}&\cot \frac{A}{2} = \sqrt {\frac{{s(s - a)}}{{(s - b)(s - c)}}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;\;= \frac{{s(s - a)}}{{\sqrt {s(s - a)(s - b)(s - c)} }} = \frac{{s(s - a)}}{\Delta } \\ \Rightarrow\quad &\;\;\;\;\;\;S = \frac{{s(s - a)(b - c) + s(s - b)(c - a) + s(s - c)(a - b)}}{\Delta } \\ \,\,\,\,\, &\qquad\;\;= \frac{{s\left( {({b^2} - {c^2}) + ({c^2} - {a^2}) + ({a^2} - {b^2})} \right)}}{\Delta } \\ \,\,\,\,\, &\qquad\;\;= 0 \\ \end{align}

Example - 51

Prove that \begin{align}\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \left( {\frac{{a + b + c}}{{b + c - a}}} \right)\cot \frac{A}{2}\end{align}

Solution:

\begin{align}& \text{LHS}=\frac{s(s-a)}{\Delta }+\frac{s(s-b)}{\Delta }+\frac{s(s-c)}{\Delta }\left\{ \begin{gathered}& \text{Using}\ \text{the}\ \text{expression}\ \text{for}\cot \frac{A}{2}\ \\ & \text{from}\ \text{the}\ \text{previous}\ \text{example} \\ \end{gathered} \right\} \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{s(3s-(a+b+c))}{\Delta } \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{{{s}^{2}}}{\Delta } \\ & \text{RHS}=\left( \frac{a+b+c}{b+c-a} \right)\cot \frac{A}{2} \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{2s}{2s-2a}\cdot \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} \\ & \,\,\,\,\,\,\,\,\,\,\,=\frac{{{s}^{2}}}{\sqrt{s(s-a)(s-b)(s-c)}} \\ & \,\,\,\,\,\,\,\,\,\,\,=\frac{{{s}^{2}}}{\Delta } \\ \end{align}

Example - 52

Consider the following statements concerning $$\Delta ABC$$  :

I. The sides a, b, c and $$\Delta$$ are rational

II. $$a,\;\tan \frac{B}{2},\;\tan \frac{C}{2}$$ are rational

III. $$a,\;\sin A,\;\sin B,\;\sin C$$ are rational

Show that I $$\Rightarrow$$  II $$\Rightarrow$$ III  $$\Rightarrow$$   I

Solution: Assuming I is true, then  \begin{align}s = \frac{{a + b + c}}{2}\end{align}  is rational. Since  $$\Delta$$ is also rational.

$\tan \frac{B}{2} = \frac{\Delta }{{s(s - b)}}\;\;{\text{and}}\;\;\tan \frac{C}{2} = \frac{\Delta }{{s(s - c)}}$

are also rational. Thus, I $$\Rightarrow$$  II.

Now, assuming II, we see that

$\sin B = \frac{{2\tan \frac{B}{2}}}{{1 + {{\tan }^2}\frac{B}{2}}}\;\;{\text{and}}\;\,\sin C = \frac{{2\tan \frac{C}{2}}}{{1 + {{\tan }^2}\frac{C}{2}}}$

are rational. Also,

$\tan \frac{A}{2} = \cot \left( {\frac{{B + C}}{2}} \right) = \frac{{1 - \tan \frac{B}{2}\tan \frac{C}{2}}}{{\tan \frac{B}{2} + \tan \frac{C}{2}}}$

becomes rational, which implies that sin A is rational. Thus, II  $$\Rightarrow$$  III

Finally, assuming III, we have

$b = a\frac{{\sin B}}{{\sin A}}\;\;{\text{and}}\;\;c = a\frac{{\sin C}}{{\sin A}}$

as rational, and further $$\Delta = \frac{1}{2}bc\;\sin A$$ , as rational. Thus, III  $$\Rightarrow$$ I.

Example - 53

Prove that  $${s^2} > 4\Delta$$

Solution: Applying the A.M – G. M. inequality on $$s,\;s - a,\;s - b\;{\text{and}}\;s - c,$$

we have

\begin{align}&\frac{{s + (s - a) + (s - b) + (s - c)}}{4} > {\left( {s(s - a)(s - b)(s - c)} \right)^{1/4}}\\ \Rightarrow\quad &\frac{s}{2} > \sqrt \Delta \quad\Rightarrow\quad {s^2} > 4\Delta \\ \end{align}

Note that equality can never hold since s cannot equal the other terms.

Example - 54

Prove that  \begin{align}(abcs)\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = {\Delta ^2}\end{align}

Solution: Using the half-angle sine formulae should yield the desired result:

\begin{align}&{\text{LHS}} = abcs\sqrt {\frac{{(s - b)(s - c)}}{{bc}}} \sqrt {\frac{{(s - c)(s - a)}}{{ca}}} \sqrt {\frac{{(s - a)(s - b)}}{{ab}}} \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \frac{{abc\;s(s - a)(s - b)(s - c)}}{{abc}} \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= {\Delta ^2} \\ \end{align}