# Examples on Sine Rule Set 1

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Example - 33

Evaluate the expression  $$S = {a^3}\sin (B - C) + {b^3}\sin (C - A) + {c^3}\sin (A - B)$$  .

Solution: Using the sine rule,

\begin{align}&\qquad\;\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = \lambda \;({\text{say}}) \\\\& \Rightarrow\quad a = \lambda \sin A,\;b = \lambda \sin B,\;c = \lambda \sin C \end{align}

Substituting this into the expression for S gives

$S = {\lambda ^3}\left\{ {{{\sin }^3}A\sin (B - C) + {{\sin }^3}B\sin (C - A) + {{\sin }^3}C\sin (A - B)} \right\}$

Let us focus on one term in S :

\begin{align}&{\sin ^3}A\sin (B - C) = {\sin ^2}A \cdot \sin A \cdot \sin (B - C) \\ \,\,\,\, &\qquad\qquad\qquad\quad\;= {\sin ^2}A\sin (B + C)\sin (B - C) & \qquad \qquad \quad({\text{Why}}\,{\text{?}}) \\ \,\,\,\, &\qquad\qquad\qquad\quad\;= {\sin ^2}A({\sin ^2}B - {\sin ^2}C) \\ \,\,\,\, &\qquad\qquad\qquad\quad\;= {\sin ^2}A{\sin ^2}B - {\sin ^2}C{\sin ^2}A\\ \end{align}

Clearly, from the other two terms in S, we’ll have similar terms generated, so that

$S = 0$

Example - 34

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

Solution:

\begin{align}&{\text{RHS}} = \left( {\frac{{\sin B - \sin \,C}}{{\sin B + \sin C}}} \right)\cot \frac{A}{2} \qquad\qquad({\text{Using}}\;{\text{the}}\;{\text{sine}}\;{\text{rule}}) \\ &\qquad= \left( {\frac{{\sin \left( {\frac{{B - C}}{2}} \right)\cos \left( {\frac{{B + C}}{2}} \right)}}{{\sin \left( {\frac{{B + C}}{2}} \right)\cos \left( {\frac{{B - C}}{2}} \right)}}} \right)\cot \frac{A}{2} \\&\qquad= \tan \left( {\frac{{B - C}}{2}} \right)\cot \left( {\frac{{B + C}}{2}} \right)\cot \frac{A}{2} \\ \end{align}

Since  \begin{align}\cot \left( {\frac{{B + C}}{2}} \right) = \cot \left( {\frac{\pi }{2} - \frac{A}{2}} \right) = \tan \frac{A}{2}\end{align}  , we are left with  \begin{align}\tan \left( {\frac{{B - C}}{2}} \right)\end{align}, proving the result.

This expression is Napier’s law of tangents, and it will be useful to commit it to memory.

Example - 35

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: Let us focus on one term in S:

\begin{align}&(b - a)\cot \frac{A}{2} = (b + c)\tan \left( {\frac{{B - C}}{2}} \right) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\{ {\text{Using}}\;{\text{Napier's}}\;{\text{law}}\} \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad= 2\lambda \left\{ {\sin \left( {\frac{{B + C}}{2}} \right)\cos \left( {\frac{{B - C}}{2}} \right)\tan \left( {\frac{{B - C}}{2}} \right)} \right\} \qquad\qquad\qquad\qquad\{ {\text{Using}}\;{\text{the}}\;{\text{sine}}\;{\text{rule}}\} \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad= 2\lambda \left\{ {{{\sin }^2}\frac{B}{2} - {{\sin }^2}\frac{C}{2}} \right\} \\ &\qquad\;\Rightarrow\quad \,\,\,S = 0 \\ \end{align}

Example - 36

In a  $$\Delta ABC,$$  if D is any point on BC such that  $$BD:DC~=~m~:~n$$,  $$\angle BAD = \alpha ,\;\angle CAD = \beta ,\;\angle CDA = \theta$$ , prove that  $$(m + n)\cot \theta = m\cot \alpha - n\cot \beta = n\cot B - m\cot C$$ .

Solution:

Using the sine rule in $$\Delta ABD$$ , we have

\begin{align}&\frac{{BD}}{{\sin \alpha }} = \frac{{AD}}{{\sin B}} = \frac{{AD}}{{\sin (\theta - \alpha )}} \qquad({\text{Why}}?) \\ \Rightarrow\quad &\frac{{BD}}{{AD}} = \frac{{\sin \alpha }}{{\sin (\theta - \alpha )}} \\ \end{align}

Similarly,                                                                                    \begin{align}\frac{{AD}}{{DC}} = \frac{{\sin (\theta + \beta )}}{{\sin \beta }} \qquad ({\text{Verify}}\,\,!)\end{align}

Now,

\begin{align}&\frac{{BD}}{{DC}} = \frac{{BD}}{{AD}} \cdot \frac{{AD}}{{DC}} = \frac{m}{n} \\\Rightarrow\quad &\frac{{\sin \alpha }}{{\sin (\theta - \alpha )}} \cdot \frac{{\sin (\theta + \beta )}}{{\sin \beta }} = \frac{m}{n} \\ \end{align}

Cross multiplying and simplifying yields

$(m + n)\cot \theta = m\cot \alpha - n\cot \beta$

To prove the other equality, we need to find expressions for  $$\cot B\;{\text{and}}\;\cot C$$  :

\begin{align} \cot B = \frac{{BE}}{{AE}} = \frac{{BD + DE}}{{AE}} = \frac{{BD}}{{AE}} + \cot \theta \\ \cot C = \frac{{CE}}{{AE}} = \frac{{CD - DE}}{{AE}} = \frac{{CD}}{{AE}} - \cot \theta \\ \end{align}

Now,

\begin{align}&\frac{{BD}}{{DC}} = \frac{{BD}}{{AE}} \cdot \frac{{AE}}{{DC}} = \frac{m}{n}\\\Rightarrow\qquad&\frac{{\left( {\cot B - \cot \theta } \right)}}{{\left( {\cot C + \cot \theta } \right)}} = \frac{m}{n} \\ \end{align}

Cross-multiplying and simplifying yields the desired result.