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.

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