# Examples on Cosine Rule Set 1

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Example - 43

In  $$\Delta ABC$$ , find the maximum values of

(a) $${S_1} = \cos A + \cos B + \cos C$$           (b)   \begin{align}{S_2} = \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\end{align}

Solution: The answer should be evident using symmetry arguments. But let us evaluate them rigorously.

(a) Using the cosine rule,  $${S_1}$$  can be written as

\begin{align}&{S_1} = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + \frac{{{c^2} + {a^2} - {b^2}}}{{2ca}} + \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\\ \,\,\,\,\,\, &\quad= \frac{1}{{2abc}}\left\{ {a({b^2} + {c^2} - {a^2}) + b({c^2} + {a^2} - {b^2}) + c({a^2} + {b^2} - {c^2})} \right\}\\\end{align}

Our approach further will be guided by a need to produce perfect square terms in the expression. To that end.

\begin{align}{S_1}&= \frac{1}{{2abc}}\left\{ {a({b^2} + {c^2} - {a^2}) + b({c^2} + {a^2} - {b^2}) + c({a^2} + {b^2} - {c^2})\underbrace { - 6abc + 6abc}_{{\text{New}}\;{\text{terms}}}} \right\}\\&= \frac{1}{{2abc}}\left\{ {a({b^2} + {c^2} - 2bc) + b({c^2} + {a^2} - 2ca) + c({a^2} + {b^2} - 2ab) - ({a^3} + {b^3} + {c^3}) + 6abc} \right\} \\&= \frac{1}{{2abc}}\left\{ {a{{(b - c)}^2} + b{{(c - a)}^2} + c{{(a - b)}^2} - ({a^3} + {b^3} + {c^3} - 3abc) + 3abc} \right\} \\&=\frac{1}{{4abc}}\left\{ {\boxed{\;(a - b - c){{(b - c)}^2} + (b - c - a){{(c - a)}^2} + (c - a - b){{(a - b)}^2}\;} + 6abc} \right\} \end{align}

You should carefully verify the last step. Finally, note that the boxed expression is always negative, since by the triangle inequality

$a > b + c,\qquad \qquad b > c + a,\qquad \qquad c > a + b$

Therefore,

${S_1} \leqslant \frac{{6\;abc}}{{4\;abc}} = \frac{3}{2}$

The maximum value of  \begin{align}\frac{3}{2}\end{align} occurs at the symmetrical situation of

$A = B = C = \frac{\pi }{3}$

(b) From a previous example, we have the identity

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos A + \cos B + \cos C = 1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\ &\Rightarrow \qquad {\left( {\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} \right)_{\max }} = \frac{1}{8} \\ \end{align}

the maximum occurring once again at   \begin{align}A = B = C = \frac{\pi }{3}\end{align} .

Example - 44

What can we say about  $$\Delta ABC$$  a where the following relation is satisfied:

$\frac{{\cos A + 2\cos C}}{{\cos A + 2\cos B}} = \frac{{\sin B}}{{\sin C}}$

Solution: Cross-multiplying and rearranging,

\begin{align}& \cos A(\sin B - \sin C) + \sin 2B - \sin 2C = 0 \\ \Rightarrow\qquad & \cos A(\sin B - \sin C) + 2\sin (B - C)\cos (B + C) = 0 \\ \Rightarrow\qquad & \cos A(\sin B - \sin C - 2\sin (B - C)) = 0 \qquad\qquad\qquad\qquad\qquad({\text{How}}?)\\ \end{align}

Expanding  $$\sin (B - C)$$  , and using the sine and cosine rule for the sine and cosine terms inside the brackets, we have,

$\cos A\left( {\lambda b - \lambda c - 2\lambda b\left( {\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \right) + 2\lambda c\left( {\frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right)} \right) = 0$

Rearranging further, we have

$\cos A(b - c)(a - 2(b + c)) = 0$

Since  $$(b + c) > a$$  , the third term is surely non-zero. So,

\begin{align} {\text{either}} \quad \cos A = 0 \qquad &\Rightarrow \qquad \angle \,A = 90\\ {\text{or}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;b = c \qquad &\Rightarrow\qquad {\text{The}}\;{\text{triangle}}\;{\text{is}}\;{\text{isosceles}}\\ \end{align}

Example - 45

Prove that \begin{align}a\cos A + b\cos B + c\cos C = \frac{{8{\Delta ^2}}}{{abc}}\end{align}

Solution: Using the sine rule for a, b, c,

\begin{align}&{\text{LHS}} = \frac{\lambda }{2}\left\{ {\sin 2A + \sin 2B + \sin 2C} \right\} \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \frac{\lambda }{2}\left\{ {4\sin A\sin B\sin C} \right\} \quad\qquad\qquad ({\text{How}}?) \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= 2a\sin B\sin C\qquad\qquad\qquad \{ {\text{Using}}\;{\text{the}}\;{\text{sine}}\;{\text{rule}}\;{\text{again}}\} \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= 2a \cdot \frac{\Delta }{{\frac{1}{2}ac}} \cdot \frac{\Delta }{{\frac{1}{2}ab}} \\ \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \frac{{8{\Delta ^2}}}{{abc}} \\ \end{align}