Question

In a triangle ABC, then find \[{a^3}\cos \left( {B - C} \right) + {b^3}\cos \left( {C - A} \right) + {c^3}\cos \left( {A - B} \right)\].
A. abc
B. 3abc
c. a + b + c
D. None of these

Answer 1

Hint: First we will use the sine rule to find the value of a, b, and c. Then substitute the value of a, b, and c in the given expression. Then we will apply trigonometry identities to simplify the given expression.

Formula used:
Sine laws:
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{b\sin B}} = \dfrac{c}{{\sin C}}\]
Trigonometry identities:
\[2\sin \left( {a + b} \right)\cos \left( {a - b} \right) = \sin 2a + \sin 2b\]
\[2\sin \left( {a + b} \right)\cos \left( {a - b} \right) = \sin 2a + \sin 2b\]
\[\sin 2\theta = 2\sin \theta \cos \theta \]
Supplementary angle of trigonometry
\[\sin \left( {\pi - \theta } \right) = \sin \theta \]

Complete step by step solution:
We know that, \[\dfrac{a}{{\sin A}} = \dfrac{b}{{b\sin B}} = \dfrac{c}{{\sin C}} = k\]
\[a = k\sin A\], \[b = k\sin B\], \[c = k\sin C\]
Putting \[a = k\sin A\], \[b = k\sin B\], \[c = k\sin C\], in the expression \[{a^3}\cos \left( {B - C} \right) + {b^3}\cos \left( {C - A} \right) + {c^3}\cos \left( {A - B} \right)\]:
\[{\left( {k\sin A} \right)^3}\cos \left( {B - C} \right) + {\left( {k\sin B} \right)^3}\cos \left( {C - A} \right) + {\left( {k\sin C} \right)^3}\cos \left( {A - B} \right)\]
\[ = {k^3}{\sin ^3}A\cos \left( {B - C} \right) + {k^3}{\sin ^3}B\cos \left( {C - A} \right) + {k^3}{\sin ^3}C\cos \left( {A - B} \right)\]
\[ = {k^3}{\sin ^3}A\cos \left( {B - C} \right) + {k^3}{\sin ^3}B\cos \left( {C - A} \right) + {k^3}{\sin ^3}C\cos \left( {A - B} \right)\]
We know that, the sum of angles of a triangle is \[\pi \].
Thus, \[A + B + C = \pi \].
\[\begin{array}{l} = {k^3}{\sin ^2}A\sin \left[ {\pi - \left( {B + C} \right)} \right]\cos \left( {B - C} \right) + {k^3}{\sin ^2}B\sin \left[ {\pi - \left( {A + C} \right)} \right]\cos \left( {C - A} \right)\\ + {k^3}{\sin ^2}C\sin \left[ {\pi - \left( {A + B} \right)} \right]\cos \left( {A - B} \right)\end{array}\]
Applying supplementary angle of triangle:
\[ = {k^3}{\sin ^2}A\sin \left( {B + C} \right)\cos \left( {B - C} \right) + {k^3}{\sin ^2}B\sin \left( {A + C} \right)\cos \left( {C - A} \right) + {k^3}{\sin ^2}C\sin \left( {A + B} \right)\cos \left( {A - B} \right)\]
Applying the formula \[2\sin \left( {a + b} \right)\cos \left( {a - b} \right) = \sin 2a + \sin 2b\]
\[ = \dfrac{{{k^3}}}{2}\left[ {{{\sin }^2}A \cdot 2\sin \left( {B + C} \right)\cos \left( {B - C} \right) + {{\sin }^2}B \cdot 2\sin \left( {A + C} \right)\cos \left( {C - A} \right) + {{\sin }^2}C \cdot 2\sin \left( {A + B} \right)\cos \left( {A - B} \right)} \right]\]\[ = \dfrac{{{k^3}}}{2}\left[ {{{\sin }^2}A\left( {\sin 2B + \sin 2C} \right) + {{\sin }^2}B\left( {\sin 2A + \sin 2C} \right) + {{\sin }^2}C\left( {\sin 2A + \sin 2B} \right)} \right]\]
Now applying \[\sin 2\theta = 2\sin \theta \cos \theta \]
\[ = \dfrac{{{k^3}}}{2}\left[ \begin{array}{l}\left( {{{\sin }^2}A \cdot 2\sin B\cos B + {{\sin }^2}A \cdot 2\sin C\cos C} \right) + \left( {{{\sin }^2}B \cdot 2\sin A\cos A + {{\sin }^2}B \cdot 2\sin C\cos C} \right)\\ + \left( {{{\sin }^2}C \cdot 2\sin A\cos A + {{\sin }^2}C \cdot 2\sin B\cos B} \right)\end{array} \right]\]
\[ = \dfrac{{{k^3}}}{2} \cdot 2\left[ \begin{array}{l}{\sin ^2}A\sin B\cos B + {\sin ^2}A\sin C\cos C + {\sin ^2}B\sin A\cos A + {\sin ^2}B\sin C\cos C\\ + {\sin ^2}C\sin A\cos A + {\sin ^2}C\sin B\cos B\end{array} \right]\]
\[ = {k^3}\left[ \begin{array}{l}\left( {{{\sin }^2}A\sin B\cos B + {{\sin }^2}B\sin A\cos A} \right) + \left( {{{\sin }^2}A\sin C\cos C + {{\sin }^2}C\sin A\cos A} \right)\\ + \left( {{{\sin }^2}B\sin C\cos C + {{\sin }^2}C\sin B\cos B} \right)\end{array} \right]\]
\[ = {k^3}\left[ \begin{array}{l}\sin A\sin B\left( {\sin A\cos B + \sin B\cos A} \right) + \sin A\sin C\left( {\sin A\cos C + \sin C\cos A} \right)\\ + \sin B\sin C\left( {\sin B\cos C + \sin C\cos B} \right)\end{array} \right]\]
Applying \[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]
\[ = {k^3}\left[ {\sin A\sin B\sin \left( {A + B} \right) + \sin A\sin C\sin \left( {A + C} \right) + \sin B\sin C\sin \left( {B + C} \right)} \right]\]
We know that, \[A + B + C = \pi \]
\[ = {k^3}\left[ {\sin A\sin B\sin \left( {\pi - C} \right) + \sin A\sin C\sin \left( {\pi - B} \right) + \sin B\sin C\sin \left( {\pi - A} \right)} \right]\]
Now applying supplementary angle \[\sin \left( {\pi - \theta } \right) = \sin \theta \]
\[ = {k^3}\left[ {\sin A\sin B\sin C + \sin A\sin C\sin B + \sin B\sin C\sin A} \right]\]
\[ = k\sin A \cdot k\sin B \cdot k\sin C + k\sin A \cdot k\sin C \cdot k\sin B + k\sin B \cdot k\sin C \cdot k\sin A\]
Putting \[a = k\sin A\], \[b = k\sin B\], \[c = k\sin C\]
\[ = abc + abc + abc\]
\[ = 3abc\]
Hence option B is the correct option.

Note: Students often do a common mistake when they put \[a = k\sin A\], \[b = k\sin B\], \[c = k\sin C\] in the given expression. They replace \[{\sin ^3}A\], \[{\sin ^3}B\], \[{\sin ^3}C\]with \[{\sin ^3}\left( {\pi - \left( {B + C} \right)} \right)\], \[{\sin ^3}\left( {\pi - \left( {A + C} \right)} \right)\], \[{\sin ^3}\left( {\pi - \left( {A + B} \right)} \right)\] respectively. But here we need convert only \[\sin A\], \[\sin B\], \[\sin C\]and the rest will be remains same.