Question

In any $\Delta ABC$ , prove that
${{a}^{3}}\sin \left( B-C \right)+{{b}^{3}}\sin \left( C-A \right)+{{c}^{3}}\sin \left( A-B \right)=0$

Answer 1

Hint: Try to simplify the left-hand side of the equation by applying the sine rule followed by the use of the identity: $\sin \left( A+B \right)\sin \left( A-B \right)={{\sin }^{2}}A-{{\sin }^{2}}B$ .

Complete step-by-step answer:
Before starting with the solution, let us draw a diagram for better visualisation.

Now let’s start with the simplification of the left-hand side of the equation given in the question.
Now we know, according to the sine rule of the triangle: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$ and in other terms, it can be written as:
$\begin{align}
  & a=k\sin A \\
 & b=k\sin B \\
 & c=k\sin C \\
\end{align}$
So, applying this to our expression, we get
${{a}^{3}}\sin \left( B-C \right)+{{b}^{3}}\sin \left( C-A \right)+{{c}^{3}}\sin \left( A-B \right)$
${{a}^{2}}ksinA\sin \left( B-C \right)+{{b}^{2}}ksinB\sin \left( C-A \right)+{{c}^{2}}ksinC\sin \left( A-B \right)$

Now as ABC is a triangle, we can say:
$\angle A+\angle B+\angle C=180{}^\circ $
$\Rightarrow \angle A=180{}^\circ -\angle C-\angle B$
So, substituting the values in our expression. On doing so, we get
\[={{a}^{2}}ksin\left( 180{}^\circ -B-C \right)\sin \left( B-C \right)+{{b}^{2}}ksin\left( 180{}^\circ -C-A \right)\sin \left( C-A \right)+{{c}^{2}}ksin\left( 180{}^\circ -A-B \right)\sin \left( A-B \right)\]

We know $\sin \left( 180{}^\circ -X \right)=\sin X$ . Using this in our expression, we get
\[={{a}^{2}}ksin\left( B+C \right)\sin \left( B-C \right)+{{b}^{2}}ksin\left( C+A \right)\sin \left( C-A \right)+{{c}^{2}}ksin\left( A+B \right)\sin \left( A-B \right)\]

Now using the identity: $\sin \left( A+B \right)\sin \left( A-B \right)={{\sin }^{2}}A-{{\sin }^{2}}B$ , we get
\[=k{{a}^{2}}\left( {{\sin }^{2}}B-{{\sin }^{2}}C \right)+k{{b}^{2}}\left( {{\sin }^{2}}C-{{\sin }^{2}}A \right)+k{{c}^{2}}\left( {{\sin }^{2}}A-{{\sin }^{2}}B \right)\]

Now we can also interpret the sine rule as:
$\begin{align}
  & a\sin B=b\sin A \\
 & b\sin C=c\sin B \\
 & c\sin A=a\sin C \\
\end{align}$
So, using this in our expression, we get
\[=k\left( {{a}^{2}}{{\sin }^{2}}B-{{a}^{2}}{{\sin }^{2}}C+{{b}^{2}}{{\sin }^{2}}C-{{b}^{2}}{{\sin }^{2}}A+{{c}^{2}}{{\sin }^{2}}A-{{c}^{2}}{{\sin }^{2}}B \right)\]
\[=k\left( {{b}^{2}}{{\sin }^{2}}A-{{a}^{2}}{{\sin }^{2}}C+{{c}^{2}}{{\sin }^{2}}B-{{b}^{2}}{{\sin }^{2}}A+{{a}^{2}}{{\sin }^{2}}C-{{c}^{2}}{{\sin }^{2}}B \right)\]

It is clearly seen that all the terms are cancelled. So, we get
\[=k\times 0=0\]

The left-hand side of the equation given in the question is equal to the right-hand side of the equation which is equal to zero. Hence, we can say that we have proved the equation given in the question.

Note: Be careful about the calculation and the signs while opening the brackets. Also, you need to learn the sine rule and the cosine rule as they are used very often. The k in the sine rule is twice the radius of the circumcircle of the triangle, i.e., sine rule can also be written as $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k=2R=\dfrac{abc}{2\Delta }$ , where $\Delta $ represents the area of the triangle.