Question

In any $\Delta ABC$, prove that $a\sin A-b\sin B=c\sin \left( A-B \right)$.

Answer 1

Hint: We will be using the solution of triangles to solve the problem. We will be using properties of a triangle to solve the problem. We will use the sine rule to find the value of side a and b of the triangle with the help of sine rule and use it to simplify the left hand side of the equation then simplify it using,
$\begin{align}
  & {{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A-B \right)\sin \left( A+B \right) \\
 & \sin \left( \pi -C \right)=\sin C \\
\end{align}$
to prove that the left hand side of the equation is equal to the right hand side.

Complete Step-by-Step solution:
We have been given a $\Delta ABC$ and we have to prove that, $a\sin A-b\sin B=c\sin \left( A-B \right)$.
So, we will first draw a $\Delta ABC$ and label its sides as a, b, c.

Now, we know that according to sine rule sides of any triangle are proportional to the sine of the angle opposite to them. So, we have,
$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k..........\left( 1 \right)$
Now, we have from equation (1)
$\begin{align}
  & a=\dfrac{\sin A}{k}..........\left( 2 \right) \\
 & b=\dfrac{\sin B}{k}..........\left( 3 \right) \\
 & c=\dfrac{\sin C}{k}..........\left( 4 \right) \\
\end{align}$
Now, we will take L.H.S of the question and prove it to be equal to the R.H.S.
In L.H.S we have,
$a\sin A-b\sin B$
Now, we will substitute the value of a and b from (2) and (3),
$a\sin A-b\sin B=\dfrac{{{\sin }^{2}}A}{k}-\dfrac{{{\sin }^{2}}B}{k}=\dfrac{1}{k}\left( {{\sin }^{2}}A-{{\sin }^{2}}B \right)$
Now, we know that, ${{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A-B \right)\sin \left( A+B \right)$ using this we have,
$a\sin A-b\sin B=\dfrac{1}{k}\left( \sin \left( A-B \right) \right)\left( \sin \left( A+B \right) \right)$
Now, we know that in a $\Delta ABC$ sum of interior angles is $180{}^\circ $. Therefore,
$\begin{align}
  & A+B+C=180{}^\circ \\
 & A+B=180{}^\circ -C \\
\end{align}$
Substituting values of A +B in $\dfrac{1}{k}\left( \sin \left( A-B \right) \right)\left( \sin \left( A+B \right) \right)$, we have,
$=\dfrac{1}{k}\left( \sin \left( A-B \right) \right)\left( \sin \left( 180{}^\circ -C \right) \right)$
Now, we know that $\sin \left( 180{}^\circ -C \right)=\sin C$. Therefore, we have,
$\begin{align}
  & =\dfrac{1}{k}\sin \left( A-B \right)\sin C \\
 & =\left( \dfrac{\sin C}{k} \right)\sin \left( A-B \right) \\
\end{align}$
Now, from equation (4) we will substitute value of $\dfrac{\sin C}{k}$, therefore,
$a\sin A-b\sin B=c\sin \left( A-B \right)$
Hence proved, since L.H.S = R.H.S.

Note: To solve these type of questions it is important to note that we have used sine rule to relate the sides of the triangle with the angle of the triangle also one must remember formulae like,
$\begin{align}
  & {{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A-B \right)\sin \left( A+B \right) \\
 & \sin \left( \pi -C \right)=\sin C \\
\end{align}$
to further simplify the problem.