Solveeit Logo
Login

Question

Question: In any \(\Delta ABC\), prove that \(a\sin \left( B-C \right)+b\sin \left( C-A \right)+c\sin \left(...

In any ΔABC\Delta ABC, prove that
asin⁡(B−C)+bsin⁡(C−A)+csin⁡(A−B)=0a\sin \left( B-C \right)+b\sin \left( C-A \right)+c\sin \left( A-B \right)=0

Explanation

Solution

Hint: Try to simplify the left-hand side of the equation given in the question using the formula of sin(X-Y) followed by the application of the sine rule of a triangle.

Complete step-by-step answer:

Before starting with the solution, let us draw a diagram for better visualisation.

Now starting with the left-hand side of the equation that is given in the question. To start the simplification, we need to use the formula sin(X-Y) = cosYsinX – sinYcosX.

asin⁡(B−C)+bsin⁡(C−A)+csin⁡(A−B)a\sin \left( B-C \right)+b\sin \left( C-A \right)+c\sin \left( A-B \right)

=a(sin⁡Bcos⁡C−cos⁡Bsin⁡C)+b(sin⁡CcosA⁡−cos⁡Csin⁡A)+c(sin⁡Acos⁡B−cos⁡Asin⁡B)=a\left( \sin B\cos C-\cos B\sin C \right)+b\left( \sin C\operatorname{cosA}-\cos C\sin A \right)+c\left( \sin A\cos B-\cos A\sin B \right)

Now we know, according to the sine rule of the triangle: asin⁡A=bsin⁡B=csin⁡C=k\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k and in other terms, it can be written as:

$\begin{aligned}

& \dfrac{a}{k}=\sin A \\

& \dfrac{b}{k}=\sin B \\

& \dfrac{c}{k}=\sin C \\

\end{aligned}$

So, applying this to our expression, we get

\dfrac{c}{k}\operatorname{cosA}-\dfrac{a}{k}\cos C \right)+c\left( \dfrac{a}{k}\cos B-\dfrac{b}{k}\cos A \right)$$ Now we will take $\dfrac{1}{k}$ common from all the terms. On doing so, we get $$=\dfrac{1}{k}\left( a\left( b\cos C-c\cos B \right)+b\left( c\operatorname{cosA}-a\cos C \right)+c\left( a\cos B-b\cos A \right) \right)$$ To further simplify the expression, we will multiply and open all the brackets. This will give us: $$=\dfrac{1}{k}\left( ab\cos C-ac\cos B+bc\operatorname{cosA}-ab\cos C+ca\cos B-bc\cos A \right)$$ In the above expression we can clearly see that each term is getting cancelled out, giving the final expression to be: $$=\dfrac{1}{k}\times 0$$ Now from the conditions of sine rule, we know that k is not equal to zero, so we can say that $\dfrac{1}{k}$ is finite. We also know that a finite number multiplied by zero gives the product as zero. Therefore, our expression is: $$=\dfrac{1}{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, remember that, infinity multiplied by zero can is not always zero; it can be a finite number or infinity as well.