Question

In $\Delta ABC,$ find $\sum a^2\frac{\sin \left(B-C\right)}{\sin B+\sin C}$.

Answer 1

$\sum \frac{a^2\sin (B-C)}{\sin B+\sin C}$

Using $a=2R\sin A,b=2R\sin B$ and $c=2R\sin C$

$=\sum \frac{{4R}^2{\sin }^{2}A\sin (B-C)}{\sin B+\sin C}$

We know that $A+B+C=\pi \Rightarrow A=\pi -(B+C)$

$=\sum \frac{{4R}^2\sin A\sin (\pi -(B+C))\sin (B-C)}{\sin B+\sin C}$

$=\sum \frac{{4R}^2\sin A\sin (B+C)\sin (B-C)}{\sin B+\sin C}$

Using $\sin (x+y)\sin (x-y)={\sin }^{2}x-{\sin }^{2}y$

$=\sum \frac{{4R}^2\sin A({\sin }^{2}B-{\sin }^{2}C)}{\sin B+\sin C}$

$=\sum \frac{{4R}^2\sin A(\sin B+\sin C)(\sin B-\sin C)}{\sin B+\sin C}$

$=\sum {4R}^2\sin A(\sin B-\sin C)$

$=4R^2\sin A(\sin B-\sin C)+4R^2\sin B(\sin C-\sin A)+4R^2\sin C(\sin A-\sin B)$

$=4R^2[\sin A\sin B-\sin A\sin C+\sin B\sin C-\sin B\sin A+\sin C\sin A-\sin C\sin B]$

$=4R^2\times 0=0$