Question

In any triangle ABC, prove the following: $b\sin B-c\sin C=a\sin (B-C)$.

Answer 1

Given: ABC is a triangle,

$[\therefore A+B+C=\pi ]$

According to sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

We have to prove: $b\sin B-c\sin C=a\sin (B-C)$.

$L.H.S=b\sin B-c\sin C$

$=k\sin B\sin B-k\sin C\sin C$

$=k({\sin }^{2}B-{\sin }^{2}C)$

$=k\left[\sin \right(B+C\left)\sin \right(B-C\left)\right]$

$[\because {\sin }^{2}A-{\sin }^{2}B=\sin (A+B\left)\sin \right(A-B\left)\right]$

$=k\left[\sin \right(\pi -A\left)\sin \right(B-C\left)\right][\because A+B+C=\pi ]$

$=k\sin A\sin (B-C)$

$=a\sin (B-C)[\because a=k\sin A]$

= R.H.S

Hence, proved.