Show that in any $△ A B C$
$a sin (B - C) + b sin (C - A) + c sin (A - B) = 0$

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Solution

Formula,

$\frac{sin a}{a} = \frac{sin b}{b} = \frac{sin c}{c} = k$

$sin a = a k, sin b = b k, sin c = c k$

LHS:

$= a sin (b - c) + b sin (c - a) + c sin (a - b)$

$= a k [b cos c - c cos b] + b k [c cos a - a cos c] + c k [a cos b - b cos a]$

$= k [a b cos c - a c cos b] + k [b c cos a - b a cos c] + k [a c cos b - b c cos a]$

$= k [0] + k [0] + k [0]$

$= 0$
'

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