Question

In $△ABC$, prove that a sin (B - C) + b sin (C - A) + c sin (A - B) = 0

Answer 1

Using the following trigonometric identity:
sin (A - B) = sin A cos B - sin B cos A
We get
a sin (B - C) + b sin (C - A) + c sin (A - B) = a sin B cos C - a sin C cos B + b sin C cos A - b sin A cos C + c sin A cos B - c sin B cos A = cos A (b sin C - c sin B)
+ cos B (c sin A - a sin C) + cos C (a sin B - b sin A)
According to the sine theorem we have
$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$
Therefrom
b sin C - c sin B = c sin A - a sin C = a sin B - b sin A = 0
And we get
a sin (B - C) + bsin (C - A) + c sin (A - B) = 0.