# Thread: prove that a^2sin(B-C) =

1. ## prove that a^2sin(B-C) =

prove that:

$a^{2}sin\left(B-C\right) = \left(b^{2} - c^{2}\right)sinA$

where $a,b,c$ are sides of triangle
and $A,B,C$ are the corresponding angles

2. I'll use the sine and cosine rules.

$a^2\sin(B-C)=a^2(\sin B\cos C-\sin C\cos B)=$

$=a^2\left(\frac{b}{2R}\cdot\frac{a^2+b^2-c^2}{2ab}-\frac{c}{2R}\cdot\frac{a^2+c^2-b^2}{2ac}\right)=$

$=\frac{2a(b^2-c^2)}{4R}=\frac{a}{2R}(b^2-c^2)=(b^2-c^2)\sin A$