Question

If R denotes circumradius of $\Delta ABC$, evaluate $\frac{b^2-c^2}{2aR}$

Answer 1

$\frac{b^2-c^2}{2aR}$
$=\frac{b^2-c^2}{2a\left(\frac{a}{2sinA}\right)}$
$=sinA\left(\frac{b^2-c^2}{a^2}\right)$
$=sinA(\frac{b-c}{a}\times \frac{b+c}{a})$
$=sinA(\frac{sin\left(\frac{B-C}{2}\right)}{cos\frac{A}{2}}\times \frac{cos\left(\frac{B-C}{2}\right)}{sin\frac{A}{2}})$
$=sinA\left(\frac{\frac{sin(B-C)}{2}}{\frac{sinA}{2}}\right)$
$=sin(B-C)$