# In any ΔABC, prove that

Question:

In any ΔABC, prove that

$\frac{(b-c)}{a} \cos \frac{A}{2}=\sin \frac{(B-C)}{2}$

Solution:

Need to prove: $\frac{(\mathrm{b}-\mathrm{c})}{\mathrm{a}} \cos \frac{\mathrm{A}}{2}=\sin \frac{(\mathrm{B}-\mathrm{C})}{2}$

We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where $R$ is the circumradius.

Therefore,

a = 2R sinA ---- (a)

Similarly, b = 2R sinB and c = 2R sinC

From Left hand side,

$=\frac{2 R \sin B-2 R \sin C}{2 R \sin A} \cos \frac{A}{2}$

$=\frac{2 \cos \left(\frac{B+C}{2}\right) \sin \left(\frac{B-C}{2}\right)}{\sin A} \cos \frac{A}{2}$

$=\frac{2 \sin \left(\frac{B-C}{2}\right) \cos \left(\frac{\pi}{2}-\frac{A}{2}\right)}{\sin A} \cos \frac{A}{2}$

$=\frac{2 \cos ^{2} \frac{A}{2} \sin \left(\frac{B-C}{2}\right)}{\sin A}$

$=\frac{\sin A \sin \left(\frac{B-C}{2}\right)}{\sin A}$

$=\sin \frac{B-C}{A}$

$=$ Right hand side. [Proved]