Question

State whether the following statement is true or false.

In $\Delta ABC$, $\sin \left(\frac{B-C}{2}\right)=\left(\frac{b-c}{a}\right)\cos \frac{A}{2}$.

• A. True
• B. False

Answer 1

The correct option is A True

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$\Rightarrow$ $a=k\sin A,$ $b=k\sin B,$ $c=k\sin C$

Now,

$\Rightarrow$ $R.H.S=\left(\frac{b-c}{a}\right)\cos \frac{A}{2}$

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

$=\frac{k(\sin B-\sin C)}{k\sin A}\cos \frac{A}{2}$

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

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

Now, $A+B+C={180}^o$

$\frac{B+C}{2}=\frac{{180}^o-A}{2}$

$\frac{B+C}{2}={90}^o-\frac{A}{2}$

$=\frac{\cos ({90}^o-\frac{A}{2})\sin \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}$

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

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

$=L.H.S$