Question

In triangle ABC, prove the following:

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

Answer 1

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ ...(1)

Consider the LHS of the equation $\left(a-b\right)\cos \frac{C}{2}=c\sin \left(\frac{A-B}{2}\right)$
$$\begin{gathered} \mathrm{LHS}=\left(a-b\right)\cos \frac{C}{2} \\ =k\left(\sin A-\sin B\right)\cos \frac{C}{2}\left(\mathrm{using}\left(1\right)\right) \\ =k\times 2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)\cos \frac{C}{2} \\ =2k\sin \left({\displaystyle \frac{A-B}{2}}\right)\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{\mathrm{\pi }-\left(\mathrm{A}+\mathrm{B}\right)}{2}\right)\left[\because A+B+C=\mathrm{\pi }\right] \\ =2k\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A+B}{2}\right) \\ =k\sin \left(\frac{A-B}{2}\right)\sin \left(A+B\right)\left[\because 2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A+B}{2}\right)=\sin \left(A+B\right)\right] \\ =k\sin \frac{A-B}{2}\sin \left(\mathrm{\pi }-\mathrm{C}\right)\left[\because A+B+C=\mathrm{\pi }\right] \\ =k\sin C\sin \left(\frac{A-B}{2}\right) \\ =C\sin \left(\frac{A-B}{2}\right)=\mathrm{RHS} \end{gathered}$$

Hence proved.