Question

In Δ ABC, prove that ${\displaystyle \cos \left(\frac{\text{A}-\text{B}}{2}\right)=\left(\frac{\text{a}+\text{b}}{\text{c}}\right)\sin \frac{\text{C}}{2}}$ .

Answer 1

By the sine rule,

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

∴ a = k sin A, b = k sin B, c = k sin C

RHS = ${\displaystyle \left(\frac{\text{a}+\text{b}}{\text{c}}\right)\sin \frac{\text{C}}{2}}$

${\displaystyle =\left(\frac{\text{k sin A}+\text{k sin B}}{\text{k sin C}}\right)\sin \frac{\text{C}}{2}}$

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

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

 ${\displaystyle =\frac{\text{sin}\frac{\text{A}+\text{B}}{2}.\cos \frac{\text{A}-\text{B}}{2}}{\cos \frac{\text{C}}{2}}}$

${\displaystyle =\frac{\sin (\frac{\pi }{2}-\frac{\text{C}}{2}).\cos \frac{\text{A}-\text{B}}{2}}{\cos \frac{\text{C}}{2}} ...[\because \text{A}+\text{B}+\text{C}=\pi ]}$

${\displaystyle =\frac{\cos \frac{\text{C}}{2}.\cos \frac{\text{A}-\text{B}}{2}}{\cos \frac{\text{C}}{2}}}$

${\displaystyle =\cos \left(\frac{\text{A}-\text{B}}{2}\right)}$

= LHS