Question
$(\text{a}-\text{b})\cos\frac{\text{C}}{2}=\text{c}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
✓
Answer
$(\text{a}-\text{b})\cos\frac{\text{C}}{2}=\text{c}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
Let $\text{a = k}\sin\text{A, b = k}\sin\text{B,c = k}\sin\text{C}$
$\text{LHS}=\text{(a}-\text{b})\cos\frac{\text{C}}{2}$
$=\text{k}(\sin\text{A}-\sin\text{B}).\cos\frac{\text{C}}{2}$
$=2\text{k}\cos\Big(\frac{\text{A + B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big).\cos\frac{\text{C}}{2}$
$=2\text{k}\cos\Big(\frac{\pi-\text{C}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big).\cos\frac{\text{C}}2{}$
$=2\text{k}\sin\Big(\frac{\text{C}}{2}\Big).\cos\frac{\text{C}}2{}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $\Big[\cos\Big(\frac{\pi}{2}-\theta\Big)=\sin\theta\Big]$
$=\text{k}\sin\text{C}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$=\text{c}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)=\text{RHS}$
Let $\text{a = k}\sin\text{A, b = k}\sin\text{B,c = k}\sin\text{C}$
$\text{LHS}=\text{(a}-\text{b})\cos\frac{\text{C}}{2}$
$=\text{k}(\sin\text{A}-\sin\text{B}).\cos\frac{\text{C}}{2}$
$=2\text{k}\cos\Big(\frac{\text{A + B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big).\cos\frac{\text{C}}{2}$
$=2\text{k}\cos\Big(\frac{\pi-\text{C}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big).\cos\frac{\text{C}}2{}$
$=2\text{k}\sin\Big(\frac{\text{C}}{2}\Big).\cos\frac{\text{C}}2{}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $\Big[\cos\Big(\frac{\pi}{2}-\theta\Big)=\sin\theta\Big]$
$=\text{k}\sin\text{C}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$=\text{c}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)=\text{RHS}$
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