Question
$\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
✓
Answer
Then,
Consider the RHS of the equation $\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
$\text{RHS}=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
$=\frac{\text{k}(\sin\text{B}-\sin\text{C})}{\text{k}\sin\text{A}}\cos\Big(\frac{\pi-(\text{B + C})}{2}\Big)$ $(\because\text{A + B + C}=\pi)$
$=\frac{2\sin\big(\frac{\text{B}-\text{C}}{2}\big)\cos\big(\frac{\text{B + C}}{2}\big)}{\sin\text{A}}$
$=\frac{\sin\big(\frac{\text{B}-\text{C}}{2}\big)2\cos\big(\frac{\text{B + C}}{2}\big)}{\sin\text{A}}\sin\Big(\frac{\text{B + C}}{2}\Big)$
$=\frac{\sin\big(\frac{\text{B}-\text{C}}{2}\big)\sin(\text{B + C})}{\sin\text{A}}$
$=\frac{\sin\big(\frac{\text{B} - \text{C}}{2}\big)\sin(\pi-\text{A})}{\sin\text{A}}$
$=\frac{\sin\text{A}\sin\big(\frac{\text{B}-\text{C}}{2}\big)}{\sin\text{A}}$
$=\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\text{LHS}$
Hence proved.
Consider the RHS of the equation $\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
$\text{RHS}=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
$=\frac{\text{k}(\sin\text{B}-\sin\text{C})}{\text{k}\sin\text{A}}\cos\Big(\frac{\pi-(\text{B + C})}{2}\Big)$ $(\because\text{A + B + C}=\pi)$
$=\frac{2\sin\big(\frac{\text{B}-\text{C}}{2}\big)\cos\big(\frac{\text{B + C}}{2}\big)}{\sin\text{A}}$
$=\frac{\sin\big(\frac{\text{B}-\text{C}}{2}\big)2\cos\big(\frac{\text{B + C}}{2}\big)}{\sin\text{A}}\sin\Big(\frac{\text{B + C}}{2}\Big)$
$=\frac{\sin\big(\frac{\text{B}-\text{C}}{2}\big)\sin(\text{B + C})}{\sin\text{A}}$
$=\frac{\sin\big(\frac{\text{B} - \text{C}}{2}\big)\sin(\pi-\text{A})}{\sin\text{A}}$
$=\frac{\sin\text{A}\sin\big(\frac{\text{B}-\text{C}}{2}\big)}{\sin\text{A}}$
$=\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\text{LHS}$
Hence proved.
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