Then, Consider the RHS of the equation ( B-C 2 )= b-c a A 2 RHS= b-c a A 2 = k( - ) k ( -(B + C) 2 ) ( + B + C= ) = 2 ( B-C 2 ) ( B + C 2 ) = ( B-C 2 )2 ( B + C 2 ) ( B + C 2 ) = ( B-C 2 ) (B + C) = ( B - C 2 ) ( -A) = ( B-C 2 ) = ( B-C 2 )=LHS Hence proved.

( B-C 2 )= b-c a A 2

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

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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.

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