a( - )=2(b-c) ^2 A 2 . - Vidyadip

Question

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

✓

Answer

Let $\text{a = k}\sin\text{A}$
$\text{a}(\cos\text{C}-\cos\text{B})=2(\text{b}-\text{c})\cos^2\frac{\text{A}}{2}$
$\text{LHS}=\text{a}(\cos\text{C}-\cos\text{B})$
$=\text{a}2.\sin\frac{\text{C + B}}{2}.\sin\frac{\text{B}-\text{C}}{2}$
$=2\text{k}\sin\text{A}\sin\frac{\pi-\text{A}}{2}.\sin\frac{\text{B}-\text{C}}{2}$
$=2\text{k}2\sin\frac{\text{A}}{2}.\cos\frac{\text{A}}{2}.\cos\frac{\text{A}}{2}.\sin\frac{\text{B}-\text{C}}{2}$
$=2\text{k}\cos^2\frac{\text{A}}{2}\Big(2\sin\frac{\text{B}-\text{C}}{2}.\sin\frac{\text{A}}{2}\Big)$
$=2\text{k}\cos^2\frac{\text{A}}{2}\Big(2\sin\frac{\text{B}-\text{C}}{2}.\sin\frac{\pi-(\text{B + C})}{2}\Big)$
$=2\text{k}\cos^2\frac{\text{A}}{2}\Big(2\sin\frac{\text{B}-\text{C}}{2}.\cos\frac{\text{B + C}}{2}\Big)$
$=2\text{k}\cos^2\frac{\text{A}}{2}(\sin\text{B}-\sin\text{C})$
$=2\cos^2\frac{\text{A}}{2}(\text{k}\sin\text{B}-\text{k}\sin\text{C})$
$=2\cos^2\frac{A}{2}(\text{b}-\text{c})=\text{RHS}$

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