Question
$\text{b}\sin\text{B}-\text{c}\sin\text{C = a}\sin(\text{B}-\text{C})$
✓
Answer
$\text{b}\sin\text{B}-\text{c}\sin\text{C = a}\sin(\text{B}-\text{C})$
$\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$
$\text{RHS}=\text{a}\sin(\text{B}-\text{C})$
$=\text{a}\sin\text{B}.\cos\text{C}-\text{a}\sin \text{C}.\cos\text{B}$
$=\text{a(bk).}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)-\text{a(ck)}.\Big(\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}\Big)$
$=\text{k}.\frac{(\text{a}^2+\text{b}^2-\text{c}^2)}{2}-\text{k}\frac{(\text{a}^2+\text{c}^2-\text{b}^2)}{2}$
$=2\text{k}.\frac{(\text{b}^2-\text{c}^2)}{2}$
$=\text{b.(kb)}-\text{c(kc)}$
$=\text{b}(\sin\text{B})-\text{c}(\sin\text{C})=\text{LHS}$
$\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$
$\text{RHS}=\text{a}\sin(\text{B}-\text{C})$
$=\text{a}\sin\text{B}.\cos\text{C}-\text{a}\sin \text{C}.\cos\text{B}$
$=\text{a(bk).}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)-\text{a(ck)}.\Big(\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}\Big)$
$=\text{k}.\frac{(\text{a}^2+\text{b}^2-\text{c}^2)}{2}-\text{k}\frac{(\text{a}^2+\text{c}^2-\text{b}^2)}{2}$
$=2\text{k}.\frac{(\text{b}^2-\text{c}^2)}{2}$
$=\text{b.(kb)}-\text{c(kc)}$
$=\text{b}(\sin\text{B})-\text{c}(\sin\text{C})=\text{LHS}$
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