Question
$\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C = 2b}\sin\text{A}\sin\text{C}$
✓
Answer
By sine rule, we know that
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k }\text{(say)}$
$\Rightarrow\text{a = k}\sin\text{A, b = k}\sin\text{B, c = k}\sin\text{C}$
Now,
$\text{LHS}=\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C}$
$=\text{k}\sin\text{A}\cos\text{A + k}\sin\text{B}\cos\text{B + k}\sin\text{C}\cos\text{C}$
$=\frac{\text{k}}{2}(2\sin\text{A}\cos\text{A}+2\sin\text{B}\cos\text{B}+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(\sin2\text{A}+\sin2\text{B}+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}\Big(2\sin\frac{2\text{A}+2\text{B}}{2}\cos\frac{2\text{A}-2\text{B}}{2}+2\sin\text{C}\cos\text{C}\Big)$
$=\frac{\text{k}}{2}(2\sin(\text{A + B})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(2\sin(\pi-\text{C})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C})$ $(\because\text{A + B + C}=\pi)$
$=\frac{\text{k}}{2}(2\sin\text{C}\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}\times2\sin\text{C}(\cos(\text{A}-\text{B})+\cos\text{C})$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\text{A}-\text{B + C}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}-\text{C}}{2}\Big)\Big)$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\pi-\text{B}-\text{B}}{2}\Big)\cos\Big(\frac{\text{B}+\text{C}-\text{A}}{2}\Big)\Big)$ $(\because\text{A + B + C}=\pi)$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\pi-2\text{B}}{2}\Big)\cos\Big(\frac{\pi-2\text{A}}{2}\Big)\Big)$ $(\because\text{A + B + C}=\pi)$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\pi}{2}-\text{B}\Big)\cos\Big(\frac{\pi}{2}-\text{A}\Big)\Big)$
$=2\text{k}\sin\text{C}(\sin\text{B}\sin\text{A})$
$=2(\text{k}\sin\text{B})\sin\text{A}\sin\text{C}$
$=2\text{b}\sin\text{A}\sin\text{C}$
$=\text{RHS}$
$\therefore\text{LHS = RHS}$
Hence, $\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C = 2 b}\sin\text{A}\sin\text{C}.$
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k }\text{(say)}$
$\Rightarrow\text{a = k}\sin\text{A, b = k}\sin\text{B, c = k}\sin\text{C}$
Now,
$\text{LHS}=\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C}$
$=\text{k}\sin\text{A}\cos\text{A + k}\sin\text{B}\cos\text{B + k}\sin\text{C}\cos\text{C}$
$=\frac{\text{k}}{2}(2\sin\text{A}\cos\text{A}+2\sin\text{B}\cos\text{B}+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(\sin2\text{A}+\sin2\text{B}+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}\Big(2\sin\frac{2\text{A}+2\text{B}}{2}\cos\frac{2\text{A}-2\text{B}}{2}+2\sin\text{C}\cos\text{C}\Big)$
$=\frac{\text{k}}{2}(2\sin(\text{A + B})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(2\sin(\pi-\text{C})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C})$ $(\because\text{A + B + C}=\pi)$
$=\frac{\text{k}}{2}(2\sin\text{C}\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}\times2\sin\text{C}(\cos(\text{A}-\text{B})+\cos\text{C})$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\text{A}-\text{B + C}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}-\text{C}}{2}\Big)\Big)$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\pi-\text{B}-\text{B}}{2}\Big)\cos\Big(\frac{\text{B}+\text{C}-\text{A}}{2}\Big)\Big)$ $(\because\text{A + B + C}=\pi)$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\pi-2\text{B}}{2}\Big)\cos\Big(\frac{\pi-2\text{A}}{2}\Big)\Big)$ $(\because\text{A + B + C}=\pi)$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\pi}{2}-\text{B}\Big)\cos\Big(\frac{\pi}{2}-\text{A}\Big)\Big)$
$=2\text{k}\sin\text{C}(\sin\text{B}\sin\text{A})$
$=2(\text{k}\sin\text{B})\sin\text{A}\sin\text{C}$
$=2\text{b}\sin\text{A}\sin\text{C}$
$=\text{RHS}$
$\therefore\text{LHS = RHS}$
Hence, $\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C = 2 b}\sin\text{A}\sin\text{C}.$
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