If $A + B + C = 180^{o}$ , show that $sin (B + 2 C) + sin (C + 2 A) + sin (A + 2 B) = 4 sin [(B - C) / 2] sin [(C - A) / 2] \cdot s i n [(A - B) / 2]$ .

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Solution

Here $A + B + C = 180^{o}$ .
$∴ B + 2 C = B + C + C = 180^{o} - A + C$
$= 180^{o} + (C - A)$
$∴ sin (B + 2 C) = - sin (C - A)$
$∴$ L.H.S. $= - [sin (C - A) + sin (A - B) + sin (B - C)]$
$= [2 sin \frac{C - A}{2} cos \frac{C - A}{2} + 2 sin \frac{A - C}{2} cos \frac{A + C - 2 B}{2}]$
$= - 2 sin \frac{C - A}{2} [cos \frac{C - A}{2} - cos \frac{A + C - 2 B}{2}]$
$= - 2 sin \frac{C - A}{2} [2 sin \frac{C - B}{2} sin \frac{A - B}{2}]$
$= 4 sin \frac{A - B}{2} sin \frac{B - C}{2} sin \frac{C - A}{2}$ .

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