For any three angles A, B, C prove that $sin A + sin B + sin C - sin (A + B + C) = 4 sin {(A + B) / 2} sin {(B + C) / 2} sin {(C + A) / 2}$ .

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Solution

Note: Here A, B, C are not angles of a triangle i.e. $A + B + C \neq 180^{o}$ .
Combining the first two and last two, we get
L.H.S. $= 2 sin \frac{A + B}{2} cos \frac{A - B}{2} + 2 cos \frac{A + B + 2 C}{2} sin (- \frac{A + B}{2})$
$= 2 sin \frac{A + B}{2} [cos \frac{A - B}{2} - cos \frac{A + B + 2 C}{2}]$
$∵ sin (- θ) = - sin θ$ .
$= 2 sin \frac{A + B}{2} [2 sin \frac{A + C}{2} sin \frac{B + C}{2}]$
$= 4 sin \frac{A + B}{2} sin \frac{B + C}{2} sin \frac{C + A}{2}$ .

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