prove
sinx+ sinb+ sinc- sina+b+c= 4sin(a+b/2) sin(b+c/2) sin(c+a/2)

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Solution

$\sin a + \sin b + \sin c - \sin (a + b + c) = 2 \sin \frac{a + b}{2} \cos \frac{a - b}{2} + 2 \cos \frac{a + b + c + c}{2} \sin \frac{- (a + b)}{2} = 2 \sin \frac{a + b}{2} \cos \frac{a - b}{2} - 2 \cos \frac{a + b + c + c}{2} \sin \frac{(a + b)}{2} (A s \sin (- a) = - \sin a) = 2 \sin \frac{a + b}{2} (\cos \frac{a - b}{2} - \cos \frac{a + b + c + c}{2}) = 2 \sin \frac{a + b}{2} (2 \sin \frac{\frac{a - b}{2} + \frac{a + b + 2 c}{2}}{2} \sin \frac{\frac{a + b + 2 c}{2} - \frac{a - b}{2}}{2}) = 2 \sin \frac{a + b}{2} (2 \sin \frac{a + c}{2} \sin \frac{b + c}{2}) = 4 \sin \frac{a + b}{2} \sin \frac{a + c}{2} \sin \frac{b + c}{2} H e n c e P r o v e d$

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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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A + B + C = π

(sin A + sin B + sin C) (- sin A + sin B + sin C) (sin A - sin B + sin C) (sin A + sin B - sin C) = 4 {sin}^{2} A {sin}^{2} B {sin}^{2} C

A + B + C = 180^{0}

cos 2 A + cos 2 B - cos 2 C = 1 - 4 sin A sin B sin C

sin A - sin B + sin C = 4 sin \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}

\frac{a^{2} sin (B - C)}{sin B + sin C} + \frac{b^{2} sin (C - A)}{sin C + sin A} + \frac{c^{2} sin (A - B)}{sin A + sin B} = 0

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