If $A + B + C = 0$ , then $sin A + sin B + sin C =$

2 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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- 2 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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- 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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Solution

The correct option is D $- 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$
As $A + B + C = 0 \Rightarrow C = - (A + B)$
$sin A + sin B + sin C = sin A + sin B + sin (- (A + B)) = sin A + sin B - sin (A + B)$
Using trigonometric identities $sin C + sin D = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})$ and $sin 2 A = 2 sin A cos A$
We get
$sin A + sin B - sin (A + B) = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2}) - 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})$
$= 2 sin (\frac{A + B}{2}) (cos (\frac{A - B}{2}) - cos (\frac{A - B}{2}))$
And using $cos C - cos D = 2 sin (\frac{C + D}{2}) sin (\frac{D - C}{2})$
$2 sin (\frac{A + B}{2}) (cos (\frac{A - B}{2}) - cos (\frac{A + B}{2})) = 2 sin (\frac{A + B}{2}) (2 sin \frac{A}{2} sin \frac{B}{2})$
Substituting $A + B = - C$ we get
$2 sin (\frac{A + B}{2}) (2 sin \frac{A}{2} sin \frac{B}{2}) = 2 sin \frac{- C}{2} (2 sin \frac{A}{2} sin \frac{B}{2}) = - 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$

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If A+B+C=0, then sinA+sinB+sinC=

If $A + B + C = 0$ , then $sin A + sin B + sin C =$