If $A + B + C = 180^{0}$ then $sin 2 A + sin 2 B + sin 2 C = ?$

sin A sin B sin C

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4 sin A sin B sin C

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3 sin A sin B sin C

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2 sin A sin B sin C

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Solution

The correct option is B $4 sin A sin B sin C$

$Solve: Given, A + B + C = 180^{\circ}$

$sin 2 A + sin 2 B + sin 2 C = [sin 2 A + sin 2 B] + sin 2 C$

$\Rightarrow sin 2 A + sin 2 B + sin 2 C = 2 sin (A + B) cos (A - B) + sin 2 C$

$[\begin{matrix} sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2}) \end{matrix}]$

$= 2 sin (180^{\circ} - c) cos (A - B) + sin 2 c$

$= 2 sin c cos (A - B) + 2 sin c cos c$

$= 2 sin c (cos (A - B) + cos c]$

$= 2 sin c [2 cos (\frac{A - B + c}{2}) cos (\frac{A - B - C}{2})]$

$= 4 sinc [cos (\frac{A + c}{2} - \frac{B}{2}) cos (\frac{A}{2} - \frac{(B + C)}{2})]$

$= 4 sin c cos (\frac{π}{2} - \frac{B}{2} - \frac{B}{2}) cos (\frac{A}{2} - \frac{π - A}{2})$

$= 4 sin c sin B cos (\frac{π}{2} - A)$

$= 4 sin A sin B sin c$

$\Rightarrow sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C$

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Similar questions

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If : a+b+c=pie

then prove that : sin2A+sin2B+sin2C= 4sinAsinBsinC

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