In any $△ A B C$ , the value of $s i n 2 A + s i n 2 B + s i n 2 C =$ ?

4 s i n A . s i n B . s i n C

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4 c o s A . c o s B . c o s C

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\frac{3}{2} s i n A . c o s B

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4 c o s A . s i n B . c o s C

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Solution

The correct option is C $4 s i n A . s i n B . s i n C$
Sum of the angle of a triangle is equal to $180^{\circ}$

$∠ A + ∠ B + ∠ C = 180^{\circ}$

$∠ B + ∠ C = 180^{\circ} - ∠ A$

$s i n (B + C) = s i n (180^{\circ} - A) = s i n A$

Use the trigonometry identities, $s i n C + s i n D = 2 s i n \frac{C + D}{2} c o s \frac{C - D}{2}$

So,

$S i n 2 A + s i n 2 B + s i n 2 C = s i n 2 A + 2 s i n (\frac{2 \times (B + C)}{2}) c o s (\frac{2 \times (B - C)}{2})$

$= s i n 2 A + 2 s i n (B + C) c o s (B - C) = s i n 2 A + 2 s i n A c o s (B - C)$

We know that $s i n 2 A = 2 s i n A \times c o s A$

Therefore,

$= 2 s i n A c o s A + 2 s i n A c o s (B - C) = 2 s i n A (c o s A + c o s (B - C)) = 2 s i n A (c o s (180^{\circ} - (B + C)) + c o s (B - C))$

$= 2 s i n A (- c o s (B + C) + c o s (B - C))$

We know that $c o s (A + B) = c o s A c o s B - s i n A s i n B$ and $c o s (A - B) = c o s A c o s B + s i n A s i n B$

$= 2 s i n A (c o s A c o s B + s i n A s i n B - (c o s A c o s B - s i n A s i n B)) = 4 s i n A \times s i n B \times s i n C$

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