$\frac{s i n 7 x + 6 s i n 5 x + 17 s i n 3 x + 12 s i n x}{s i n 6 x + 5 s i n 4 x + 12 s i n 2 x}$ is equal to

cos x

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2 cos x

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sin x

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2 sin x

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Solution

The correct option is D 2 cos x
$\frac{sin 7 x + 6 sin 5 x + 17 sin 3 x + 12 sin x}{sin 6 x + 5 sin 4 x + 12 sin 2 x}$
$= \frac{sin 7 x + sin 5 x + 5 sin 5 x + 5 sin 3 x + 12 sin 3 x + 12 sin x}{sin 6 x + 5 sin 4 x + 12 sin 2 x}$

we know that $sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}$
$= \frac{2 sin 6 x cos x + 10 sin 4 x cos x + 24 sin 2 x cos x}{sin 6 x + 5 sin 4 x + 12 sin 2 x}$
$= \frac{2 cos x [sin 6 x + 5 sin 4 x + 12 sin 2 x]}{sin 6 x + 5 sin 4 x + 12 sin 2 x}$
$= 2 cos x$

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