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$\frac{\sin 8 x + 7 \sin 6 x + 18 \sin 4 x + 12 \sin 2 x}{\sin 7 x + 6 \sin 5 x + 12 \sin 3 x} =$

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a

sinx

b

cosx

c

2sinx

d

2cosx

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detailed solution

Correct option is D

$\sin 8 x + 7 \sin 6 x + 18 \sin 4 x + 12 \sin 2 x = \sin 8 x + \sin 6 x + 6 (\sin 6 x + \sin 4 x) + 12 (\sin 4 x + \sin 2 x) = 2 \sin 7 x \cos x + 6.2 \sin 5 x \cos x + 12.2 \sin 3 x \cos x = 2 \cos x (\sin 7 x + 6 \sin 5 x + 12 \sin 3 x) ∵ T h e g i v e n f r a c t i o i n i s 2 \cos x$

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