Trigonometric Identities Question 176
Question: The value of $ \frac{\sin 8x+7\sin 6x+18\sin 4x+12\sin 2x}{\sin 7x+6\sin 5x+12\sin 3x} $ equal to?
Options:
A) $ 2\cos x $
B) $ \cos x $
C) $ 2sinx $
D) $ sinx $
Show Answer
Answer:
Correct Answer: A
Solution:
Numerator of given expression $ =(\sin 8x+\sin 6x)+6(\sin 6x+\sin 4x) $ $ +12(\sin 4x+\sin 2x) $ $ =2\sin 7x\cos x+12\sin 5x\cos x+24\sin 3x\cos x $ $ =2\cos x(\sin 7x+6\sin 5x+12\sin 3x) $ Hence, given expression $ =\frac{2\cos x(\sin 7x+6\sin 5x+12sin3x)}{(\sin 7x+6\sin 5x+12\sin 3x)}=2\cos x $
BETA
