Trigonometric Identities Question 294
Question: If $ \sin 6\theta =32{{\cos }^{5}}\theta \sin \theta -32{{\cos }^{3}}\theta \sin \theta +3x, $ then $ x= $
[EAMCET 2003]
Options:
A) $ \cos \theta $
B) $ \cos 2\theta $
C) $ \sin \theta $
D) $ \sin 2\theta $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \sin 6\theta =2\sin 3\theta \cos 3\theta $ $ =2,[3\sin \theta -4{{\sin }^{3}}\theta ],[4{{\cos }^{3}}\theta -3\cos \theta ] $ =24sinqcosq(sin2q+cos2q) -18sinqcosq - 32sin2qcos2q $ =32{{\cos }^{5}}\theta \sin \theta -32{{\cos }^{3}}\theta \sin \theta +3\sin 2\theta $ On comparing, $ x=\sin 2\theta . $
BETA
