Differentiation Question 348
Question: If $ y=\sin x\sin 3x, $ then $ y _{n}= $
Options:
A) $ \frac{1}{2}[ \cos ( 2x+n\frac{\pi }{2} )-\cos ( 4x+n\frac{\pi }{2} ) ] $
B) $ \frac{1}{2}[ {2^{n}}\cos ( 2x+n\frac{\pi }{2} )-4^{n}\cos ( 4x+n\frac{\pi }{2} ) ] $
C) $ \frac{1}{2}[ 4^{n}\cos ( 4x+n\frac{\pi }{2} )-2^{n}\cos ( 2x+n\frac{\pi }{2} ) ] $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \sin x\sin 3x=\frac{1}{2}[\cos 2x-\cos 4x] $ .
BETA
