Trigonometric-Identities Question 395
Question: $ 1+\cos 2x+\cos 4x+\cos 6x= $
[Roorkee 1974]
Options:
A) $ 2\cos x\cos 2x\cos 3x $
B) $ 4\sin x,\cos 2x\cos 3x $
C) $ 4\cos x\cos 2x\cos 3x $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ 1+\cos 2x+\cos ,4x+\cos ,6x $ $ =(1+\cos ,6x)+(\cos ,2x+\cos ,4x) $ $ =2,{{\cos }^{2}}3x+2,\cos ,3x,\cos x=2,\cos ,3x,(\cos ,3x+\cos ,x) $ $ =4,\cos x,\cos ,2x,\cos ,3x $ .
BETA
