Question

$4\cos 6\theta \cos 4\theta \cos 2\theta$ is equal to

• A. $\cos 12\theta +\cos 8\theta +\cos 4\theta +1$
• B. $\cos 12\theta +\cos 8\theta -\cos 4\theta +1$
• C. $\cos 12\theta -\cos 8\theta +\cos 4\theta +1$
• D. $\cos 12\theta -\cos 8\theta -\cos 4\theta -1$

Answer 1

The correct option is A $\cos 12\theta +\cos 8\theta +\cos 4\theta +1$

Let $4\cos 6\theta \cos 4\theta \cos 2\theta =I$
Using $2\cos A\cos B=\cos (A-B)+\cos (A+B)$
$I=2(\cos 2\theta +\cos 10\theta )\cos 2\theta$

$=2{\cos }^{2}2\theta +2\cos 10\theta \cos 2\theta$
Again using $\cos 2A=2{\cos }^{2}A-1$
$I=cos4\theta +1+\cos 8\theta +\cos 12\theta$