Question

For all values of $\theta ,$ cos $2\theta$ (cos $6\theta$ +cos$2\theta )$ is

• A. $\ge -\frac{1}{4}$
• B. $\ge \frac{1}{8}$
• C. $\ge \frac{7}{8}$
• D. $\ge -\frac{1}{8}$

Answer 1

The correct option is A $\ge -\frac{1}{4}$

$\cos 2\theta (\cos 6\theta +\cos 2\theta )$

$=\frac{1}{2}\times 2\times \cos 2\theta (\cos 6\theta +\cos 2\theta )$

$=\frac{1}{2}[\cos 8\theta +\cos 4\theta +2{\cos }^{2}2\theta ]$

$=\frac{1}{2}[\cos 8\theta +\cos 4\theta +\cos 4\theta +1]$

$=\frac{1}{2}[2{\cos }^{2}4\theta -1+2\cos 4\theta +1]$

$={\cos }^{2}4\theta +\cos 4\theta$

$={\cos }^{2}4\theta +2\cdot \frac{1}{2}\cdot \cos 4\theta +\frac{1}{4}-\frac{1}{4}$

$=(\cos 4\theta +\frac{1}{2}{)}^2-\frac{1}{4}$

$(\cos 4\theta +\frac{1}{2}{)}^2$ is always $\ge 0$

So, $\cos 2\theta (\cos 6\theta +\cos 2\theta )\ge \frac{-1}{4}$