Question

The value of ${\cos }^{2}x+{\cos }^2\left(x+\frac{\pi }{3}\right)+{\cos }^2\left(x-\frac{\pi }{3}\right)$ is equal to

• A. $\frac{1}{2}$
• B. 2
• C. 3
• D. None of the above

Answer 1

Answer: (D)

We have, ${\cos }^{2}x+{\cos }^2\left(x+\frac{\pi }{3}\right)+{\cos }^2\left(x-\frac{\pi }{3}\right)$
$=\frac{1+\cos 2x}{2}+\frac{1+\cos \left(2x+\frac{2\pi }{3}\right)}{2}+\frac{1+\cos \left(2x-\frac{2\pi }{3}\right)}{2}$
$=\frac{1}{2}\left[3+\cos 2x+\cos \left(2x+\frac{2\pi }{3}\right)+\cos \left(2x-\frac{2\pi }{3}\right)\right]$
$=\frac{1}{2}\left[3+\cos 2x+2\cos 2x\cos \frac{2\pi }{3}\right]$
$=\frac{1}{2}[3+\cos 2x+2\cos 2x\cos \left(\pi -\frac{2\pi }{3}\right)$
$=\frac{1}{2}\left[3+\cos 2x-2\cos 2x\cos \frac{\pi }{3}\right]$
$=\frac{1}{2}[3+\cos 2x-\cos 2x]=\frac{3}{2}$