# The value of cos

Question:

The value of $\cos ^{2}\left(\frac{\pi}{6}+x\right)-\sin ^{2}\left(\frac{\pi}{6}-x\right)$ is

(a) $\frac{1}{2} \cos 2 x$

(b) 0

(c) $-\frac{1}{2} \cos 2 x$

(d) $\frac{1}{2}$

Solution:

(a) $\frac{1}{2} \cos 2 x$

We have,

$\cos ^{2}\left(\frac{\pi}{6}+x\right)-\sin ^{2}\left(\frac{\pi}{6}-x\right)$

$=\cos ^{2}\left(\frac{\pi}{6}+x\right)-\cos ^{2}\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]$

$=\cos ^{2}\left(\frac{\pi}{6}+x\right)-\cos ^{2}\left(\frac{\pi}{3}+x\right)$

$=\left[\cos \left(\frac{\pi}{6}+x\right)+\cos \left(\frac{\pi}{3}+x\right)\right]\left[\cos \left(\frac{\pi}{6}+x\right)-\cos \left(\frac{\pi}{3}+x\right)\right]$

$=2 \cos \left(\frac{\frac{\pi}{6}+x+\frac{\pi}{3}+x}{2}\right) \cos \left(\frac{\frac{\pi}{6}+x-\frac{\pi}{3}-x}{2}\right) 2 \sin \left(\frac{\frac{\pi}{6}+x+\frac{\pi}{3}+x}{2}\right) \sin \left(\frac{\frac{\pi}{3}+x-\frac{\pi}{6}-x}{2}\right)$

$=4 \cos \left(\frac{\pi}{4}+x\right) \cos \left(-\frac{\pi}{12}\right) \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{12}\right)$

$=4 \cos \left(\frac{\pi}{4}+x\right) \cos \left(\frac{\pi}{12}\right) \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{12}\right)$

$=\left[2 \sin \left(\frac{\pi}{4}+x\right) \cos \left(\frac{\pi}{4}+x\right)\right]\left[2 \sin \left(\frac{\pi}{12}\right) \cos \left(\frac{\pi}{12}\right)\right]$

$=\sin \left(\frac{\pi}{2}+2 x\right) \sin \frac{\pi}{6}$

$=\cos 2 x \times \frac{1}{2}$

$=\frac{1}{2} \cos 2 x$