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) $\frac{1}{2} \cos 2 x$
$\cos ^{2}\left(\frac{\pi}{6}+x\right)-\sin ^{2}\left(\frac{\pi}{6}-x\right)$
$=\cos \left(\frac{\pi}{6}+x+\frac{\pi}{6}-x\right) \cos \left(\frac{\pi}{6}+x-\frac{\pi}{6}+x\right) \quad\left[\right.$ Using $\left.\cos (A+B) \cos (A-B)=\cos ^{2} A-\sin ^{2} B\right]$
$=\cos \frac{2 \pi}{6} \cos 2 x$
$=\frac{1}{2} \cos 2 x \quad\left[\right.$ As $\left.\cos \frac{\pi}{3}=\frac{1}{2}\right]$