# 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) $\frac{1}{2} \cos 2 x$

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

$=\cos \left(\frac{\pi}{6}+x+\frac{\pi}{6}-x\right) \cos \left(\frac{\pi}{6}+x-\frac{\pi}{6}+x\right) \quad\left[\right.$ 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 $\left.\cos \frac{\pi}{3}=\frac{1}{2}\right]$