The Value of \( \cos^2\left(\frac{\pi}{6}+x\right)-\sin^2\left(\frac{\pi}{6}-x\right) \)
Question
Find the value of
\[ \cos^2\left(\frac{\pi}{6}+x\right) – \sin^2\left(\frac{\pi}{6}-x\right) \]
(a) \(\frac{1}{2}\cos2x\)
(b) \(0\)
(c) \(-\frac{1}{2}\cos2x\)
(d) \(\frac{1}{2}\)
Solution
Use the identities
\[ \cos^2\theta=\frac{1+\cos2\theta}{2} \]
and
\[ \sin^2\theta=\frac{1-\cos2\theta}{2} \]
Therefore,
\[ \cos^2\left(\frac{\pi}{6}+x\right) = \frac{1+\cos\left(\frac{\pi}{3}+2x\right)}{2} \]
\[ \sin^2\left(\frac{\pi}{6}-x\right) = \frac{1-\cos\left(\frac{\pi}{3}-2x\right)}{2} \]
Subtracting,
\[ \frac{\cos\left(\frac{\pi}{3}+2x\right) +\cos\left(\frac{\pi}{3}-2x\right)}{2} \]
Using
\[ \cos C+\cos D = 2\cos\frac{C+D}{2} \cos\frac{C-D}{2} \]
we get
\[ = \frac{ 2\cos\frac{2\pi/3}{2} \cos\frac{4x}{2} }{2} \]
\[ = \cos\frac{\pi}{3}\cos2x \]
Since
\[ \cos\frac{\pi}{3} = \frac{1}{2} \]
\[ = \frac{1}{2}\cos2x \]
Final Answer
\[ \boxed{\frac{1}{2}\cos2x} \]
Hence, the correct option is (a) \(\frac{1}{2}\cos2x\).