Find the Value of cos²(π/6 + x) − sin²(π/6 − x)

Solution

Use the identities:

\[ \cos^2\theta=\frac{1+\cos2\theta}{2} \]

and

\[ \sin^2\theta=\frac{1-\cos2\theta}{2} \]

Therefore,

\[ \cos^2\left(\frac{\pi}{6}+x\right) = \frac{ 1+\cos\left(\frac{\pi}{3}+2x\right) }{2} \]

Also,

\[ \sin^2\left(\frac{\pi}{6}-x\right) = \frac{ 1-\cos\left(\frac{\pi}{3}-2x\right) }{2} \]

Subtracting,

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

\[ = \frac{ 1+\cos\left(\frac{\pi}{3}+2x\right) }{2} – \frac{ 1-\cos\left(\frac{\pi}{3}-2x\right) }{2} \]

\[ = \frac{ \cos\left(\frac{\pi}{3}+2x\right) + \cos\left(\frac{\pi}{3}-2x\right) }{2} \]

Using the identity:

\[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]

we get

\[ = \frac{ 2\cos\frac{2\pi/3}{2}\cos\frac{4x}{2} }{2} \]

\[ = \cos\frac{\pi}{3}\cos2x \]

Since

\[ \cos\frac{\pi}{3}=\frac{1}{2} \]

Therefore,

\[ \boxed{ \cos^2\left(\frac{\pi}{6}+x\right) – \sin^2\left(\frac{\pi}{6}-x\right) = \frac{1}{2}\cos2x } \]

Final Answer

\[ \boxed{ \cos^2\left(\frac{\pi}{6}+x\right) – \sin^2\left(\frac{\pi}{6}-x\right) = \frac{1}{2}\cos2x } \]

Correct Option: (a)