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The value of ...

Question

The value of $\cos^{2} (\frac{π}{6} + x) - \sin^{2} (\frac{π}{6} - x)$ is
(a) $\frac{1}{2} \cos 2 x$

(b) 0

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

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

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Solution

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

$We have, \cos^{2} (\frac{π}{6} + x) - \sin^{2} (\frac{π}{6} - x) = \cos^{2} (\frac{π}{6} + x) - \cos^{2} [\frac{π}{2} - (\frac{π}{6} - x)] = \cos^{2} (\frac{π}{6} + x) - \cos^{2} (\frac{π}{3} + x) = [\cos (\frac{π}{6} + x) + \cos (\frac{π}{3} + x)] [\cos (\frac{π}{6} + x) - \cos (\frac{π}{3} + x)] = 2 \cos (\frac{\frac{π}{6} + x + \frac{π}{3} + x}{2}) \cos (\frac{\frac{π}{6} + x - \frac{π}{3} - x}{2}) 2 \sin (\frac{\frac{π}{6} + x + \frac{π}{3} + x}{2}) \sin (\frac{\frac{π}{3} + x - \frac{π}{6} - x}{2}) = 4 \cos (\frac{π}{4} + x) \cos (- \frac{π}{12}) \sin (\frac{π}{4} + x) \sin (\frac{π}{12}) = 4 \cos (\frac{π}{4} + x) \cos (\frac{π}{12}) \sin (\frac{π}{4} + x) \sin (\frac{π}{12}) = [2 \sin (\frac{π}{4} + x) \cos (\frac{π}{4} + x)] [2 \sin (\frac{π}{12}) \cos (\frac{π}{12})] = \sin (\frac{π}{2} + 2 x) \sin \frac{π}{6} = \cos 2 x \times \frac{1}{2} = \frac{1}{2} \cos 2 x$

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