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Q. The maximum value of $\cos ^2\left(\frac{\pi}{3}-x\right)-\cos ^2$ $\left(\frac{\pi}{3}+x\right)$ is

Trigonometric Functions

Solution:

$\cos ^2\left(\frac{\pi}{3}-x\right)-\cos ^2\left(\frac{\pi}{3}+x\right) $
$=\sin \left(\frac{\pi}{3}-x+\frac{\pi}{3}+x\right) \sin \left(\frac{\pi}{3}+x-\frac{\pi}{3}+x\right) $
$=\sin \frac{2 \pi}{3} \sin 2 x=\frac{\sqrt{3}}{2} \sin 2 x$
(since, maximum value of $\sin 2 x$ is 1 )
Therefore, its maximum value is $\frac{\sqrt{3}}{2}$.