The maximum value of cos^2 left( frac{pi}{3} | vedclass

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(C) Using the identity $\cos^2 A - \cos^2 B = \sin(B - A) \sin(B + A)$:Let $A = \frac{\pi}{3} - x$ and $B = \frac{\pi}{3} + x$.Then $B - A = (\frac{\p

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$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(C) Using the identity $\cos^2 A - \cos^2 B = \sin(B - A) \sin(B + A)$:Let $A = \frac{\pi}{3} - x$ and $B = \frac{\pi}{3} + x$.Then $B - A = (\frac{\pi}{3} + x) - (\frac{\pi}{3} - x) = 2x$.And $B + A = (\frac{\pi}{3} + x) + (\frac{\pi}{3} - x) = \frac{2\pi}{3}$.Therefore,$\cos^2 \left( \frac{\pi}{3} - x ight) - \cos^2 \left( \frac{\pi}{3} + x ight) = \sin(2x) \sin\left( \frac{2\pi}{3} ight)$.Since $\sin\left( \frac{2\pi}{3} ight) = \frac{\sqrt{3}}{2}$,the expression becomes $\frac{\sqrt{3}}{2} \sin(2x)$.The maximum value of $\sin(2x)$ is $1$.Thus,the maximum value of the expression is $\frac{\sqrt{3}}{2} 	imes 1 = \frac{\sqrt{3}}{2}$.

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

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