If f(x) = sin^2 left( frac{pi}{8} + frac{x}{2 | vedclass

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(D) We use the trigonometric identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$.Let $A = \frac{\pi}{8} + \frac{x}{2}$ and $B = \frac{\pi}{8} - \frac

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$2\pi$

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(D) We use the trigonometric identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$.Let $A = \frac{\pi}{8} + \frac{x}{2}$ and $B = \frac{\pi}{8} - \frac{x}{2}$.Then $A+B = \frac{\pi}{8} + \frac{x}{2} + \frac{\pi}{8} - \frac{x}{2} = \frac{\pi}{4}$.And $A-B = \frac{\pi}{8} + \frac{x}{2} - \left( \frac{\pi}{8} - \frac{x}{2} \right) = x$.Thus,$f(x) = \sin \left( \frac{\pi}{4} \right) \sin(x) = \frac{1}{\sqrt{2}} \sin(x)$.The period of $\sin(x)$ is $2\pi$.Therefore,the period of $f(x)$ is $2\pi$.

If $f(x) = \sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right)$,then the period of $f$ is

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