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

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

Vedclass · Accepted Answer

$2\pi$

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(D) We use the identity $\sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B)$.Here,$A = \frac{\pi}{8} + \frac{x}{2}$ 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 $A - B = \frac{\pi}{8} + \frac{x}{2} - \left(\frac{\pi}{8} - \frac{x}{2}\right) = x$.So,$f(x) = \sin\left(\frac{\pi}{4}\right) \sin(x) = \frac{1}{\sqrt{2}} \sin(x)$.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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