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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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Verified Answer
The correct answer is:
$2 \pi$
We have,
$$
\begin{aligned}
& f(x)=\sin ^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin ^2\left(\frac{\pi}{8}-\frac{x}{2}\right) \\
& f(x)=\sin \left(\frac{\pi}{8}+\frac{x}{2}+\frac{\pi}{8}-\frac{x}{2}\right) \sin \left(\frac{\pi}{8}+\frac{x}{2}-\frac{\pi}{8}+\frac{x}{2}\right)
\end{aligned}
$$
$$
=\sin \frac{\pi}{4} \sin x=\frac{1}{\sqrt{2}} \sin x
$$
$\therefore$ Period of $f(x)$ is $2 \pi$
[ $\because$ Period of $\sin x$ is $2 \pi$ ]
$$
\begin{aligned}
& f(x)=\sin ^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin ^2\left(\frac{\pi}{8}-\frac{x}{2}\right) \\
& f(x)=\sin \left(\frac{\pi}{8}+\frac{x}{2}+\frac{\pi}{8}-\frac{x}{2}\right) \sin \left(\frac{\pi}{8}+\frac{x}{2}-\frac{\pi}{8}+\frac{x}{2}\right)
\end{aligned}
$$
$$
=\sin \frac{\pi}{4} \sin x=\frac{1}{\sqrt{2}} \sin x
$$
$\therefore$ Period of $f(x)$ is $2 \pi$
[ $\because$ Period of $\sin x$ is $2 \pi$ ]
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