A) \[\pi \]
B) \[2\pi \]
C) \[\frac{\pi }{2}\]
D) None of these\[\]
Correct Answer: A
Solution :
\[{{\sin }^{2}}x=\frac{1-\cos 2x}{2}\]Þ Period\[=\frac{2\pi }{2}=\pi \].
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