A) \[{{\pi }^{2}}\]
B) \[\pi \]
C) \[2\pi \]
D) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
\[{{\sin }^{2}}\theta =\frac{1-\cos 2\theta }{2}\]Period \[=\frac{2\pi }{2}=\pi \]You need to login to perform this action.
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