A) \[{{\pi }^{2}}\]
B) \[\pi \]
C) \[2\pi \]
D) \[\pi /2\]
Correct Answer: B
Solution :
Period of \[\sin \theta \] and \[\cos \theta \] is \[2\pi \]. Since, \[{{\sin }^{2}}\theta =\frac{1-\cos 2\theta }{2}=\frac{1}{2}-\frac{1}{2}\cos 2\theta \] \[\therefore \] Period of \[{{\sin }^{2}}\theta =\frac{2\pi }{2}=\pi \]
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