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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2005

  • question_answer
        The period of\[{{\sin }^{2}}\theta \]is:

    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 \[=\frac{2\pi }{2}=\pi \]


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