CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer If\[n=1,\text{ }2,\text{ }3,\text{ }....\]then\[\cos \alpha \cos 2\alpha \cos 4\alpha .....\cos {{2}^{n-1}}\alpha \] is equal to:

    A)  \[\frac{\sin 2n\alpha }{2n\sin \alpha }\]                               

    B)  \[\frac{\sin {{2}^{n}}\alpha }{{{2}^{n}}\sin {{2}^{n-1}}\alpha }\]

    C)         \[\frac{\sin {{4}^{n-1}}\alpha }{{{4}^{n-1}}\sin \alpha }\]              

    D)         \[\frac{\sin {{2}^{n}}\alpha }{{{2}^{n}}\sin \alpha }\]

    E)  none of these

    Correct Answer: D

    Solution :

    \[\cos \alpha \,\cos 2\alpha \cos 4\alpha ....\cos {{2}^{n-1}}\alpha =\frac{\sin {{2}^{n}}\alpha }{{{2}^{n}}\sin \alpha }\]

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