CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    \[\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{{{(\sin \theta +i\cos \theta )}^{5}}}\]is equal to:

    A)  \[\cos \theta -i\sin \theta \]

    B)         \[\sin \theta -i\cos \theta \]

    C)  \[\cos 9\theta -i\sin 9\theta \]

    D)         \[\sin 9\theta -i\cos 9\theta \]

    E)  \[\cos \theta +i\sin \theta \]

    Correct Answer: D

    Solution :

    \[\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{{{(\sin \theta +i\cos \theta )}^{5}}}=\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{{{i}^{5}}{{(\cos \theta -i\sin \theta )}^{5}}}\] \[=-i{{(\cos \theta +i\sin \theta )}^{4}}{{(\cos \theta +i\sin \theta )}^{5}}\] \[=-i(\cos 9\theta +i\sin 9\theta )\] \[=\sin 9\theta +i\cos 9\theta \]


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