CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    If we express\[\frac{{{(cos2\theta -i\sin 2\theta )}^{4}}{{(cos4\theta +i\sin 4\theta )}^{-5}}}{{{(cos3\theta +i\sin 3\theta )}^{-2}}{{(cos3\theta -i\sin 3\theta )}^{-9}}}\]in the form of\[x+iy,\]we get:

    A)  \[\cos 49\theta -i\sin 49\theta \]

    B)  \[\cos 23\theta +i\sin 23\theta \]

    C)  \[\cos 49\theta +i\sin 49\theta \]

    D)  \[\cos 21\theta +i\sin 21\theta \]

    Correct Answer: A

    Solution :

     \[\frac{{{(\cos 2\theta -i\sin 2\theta )}^{4}}{{(\cos 4\theta +i\sin 4\theta )}^{-5}}}{{{(\cos 3\theta +i\sin 3\theta )}^{-2}}{{(\cos 3\theta -i\sin 3\theta )}^{-9}}}\] \[=\frac{{{(\cos \theta -i\sin \theta )}^{8}}{{(\cos \theta +i\sin \theta )}^{-20}}}{{{(\cos \theta +i\sin \theta )}^{-6}}{{(\cos \theta -i\sin \theta )}^{-27}}}\] \[=\frac{{{(\cos \theta +i\sin \theta )}^{-8}}{{(\cos \theta +i\sin \theta )}^{-20}}}{{{(\cos \theta +i\sin \theta )}^{-6}}{{(\cos \theta +i\sin \theta )}^{27}}}\] \[={{(cos\theta +sin\theta )}^{-27-8-20+6}}\] \[={{(\cos \theta +i\sin \theta )}^{-49}}=\cos 49-i\sin 49\]


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