RAJASTHAN ­ PET Rajasthan PET Solved Paper-2004

  • question_answer
    \[{{(1+\cos \theta +i\sin \theta )}^{3}}+{{(1-\cos \theta -i\sin \theta )}^{3}}\]is equal to

    A)  \[{{2}^{4}}{{\cos }^{3}}\left( \frac{\theta }{2} \right)\cos \left( \frac{3\theta }{2} \right)\]

    B)  \[{{2}^{3}}{{\cos }^{3}}\left( \frac{\theta }{2} \right)\cos \left( \frac{3\theta }{2} \right)\]

    C)  \[{{2}^{3}}{{\cos }^{3}}\left( \frac{\theta }{2} \right)\sin \left( \frac{3\theta }{2} \right)\]

    D)  \[{{2}^{4}}{{\cos }^{3}}\left( \frac{\theta }{2} \right)\sin \left( \frac{3\theta }{2} \right)\]

    Correct Answer: A

    Solution :

     \[{{(1+\cos \theta +i\sin \theta )}^{3}}+{{(1+\cos \theta -i\sin \theta )}^{3}}\] \[={{\left[ 1+2{{\cos }^{2}}\frac{\theta }{2}-1+i2\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right]}^{3}}\] \[+{{\left[ 1+2{{\cos }^{2}}\frac{\theta }{2}-1-i2\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right]}^{3}}\] \[={{\left[ 2\cos \frac{\theta }{2}\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right) \right]}^{3}}\] \[+{{\left[ 2\cos \frac{\theta }{2}\left( \cos \frac{\theta }{2}-i\sin \frac{\theta }{2} \right) \right]}^{3}}\] \[=8{{\cos }^{3}}\frac{\theta }{2}\left[ \cos \frac{3\theta }{2}+i\sin \frac{3\theta }{2}+\cos \frac{3\theta }{2}-i\sin \frac{3\theta }{2} \right]\] \[=8{{\cos }^{3}}\frac{\theta }{2}\left[ 2\cos \frac{3\theta }{2} \right]\] \[=16{{\cos }^{3}}\frac{\theta }{2}.\cos \frac{3\theta }{2}\] \[={{2}^{4}}{{\cos }^{3}}\frac{\theta }{2}.\cos \frac{3\theta }{2}\]


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