BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    If\[{{x}_{r}}=\cos \left( \frac{\pi }{{{3}^{r}}} \right)-i\sin \left( \frac{\pi }{{{3}^{r}}} \right),\](where\[i=\sqrt{-1}\]),  then the value of\[{{x}_{1}}.{{x}_{2}}...\infty ,\]is

    A)  1                                            

    B)  \[-1\]                   

    C)  \[-\,i\]                 

    D)         \[\,i\]

    Correct Answer: C

    Solution :

    Since, \[{{x}_{f}}=\cos \left( \frac{\pi }{{{3}^{r}}} \right)-i\sin \left( \frac{\pi }{3} \right)\] \[{{x}_{1}}.{{x}_{2}}.{{x}_{3}}...\infty =\cos \left( \frac{\pi }{{{3}^{1}}}+\frac{\pi }{{{3}^{2}}}+\frac{\pi }{{{3}^{3}}}+...\infty  \right)\] \[-i\sin \left( \frac{\pi }{{{3}^{1}}}+\frac{\pi }{{{3}^{2}}}+\frac{\pi }{{{3}^{3}}}+...\infty  \right)\] \[=\cos \left( \frac{\frac{\pi }{3}}{1-\frac{1}{3}} \right)-i\sin \left( \frac{\frac{\pi }{3}}{1-\frac{1}{3}} \right)\] \[=\cos \left( \frac{\pi }{2} \right)-i\sin \left( \frac{\pi }{2} \right)=-i\]


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