CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    If\[a=\frac{\pi }{18}\]rad then\[\cos a+\cos 2a+...+\cos 18a\] is equal to:

    A)  0                            

    B)         \[-1\]                   

    C)  1                            

    D)         \[\pm 1\]

    E)  none of these

    Correct Answer: B

    Solution :

    \[\cos a+\cos 2a+....+\cos 18a\] \[=\cos \frac{\pi }{18}+\cos \frac{\pi }{18}+.....\]                                 \[+\cos \frac{16\pi }{18}+\cos \frac{17\pi }{18}+\cos \pi \] \[=\cos \frac{\pi }{18}+\cos \frac{2\pi }{18}+....\]                                 \[-\cos \frac{2\pi }{18}-\cos \frac{\pi }{18}+\cos \pi \] \[=\cos \pi \] \[=-1\]


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