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anonymous

  • 5 years ago

\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]

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  1. anonymous
    • 5 years ago
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    \sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]

  2. anonymous
    • 5 years ago
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    \[\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]\]

  3. anonymous
    • 5 years ago
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    \[\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]\]

  4. anonymous
    • 5 years ago
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    I guess it doesn't work when you are entering your question. So do you know what \[\sum_{k=1}^{10} [\sin (2k \pi/11)\] is?

  5. anonymous
    • 5 years ago
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    yeah that should be zero right?

  6. anonymous
    • 5 years ago
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    Yeah. So what is \[\sum_{k=1}^{10} [\cos (2k \pi/11)\]

  7. anonymous
    • 5 years ago
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    10?

  8. anonymous
    • 5 years ago
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    It is cos(2pi/11) + cos(4pi/11) + ... + cos(20pi/11) which I don't like at all. Are you sure you have the right question?

  9. anonymous
    • 5 years ago
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    yeah.. isn't cos(2kpi) =1?? where k is any constant

  10. anonymous
    • 5 years ago
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    Apparently it is -10 I plugged it into MAthematica to be honest. So 0 is the real part and -10 is the imaginary part. I'm not aware of any easy way of solving the summation of cos(2kpi/11)

  11. anonymous
    • 5 years ago
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    Uh quick question. Is the /11 outside or inside the cosine?

  12. anonymous
    • 5 years ago
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    you know what my book says the correct answer is -i. i dont know how the hell that shows up!

  13. anonymous
    • 5 years ago
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    inside. \[\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]\]

  14. anonymous
    • 5 years ago
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    Weird, mathematica is telling me otherwise. Sorry.

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