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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah that should be zero right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. So what is \[\sum_{k=1}^{10} [\cos (2k \pi/11)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah.. isn't cos(2kpi) =1?? where k is any constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Apparently 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Uh quick question. Is the /11 outside or inside the cosine?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you know what my book says the correct answer is i. i dont know how the hell that shows up!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0inside. \[\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Weird, mathematica is telling me otherwise. Sorry.
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