anonymous
  • anonymous
\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]
anonymous
  • anonymous
\[\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]\]
anonymous
  • anonymous
\[\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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?
anonymous
  • anonymous
yeah that should be zero right?
anonymous
  • anonymous
Yeah. So what is \[\sum_{k=1}^{10} [\cos (2k \pi/11)\]
anonymous
  • anonymous
10?
anonymous
  • anonymous
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?
anonymous
  • anonymous
yeah.. isn't cos(2kpi) =1?? where k is any constant
anonymous
  • anonymous
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)
anonymous
  • anonymous
Uh quick question. Is the /11 outside or inside the cosine?
anonymous
  • anonymous
you know what my book says the correct answer is -i. i dont know how the hell that shows up!
anonymous
  • anonymous
inside. \[\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]\]
anonymous
  • anonymous
Weird, mathematica is telling me otherwise. Sorry.

Looking for something else?

Not the answer you are looking for? Search for more explanations.