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anonymous

  • 5 years ago

sum of (sin 2kx/11 + i cos2kx/11) from k=1 to k=10 where i=square root of negative one

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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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    what do you want to find here? ^_^

  3. anonymous
    • 5 years ago
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    the value of this summation=)

  4. anonymous
    • 5 years ago
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    Well. First take 1/11 as a common factor and cut the summation to two parts.

  5. anonymous
    • 5 years ago
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    sorry common factor??

  6. anonymous
    • 5 years ago
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    You will see that the sum of (sin2k pi) is always zero for any value for k=1,2,3,..

  7. anonymous
    • 5 years ago
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    Yeah.

  8. anonymous
    • 5 years ago
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    sin (2k pi) is always zero for k=1,2,3,..*

  9. anonymous
    • 5 years ago
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    yes.. so the sine part comes out to be zero. what about the cosine part?

  10. anonymous
    • 5 years ago
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    on the contract cos (2k pi) is one for k=1,2,..

  11. anonymous
    • 5 years ago
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    So the summation of cos (2k pi) is from k=1 to k=10 is equal to 10. right?

  12. anonymous
    • 5 years ago
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    Are you following? :)

  13. anonymous
    • 5 years ago
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    well the answer is supposed to be -i

  14. anonymous
    • 5 years ago
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    \[{1 \over 11}\sum_{k=1}^{10}\sin(2k \pi)+{i \over 11}\sum_{k=1}^{10}\cos (2k \pi)=0+{i \over 11}(10)\]

  15. anonymous
    • 5 years ago
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    Are you sure?

  16. anonymous
    • 5 years ago
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    wait let me check again.

  17. anonymous
    • 5 years ago
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    Is 11 part of the angle?

  18. anonymous
    • 5 years ago
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    1/11 is not multiplied to sin but its multiplied to its angle

  19. anonymous
    • 5 years ago
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    isn't cos(20 pi) = -1? which will give you -i/11?

  20. anonymous
    • 5 years ago
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    LOL

  21. anonymous
    • 5 years ago
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    what x_x

  22. anonymous
    • 5 years ago
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    and 1/11 = not part of the angle

  23. anonymous
    • 5 years ago
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    You should have put that in parenthesis >.<

  24. anonymous
    • 5 years ago
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    why didn't you put it in the angle =_=...

  25. anonymous
    • 5 years ago
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    Well, you got a solution for "a" problem :P

  26. anonymous
    • 5 years ago
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    hey anwar, aren't you supposed to find the summation of both sin and cos from 1 - 10?

  27. anonymous
    • 5 years ago
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    sorry about that guys

  28. anonymous
    • 5 years ago
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    Yeah sstarica

  29. anonymous
    • 5 years ago
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    It's OK.. But, I am not sure if I have enough time to do the problem. Just let me think about it quickly.

  30. anonymous
    • 5 years ago
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    okay anwar

  31. anonymous
    • 5 years ago
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    is it possible to solve using euler's representation of complex no.?

  32. anonymous
    • 5 years ago
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    like this : \[[\sum_{k=1}^{10}\sin(\frac{2 \pi}{11}) + i \cos(\frac{2\pi}{11}) ] + [\sum_{k=1}^{10}\sin(\frac{2(10)\pi}{11}) + i \cos(\frac{2(10) \pi}{11})]\] ? .-. right

  33. anonymous
    • 5 years ago
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    That's what I am thinking about right now.

  34. anonymous
    • 5 years ago
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    lol

  35. anonymous
    • 5 years ago
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    yeah all through 2,3,4 to 10

  36. anonymous
    • 5 years ago
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    @sstarica: you just substitute for k=1,2,3,.. and add them up.

  37. anonymous
    • 5 years ago
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    no need, there's a faster way to compute the sum without going through the 1 2 3 stages :)

  38. anonymous
    • 5 years ago
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    take first term and add it with last term

  39. anonymous
    • 5 years ago
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    and whats that?

  40. anonymous
    • 5 years ago
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    something like that?

  41. anonymous
    • 5 years ago
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    I don't know lol

  42. anonymous
    • 5 years ago
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    Well yeah.. You just need to see a pattern.

  43. anonymous
    • 5 years ago
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    lol..

  44. anonymous
    • 5 years ago
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    my prof did that last time, he took the first term and added it up with the last term, then got the answer, or it's prolly one of my hallucinations? I just remembered he did something like that

  45. anonymous
    • 5 years ago
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    z_z oh nvm me proceed.

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

  47. anonymous
    • 5 years ago
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    i dont know how to proceed any further

  48. anonymous
    • 5 years ago
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    that's right, but there is an i in the power.

  49. anonymous
    • 5 years ago
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    there is?

  50. anonymous
    • 5 years ago
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    There should be :)

  51. anonymous
    • 5 years ago
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    o.o

  52. anonymous
    • 5 years ago
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    what?!

  53. anonymous
    • 5 years ago
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    yea there should be.. otherwise its not a correct representation of the complex number

  54. anonymous
    • 5 years ago
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    o..o

  55. anonymous
    • 5 years ago
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    \[i \sum_{k=1}^{10} (e ^{-2k \pi i/11})\] should look like this

  56. anonymous
    • 5 years ago
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    =.=

  57. anonymous
    • 5 years ago
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    hey star u workin on it?

  58. anonymous
    • 5 years ago
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    I gtg now.. I'll BRB

  59. anonymous
    • 5 years ago
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    bye ^_^

  60. anonymous
    • 5 years ago
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    bye

  61. anonymous
    • 5 years ago
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    I am back, but you're gone. :(

  62. anonymous
    • 5 years ago
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    I think I am always getting the idea.

  63. anonymous
    • 5 years ago
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    not always, almost :)

  64. anonymous
    • 5 years ago
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    giv it a try..=)

  65. anonymous
    • 5 years ago
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    sin(2 pi/11)+sin(20 pi/11)=?

  66. anonymous
    • 5 years ago
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    Hello, anyone there?

  67. anonymous
    • 5 years ago
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    sin x+sin y=2sin((x+y)/2)+....., right?

  68. anonymous
    • 5 years ago
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    I have no idea lol

  69. anonymous
    • 5 years ago
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    I'm trying to fix the DAMN PRINTER! >_<

  70. anonymous
    • 5 years ago
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    Apply it to sin(2 pi/11)+sin(20 pi/11)

  71. anonymous
    • 5 years ago
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    excuse my language ^^"

  72. anonymous
    • 5 years ago
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    sin((2 * pi) / 11) + sin((20 * pi) / 11) = 0

  73. anonymous
    • 5 years ago
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    It will give you 0. Same with sin(4 pi/11)+sin(18 pi/11), and so on.

  74. anonymous
    • 5 years ago
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    yup

  75. anonymous
    • 5 years ago
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    Do you see that the "sin" part of the summation will be 0?

  76. anonymous
    • 5 years ago
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    agreed... wat abt the cosine part

  77. anonymous
    • 5 years ago
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    trying it :).. just give me few minutes.

  78. anonymous
    • 5 years ago
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    1,2,3,4,5,6,7,8,9........60

  79. anonymous
    • 5 years ago
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    yup sure

  80. anonymous
    • 5 years ago
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    what are you saying starica?

  81. anonymous
    • 5 years ago
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    I'm counting ^_^

  82. anonymous
    • 5 years ago
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    3 mins have passed

  83. anonymous
    • 5 years ago
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    its still "few"

  84. anonymous
    • 5 years ago
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    lol yeah~

  85. anonymous
    • 5 years ago
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    =) arent you trying?

  86. anonymous
    • 5 years ago
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    Don't count -.-

  87. anonymous
    • 5 years ago
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    cosx+cosy=2cos((x+y)/2)cos((x-y)/2)

  88. anonymous
    • 5 years ago
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    I'm too tired to try and I love counting lol =P

  89. anonymous
    • 5 years ago
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    it looks simple though u_u

  90. anonymous
    • 5 years ago
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    well you should know..:P

  91. anonymous
    • 5 years ago
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    If I try, I'll figure it out, but I'm tired :( ...

  92. anonymous
    • 5 years ago
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    why dont you go take a nap for a change..

  93. anonymous
    • 5 years ago
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    I'm in campus, and I can't sleep during the day

  94. anonymous
    • 5 years ago
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    r u an undergrad student?

  95. anonymous
    • 5 years ago
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    mhm, a first year ^_^

  96. anonymous
    • 5 years ago
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    gr8.. ehat are you majoring in?

  97. anonymous
    • 5 years ago
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    *what

  98. anonymous
    • 5 years ago
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    IT , computer engineering hopefully.

  99. anonymous
    • 5 years ago
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    I am tired :)

  100. anonymous
    • 5 years ago
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    You said you want to go to chemical engineering, right?

  101. anonymous
    • 5 years ago
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    isnt workin out anwar?:-(

  102. anonymous
    • 5 years ago
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    Not really. My friend is distracting me :P .. He wants me to go with him get lunch :)

  103. anonymous
    • 5 years ago
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    lol.. anyways i'd better leave now.. thanx for your effort anwar and sstarica..=)

  104. anonymous
    • 5 years ago
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    No problem. If I get the idea, I will just post the solution here.

  105. anonymous
    • 5 years ago
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    okay buddy.. bye

  106. anonymous
    • 5 years ago
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    Bye!!

  107. anonymous
    • 5 years ago
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    not really, I'm staying in IT

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