A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ParthKohli

  • one year ago

@ganeshie8 @Miracrown @dan815 @UsukiDoll

  • This Question is Closed
  1. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[\cot(\pi/15) - 4\sin(\pi/15) = \sqrt{15}\]

  2. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    My work so far. This expression is equivalent to \(\cot(16\pi/15) + 4\sin(16\pi/15)\). \[x := e^{2\pi i /11}\]Then:\[\cot(16\pi/15) = \frac{\cos(16\pi/15)}{\sin(16\pi/15) } = \dfrac{(x^4 + \overline{x^4})/2}{(x^4 - \overline{x^4})/2i}\]\[\Rightarrow i\cot(16\pi/15) = \dfrac{x^{11}+ x^4}{x^{11} - x^4} = \dfrac{x^7+1}{x^7 - 1}\]

  3. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Similarly,\[4i \sin(16\pi/15) = 2(x^4 - x^{11})\]

  4. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I quit... what is this?!

  5. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Complex numbers.

  6. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    like complex analysis?

  7. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    If we add those two, we get\[i \cot(16\pi/15) + 4i \sin(16\pi/15) = \frac{x^7+1}{x^7 - 1} + 2(x^4 - x^{11})\]Somehow, I want to make that fraction a polynomial using properties of roots of unity. Any ideas?

  8. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    @UsukiDoll No, I'm just trying to prove a trigonometric identity using complex numbers. Specifically, I mean roots of unity. Is that complex analysis? I don't know.

  9. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm still a noob when it comes to proofs D:

  10. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Also, we have to show that the right-hand side is \(i\sqrt{15}\) and we'll be done.

  11. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    I know I can write \(1\) as \(x^{15}\) or \(x^{30}\) or so on. Which exponent should I choose so as to make it divisible by \(x^7 + 1\)?

  12. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \(x^7 - 1\) actually, which means that we're also looking at a seventh root of unity.

  13. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    @ganeshie8 Speak out. :P

  14. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1433849058768:dw|

  15. Phebe
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    LOL

  16. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[x^7 - 1 \mid x^{15n}+x^7\]

  17. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im still going through... you meant \(x := e^{2\pi i /15}\) right

  18. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Yes, that's what I meant. Whoops.

  19. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let me grab pen and paper

  20. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it's skully!

  21. skullpatrol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    A pencil is preferable :P

  22. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1433850807361:dw| pacman

  23. skullpatrol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :D

  24. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Dan just made me realise that this problem is lame and we shouldn't be working on it as much as we are right now.

  25. Phebe
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    lolzz

  26. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    This problem is also really hard.

  27. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if it's hard then why are we viewing it? rofl xD

  28. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    To help me solve it.

  29. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but we're not pros at this D:

  30. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Then why are you viewing it?

  31. skullpatrol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Try MSE

  32. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Know what, this problem was asked at MSE, and is currently unanswered.

  33. skullpatrol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :O

  34. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1433851540716:dw|

  35. skullpatrol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Try MO

  36. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    This isn't research-level.

  37. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Oh, it was answered. Lookie-lookie. http://math.stackexchange.com/questions/1317960/how-prove-this-cot-pi-15-4-sin-pi-15-sqrt15

  38. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    OK, so he used the 30th cyclotomic polynomial. Cool.

  39. skullpatrol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Carlos?

  40. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Nah, that's not me.

  41. skullpatrol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    icic

  42. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    I still want to use complex numbers for this - that should improve my currently zero understanding of them.

  43. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    It's known that\[\sin(\pi/5) = \dfrac{\sqrt{10 - 2\sqrt 5}}{4} = 3\sin^3 (\pi/15) - 4\sin(\pi/15)\]We can simply solve for trig-ratios of \(\pi/15\) like this, but it's stupid and time-consuming.

  44. skullpatrol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    math consumes time

  45. Phebe
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    TRUE

  46. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    deep

  47. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    il get back to this later, my head is spinning right nw

  48. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Same, I'm sleepy even though I woke up at 1 PM.

  49. Miracrown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I need sleep. G'night

  50. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Dan kisses you good night.

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.