ParthKohli
  • ParthKohli
@ganeshie8 @Miracrown @dan815 @UsukiDoll
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ParthKohli
  • ParthKohli
\[\cot(\pi/15) - 4\sin(\pi/15) = \sqrt{15}\]
ParthKohli
  • ParthKohli
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:\[\cot(16\pi/15) = \frac{\cos(16\pi/15)}{\sin(16\pi/15) } = \dfrac{(x^4 + \overline{x^4})/2}{(x^4 - \overline{x^4})/2i}\]\[\Rightarrow i\cot(16\pi/15) = \dfrac{x^{11}+ x^4}{x^{11} - x^4} = \dfrac{x^7+1}{x^7 - 1}\]
ParthKohli
  • ParthKohli
Similarly,\[4i \sin(16\pi/15) = 2(x^4 - x^{11})\]

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More answers

UsukiDoll
  • UsukiDoll
I quit... what is this?!
ParthKohli
  • ParthKohli
Complex numbers.
UsukiDoll
  • UsukiDoll
like complex analysis?
ParthKohli
  • ParthKohli
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?
ParthKohli
  • ParthKohli
@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.
UsukiDoll
  • UsukiDoll
I'm still a noob when it comes to proofs D:
ParthKohli
  • ParthKohli
Also, we have to show that the right-hand side is \(i\sqrt{15}\) and we'll be done.
ParthKohli
  • ParthKohli
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 \(x^7 + 1\)?
ParthKohli
  • ParthKohli
\(x^7 - 1\) actually, which means that we're also looking at a seventh root of unity.
ParthKohli
  • ParthKohli
@ganeshie8 Speak out. :P
UsukiDoll
  • UsukiDoll
|dw:1433849058768:dw|
Phebe
  • Phebe
LOL
ParthKohli
  • ParthKohli
\[x^7 - 1 \mid x^{15n}+x^7\]
ganeshie8
  • ganeshie8
im still going through... you meant \(x := e^{2\pi i /15}\) right
ParthKohli
  • ParthKohli
Yes, that's what I meant. Whoops.
ganeshie8
  • ganeshie8
let me grab pen and paper
UsukiDoll
  • UsukiDoll
it's skully!
skullpatrol
  • skullpatrol
A pencil is preferable :P
UsukiDoll
  • UsukiDoll
|dw:1433850807361:dw| pacman
skullpatrol
  • skullpatrol
:D
ParthKohli
  • ParthKohli
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.
Phebe
  • Phebe
lolzz
ParthKohli
  • ParthKohli
This problem is also really hard.
UsukiDoll
  • UsukiDoll
if it's hard then why are we viewing it? rofl xD
ParthKohli
  • ParthKohli
To help me solve it.
UsukiDoll
  • UsukiDoll
but we're not pros at this D:
ParthKohli
  • ParthKohli
Then why are you viewing it?
skullpatrol
  • skullpatrol
Try MSE
ParthKohli
  • ParthKohli
Know what, this problem was asked at MSE, and is currently unanswered.
skullpatrol
  • skullpatrol
:O
UsukiDoll
  • UsukiDoll
|dw:1433851540716:dw|
skullpatrol
  • skullpatrol
Try MO
ParthKohli
  • ParthKohli
This isn't research-level.
ParthKohli
  • ParthKohli
Oh, it was answered. Lookie-lookie. http://math.stackexchange.com/questions/1317960/how-prove-this-cot-pi-15-4-sin-pi-15-sqrt15
ParthKohli
  • ParthKohli
OK, so he used the 30th cyclotomic polynomial. Cool.
skullpatrol
  • skullpatrol
Carlos?
ParthKohli
  • ParthKohli
Nah, that's not me.
skullpatrol
  • skullpatrol
icic
ParthKohli
  • ParthKohli
I still want to use complex numbers for this - that should improve my currently zero understanding of them.
ParthKohli
  • ParthKohli
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.
skullpatrol
  • skullpatrol
math consumes time
Phebe
  • Phebe
TRUE
ParthKohli
  • ParthKohli
deep
ganeshie8
  • ganeshie8
il get back to this later, my head is spinning right nw
ParthKohli
  • ParthKohli
Same, I'm sleepy even though I woke up at 1 PM.
Miracrown
  • Miracrown
I need sleep. G'night
ParthKohli
  • ParthKohli
Dan kisses you good night.

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