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\[\cot(\pi/15) - 4\sin(\pi/15) = \sqrt{15}\]
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:\[\cot(16\pi/15) = \frac{\cos(16\pi/15)}{\sin(16\pi/15) } = \dfrac{(x^4 + \overline{x^4})/2}{(x^4 - \overline{x^4})/2i}\]\[\Rightarrow i\cot(16\pi/15) = \dfrac{x^{11}+ x^4}{x^{11} - x^4} = \dfrac{x^7+1}{x^7 - 1}\]
Similarly,\[4i \sin(16\pi/15) = 2(x^4 - x^{11})\]

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Other answers:

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

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