@ganeshie8 @Miracrown @dan815 @UsukiDoll

- ParthKohli

@ganeshie8 @Miracrown @dan815 @UsukiDoll

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- ParthKohli

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

- 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

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

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## More answers

- UsukiDoll

I quit... what is this?!

- ParthKohli

Complex numbers.

- UsukiDoll

like complex analysis?

- 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

@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

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

- ParthKohli

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

- 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

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

- ParthKohli

@ganeshie8 Speak out. :P

- UsukiDoll

|dw:1433849058768:dw|

- Phebe

LOL

- ParthKohli

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

- ganeshie8

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

- ParthKohli

Yes, that's what I meant. Whoops.

- ganeshie8

let me grab pen and paper

- UsukiDoll

it's skully!

- skullpatrol

A pencil is preferable :P

- UsukiDoll

|dw:1433850807361:dw| pacman

- skullpatrol

:D

- 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

lolzz

- ParthKohli

This problem is also really hard.

- UsukiDoll

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

- ParthKohli

To help me solve it.

- UsukiDoll

but we're not pros at this D:

- ParthKohli

Then why are you viewing it?

- skullpatrol

Try MSE

- ParthKohli

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

- skullpatrol

:O

- UsukiDoll

|dw:1433851540716:dw|

- skullpatrol

Try MO

- ParthKohli

This isn't research-level.

- 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

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

- skullpatrol

Carlos?

- ParthKohli

Nah, that's not me.

- skullpatrol

icic

- ParthKohli

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

- 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

math consumes time

- Phebe

TRUE

- ParthKohli

deep

- ganeshie8

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

- ParthKohli

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

- Miracrown

I need sleep. G'night

- ParthKohli

Dan kisses you good night.

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