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

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

I quit... what is this?!

Complex numbers.

like complex analysis?

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.

\(x^7 - 1\) 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

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.

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

Carlos?

Nah, that's not me.

icic

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.