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They're the roots of \(x^n - 1\) so... using Vieta's Formulas?

Clever! that is still algebra...

this is also helpful :O
\(\Large e^{i\theta}=\cos \theta+i \sin \theta \)

It's pretty easy when you geometrically see it.

@Nishant_Garg au can and its also work

sure euler formula works for ALL \(\theta\), need not be an integer..

tell me about the geometric method @ParthKohli

Eh, so far, it's only about seeing. It's not a strict proof. |dw:1439296375015:dw|

These can be looked at as vectors and rearranged to give you a regular polygon.

I wish that n was not there XD

For some weird reason, `\frac` isn't working but `\dfrac` does.

working for me, I use it like this:
frac{3}{4}
output:
\[\frac{3}{4}\]

|dw:1439297117562:dw|

Feel free to continue, and here I'm trying to study my course and getting distracted!!

OK, so try seeing it this way: place 2 at the ending of 1, place 3 at the ending of 2, and so on.

what about
\[\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots\]

that's my point Garg :3

yeah, that's a polynomial now isnt it :P

just a very long one

That's a series, not a polynomial. A polynomial has a defined degree.

So it's not a polynomial as it goes on infinitely?

its a polynomial series or as we love to say infinite polynomial function :3

i'll go anyway, bbye.

lol xD

P(x) = 0 has infinite roots

It's of course a regular polygon as all sides are equal.

And so are all interior angles.

Has this not been proved?? like is it an unsolved question?

What kind of proof do you have with you? @ganeshie8

|dw:1439298571809:dw|

Yeah nice.

same idea as exterior angles adding up to 360 in a regular polygon

I've the best proof... multiply both sides by...
ZERO XD

i kno you're just kidding, but i wanted to clarify it anyways :)