Recent content by Duckbeam

I do believe it! Because when x=1 the absolute values of the brackets on the RHS is basically the value of each distance and hence when multiplied together equals to n. But i still don't understand how to prove it. i don't know if I am just stupid here.. but i can't see it at all.. :(btw, on...

Man, i can't get it for general term n.. i don't know why Any other hint that can help help me. This is what i have found for n=3,4,5 n=3 : 4sin^2(π/3) .. n=4 : 8sin^2(π/4).. n=5 : 16sin^2(π/5)sin^2(2π/5) I noticed this pattern: 2^n-1 sin^2(π/n) sin^2(2π/n)... sin^2(mπ/n) where m is the...

I think i get it! I appreciate it! :D

Yes yes i have done this! When n=3 --> then x^2+x+1 = x^2-x(2cos(3π/2))+1 Pluggin in one --> 3 = 2-2cos(3π/2) Using DA formula--> 3 = 2(1-(1-sin^2(π/3)) So 3 = 4sin^2 (π/3) Now in my assignment i have already shown that | 1 - e^\frac{2 \pi i}{3} | = 2 sin(\frac{\pi}{3}) .. A quick...

I really need help with this ASAP! This proof is only PART of my huge assignment that is due in two days and i can't get it down. I really need some big time help here.

:( -- I am not completely familiar with all the notation here. Especially the (exp) and the imaginary part. I am not sure how it all works together. Any chance you can elaborate a little? Also.. in the product it goes from k=1 to k=n-1 right? Thats not what you have written above -- is...

Yes i see somewhat of a connection here. If i do it for n=3,4,5: n=3 ---> (x-1) (x^2-x(2cos(2π/3)+1) n=4 ---> (x-1)(x+1)(x^2-x(2cos(π/2)+1) n=5---> (x-1)(x^2-x(2cos(2π/5)+1)(x^2-x(2cos(4π/5)+1 The connection is see is that when x=1 then if (x-1) is excluded it equals to n. I am not...

Anyone willing to help here?

I hope you didnt give up on me :-p

--> 2sin(\pi/3)2sin(2\pi/3)= 3 Did i write it wrong in my first post since you forgot the two in front of sin(2\pi/3)?

What if k=1 --- this point isn't a root of z^n= 1? As far as i understand? __ Now to your question I am also a little confused. I take z^3=1 as an example. It has the roots e^{ik2\pi/3}. If I exclude the root at (1,0) I have two imaginary parts left: i\sin2\pi/3 and i\sin4\pi/3. So...

1. The problem statement, all variables and given/kno(It n data I need to prove that \prod_{k=1}^{n-1}2\sin\tfrac{k\pi}{n}=n I just don't know where to start and what type of proof to do. Is there any help to get? :D Homework Equations I know this has nothing to do with an equation...