Question

How do you apply vieta's formula to \[1+z+...+z^9=0?~ z^{10}=1~ z\not=1\]

Answer 1

\[1+z+z^2+...+z^{10}=1\] making one solution \(z=0\) and the other the tenth roots of 1 i think

Answer 2

well, the prob is i have (z-1)(1+z+...z^9)=z^10-1

Answer 3

i guess this is not viete's formula

Answer 4

but I don't really understand the theorem

Answer 5

http://en.wikipedia.org/wiki/Vieta's_formulas

Answer 6

I mean I can make that relation

Answer 7

not sure what you are supposed to do with them, they are the elementary symmetric functions
it does not solve the equation for you though

Answer 8

ugh, grrr I'm hung up on the pf, for tan pi/10

Answer 9

i wanted to use the complex way, but I don't understand how they went from step to step in the pfs you supplied

Answer 10

did you get any help for zakon?

Answer 11

nope, he said you were rigth

Answer 12

*from

Answer 13

lol great

Answer 14

yup haha

Answer 15

do you think you could explain the pf? or at least help with the steps, I just don't get it

Answer 16

ok lets see if we can figure it out

Answer 17

it's this If 10x=π sin2x=cos3x as 2x+3x=5x=π2 ⟹2sinxcosx=4cos3x−3cosx
⟹2sinx=4cos2x−3 as cosx≠0
If sinx=t,2t=4(1−t2)−3⟹4t2+2t−1=0
⟹t=−1±5‾‾√4
, but sinx>0 as 0<x<π
sinπ10=5‾‾√−14
(1)So,
cosπ10=1−(sinπ10)2‾‾‾‾‾‾‾‾‾‾‾‾‾√=10+25‾‾√‾‾‾‾‾‾‾‾‾‾√4
So,
tanπ10=5√−1410+25√√4=5‾‾√−110+25‾‾√‾‾‾‾‾‾‾‾‾‾√=5‾‾√−110+25‾‾√‾‾‾‾‾‾‾‾‾‾√
=(5‾‾√−1)210+25‾‾√‾‾‾‾‾‾‾‾‾‾‾√=3−5‾‾√5‾‾√(5‾‾√+1)‾‾‾‾‾‾‾‾‾‾‾‾‾√=(3−5‾‾√)(5‾‾√−1)5‾‾√(5‾‾√+1)(5‾‾√−1)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=5‾‾√−25‾‾√‾‾‾‾‾‾‾‾√
Or(2)
cosπ5=1−2(5‾‾√−14)2=5‾‾√+14
We know
cos2y=1−tan2y1+tan2y⟹tan2y=1−cos2y1+cos2y
So,
tan2π10=1−5√+141+5√+14=3−5‾‾√5‾‾√(5‾‾√+1)
which we have already encountered in (1).

Answer 18

so first question, how did he do the 10x=pi

Answer 19

ok hold the phone, maybe there is an easier way

Answer 20

ok
i'll wait

Answer 21

ooh nvm! got it!

Answer 22

Thanks sat!

Answer 23

oh, maybe this is snappy enough
to do it the other way, you make \(\sin(5x)\) in to a quadratic equation in \(\sin(x)\)
oh ok good because although this is a gimmick it is pretty snappy right?

Answer 24

it is it is haha

Answer 25

young snappy snapper

Answer 26

lol