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Viete's relations problem

by ehrenfest
Tags: relations, solved, viete
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ehrenfest
#1
May21-08, 08:37 PM
P: 1,996
1. The problem statement, all variables and given/known data
The zeros of the polynomial P(x) = x^3 -10x+11 are u,v,and w. Determine the value of arctan u +arctan v+ arctan w.


2. Relevant equations
http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas

3. The attempt at a solution
I must admit I have no idea how to do this problem. I usually try to show some work but I have no clue how Vietes relations allow you to connect u,v,and w to the inverse trig functions. Sorry. Someone please just give me a nudge in the right direction.
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rock.freak667
#2
May21-08, 09:04 PM
HW Helper
P: 6,202
Let [itex]P(x)=x^3 -10x+11=0[/itex]

what is u+v+w,uv+uw+vw and uvw ?

and what you want to get is [itex]tan^{-1}(u)+tan^{-1}(v)+tan^{-1}(w)[/itex]

if we let [itex]A=tan^{-1}(u);B=tan^{-1}(v);C=tan^{-1}(w)[/itex]

thus making tanA=u;tanB=v and tanC=w, are able with some trig identity to help you?
ehrenfest
#3
May21-08, 09:29 PM
P: 1,996
Quote Quote by rock.freak667 View Post
Let [itex]P(x)=x^3 -10x+11=0[/itex]

what is u+v+w,uv+uw+vw and uvw ?

and what you want to get is [itex]tan^{-1}(u)+tan^{-1}(v)+tan^{-1}(w)[/itex]

if we let [itex]A=tan^{-1}(u);B=tan^{-1}(v);C=tan^{-1}(w)[/itex]

thus making tanA=u;tanB=v and tanC=w, are able with some trig identity to help you?
u+v+w = 0

uv+uw+vw = -10

uvw = -11

The only relevant identity I can think of is

[tex]\tan(a\pm b) = \frac{\tan a \pm \tan b}{1\mp \tan a \tan b} [/tex]

which does not seem very useful here...

We can of course obtain that

[tex]\tan^3 A +\tan^3 B + \tan ^3 C = 33 [/tex]

rock.freak667
#4
May21-08, 09:32 PM
HW Helper
P: 6,202
Viete's relations problem

Quote Quote by ehrenfest View Post
u+v+w = 0

uv+uw+vw = -10

uvw = -11
Good.

Now onto a new part.

Quote Quote by ehrenfest View Post
The only relevant identity I can think of is

[tex]\tan(a\pm b) = \frac{\tan a \pm \tan b}{1\mp \tan a \tan b} [/tex]

which does not seem very useful here...
It is relevant but what you need to do is extend it to three things inside the brackets.

i.e. tan(A+B+C)=tan(A+(B+C))
ehrenfest
#5
May21-08, 09:44 PM
P: 1,996
Quote Quote by rock.freak667 View Post
Good.

Now onto a new part.



It is relevant but what you need to do is extend it to three things inside the brackets.

i.e. tan(A+B+C)=tan(A+(B+C))
Wow that worked out really really nicely.

I get tan(A+B+C) = 1 which implies that the desired sum is [itex]\frac{\pi}{4}[/itex]. Thanks.
ehrenfest
#6
May21-08, 09:53 PM
P: 1,996
On second thought, do we know that the sum is not 5 pi/4 or -3 pi/4?
ehrenfest
#7
May21-08, 10:29 PM
P: 1,996
Never mind, I got it.


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