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Viete's relations problem

  • Thread starter ehrenfest
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  • #1
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[SOLVED] viete's relations problem

Homework Statement


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.


Homework Equations


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

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.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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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?
 
  • #3
2,012
1
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]
 
Last edited:
  • #4
rock.freak667
Homework Helper
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u+v+w = 0

uv+uw+vw = -10

uvw = -11
Good.

Now onto a new part.

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))
 
  • #5
2,012
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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.
 
  • #6
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On second thought, do we know that the sum is not 5 pi/4 or -3 pi/4?
 
  • #7
2,012
1
Never mind, I got it.
 

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