# [SOLVED] viete's relations problem

by ehrenfest
Tags: relations, solved, viete
 P: 1,998 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.
 HW Helper P: 6,193 Let $P(x)=x^3 -10x+11=0$ what is u+v+w,uv+uw+vw and uvw ? and what you want to get is $tan^{-1}(u)+tan^{-1}(v)+tan^{-1}(w)$ if we let $A=tan^{-1}(u);B=tan^{-1}(v);C=tan^{-1}(w)$ thus making tanA=u;tanB=v and tanC=w, are able with some trig identity to help you?
P: 1,998
 Quote by rock.freak667 Let $P(x)=x^3 -10x+11=0$ what is u+v+w,uv+uw+vw and uvw ? and what you want to get is $tan^{-1}(u)+tan^{-1}(v)+tan^{-1}(w)$ if we let $A=tan^{-1}(u);B=tan^{-1}(v);C=tan^{-1}(w)$ 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

$$\tan(a\pm b) = \frac{\tan a \pm \tan b}{1\mp \tan a \tan b}$$

which does not seem very useful here...

We can of course obtain that

$$\tan^3 A +\tan^3 B + \tan ^3 C = 33$$

HW Helper
P: 6,193

## [SOLVED] viete's relations problem

 Quote by ehrenfest u+v+w = 0 uv+uw+vw = -10 uvw = -11
Good.

Now onto a new part.

 Quote by ehrenfest The only relevant identity I can think of is $$\tan(a\pm b) = \frac{\tan a \pm \tan b}{1\mp \tan a \tan b}$$ 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))
P: 1,998
 Quote by rock.freak667 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 $\frac{\pi}{4}$. Thanks.
 P: 1,998 On second thought, do we know that the sum is not 5 pi/4 or -3 pi/4?
 P: 1,998 Never mind, I got it.

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