
#1
May2108, 08:37 PM

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. 



#2
May2108, 09:04 PM

HW Helper
P: 6,214

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
May2108, 09:29 PM

P: 1,998

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] 



#4
May2108, 09:32 PM

HW Helper
P: 6,214

[SOLVED] viete's relations problemNow onto a new part. i.e. tan(A+B+C)=tan(A+(B+C)) 



#5
May2108, 09:44 PM

P: 1,998

I get tan(A+B+C) = 1 which implies that the desired sum is [itex]\frac{\pi}{4}[/itex]. Thanks. 



#6
May2108, 09:53 PM

P: 1,998

On second thought, do we know that the sum is not 5 pi/4 or 3 pi/4?




#7
May2108, 10:29 PM

P: 1,998

Never mind, I got it.



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