[SOLVED] viete's relations problem 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/Viète's_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.
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]
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.