Solving Viete's Relations Problem Using Inverse Trig Functions

[ehrenfest]

[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.

 

[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?

 

[ehrenfest]

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

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

 

[rock.freak667]

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

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

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))

 

[ehrenfest]

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.

 

[ehrenfest]

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

 

[ehrenfest]

Never mind, I got it.