Results 1 to 4 of 4

Math Help - Trigo-system

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    534

    Trigo-system

    Solve in :


    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Let \tan x=a, \ \tan y=b

    Then, \left\{\begin{array}{ll}a^2+b^2=6\\a^2+b^2=-6ab\end{array}\right.

    Now, let s=a+b, \ p=ab

    \Rightarrow\left\{\begin{array}{ll}s^2-2p=6\\s^2+4p=0\end{array}\right.\Rightarrow s=2, \ p=-1 or s=-2, \ p=-1

    If s=2, \ p=-1 then a=1+\sqrt{2},b=1-\sqrt{2} or a=1-\sqrt{2},b=1+\sqrt{2}

    If s=-2,p=-1 then a=-1+\sqrt{2},b=-1-\sqrt{2} or a=-1-\sqrt{2},b=-1+\sqrt{2}

    Now replace a and b and find x and y.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,711
    Thanks
    630
    Hello, dhiab!

    \text{Solve in }\mathbb{R}^2\quad\begin{array}{cccc}\tan^2\!x + \tan^2\!y &=& 6 & {\color{blue}[1]} \\ \\[-4mm]\dfrac{\tan x}{\tan y} + \dfrac{\tan y}{\tan x} &=& \text{-}6 & {\color{blue}[2]} \end{array}

    From {\color{blue}[2]} we have: . \frac{\tan^2\!x + \tan^2\!y}{\tan x\tan y} \:=\:-6

    Substitute {\color{blue}[1]}: . \frac{6}{\tan x\tan y} \:=\:-6 \quad\Rightarrow\quad \tan x\tan y \:=\:-1 \quad\Rightarrow\quad \tan y \:=\:\frac{-1}{\tan x}\;\;{\color{blue}[3]}

    Substitute into {\color{blue}[1]}:\quad \tan^2\!x + \left(\frac{-1}{\tan x}\right)^2 \:=\:6 \quad\Rightarrow\quad \tan^2\!x + \frac{1}{\tan^2\!x} \:=\:6

    . . \tan^4\!x + 1 \:=\:6\tan^2\!x \quad\Rightarrow\quad \tan^4\!x - 6\tan^2\!x + 1 \:=\:0 . . . a quadratic in \tan^2\!x


    Quadratic Formula: . \tan^2\!x \;=\;\frac{6 \pm\sqrt{32}}{2} \:=\:3 \pm 2\sqrt{2} \:=\:(1 \pm\sqrt{2})^2

    . . Hence: . \tan x \:=\:\pm(1 \pm\sqrt{2})


    Substitute into {\color{blue}[3]}\!:\quad \tan y \:=\:\frac{-1}{\pm(1\pm\sqrt{2})}

    . . Hence: . \tan y \;=\;\pm(1\mp\sqrt{2})


    Therefore: . \begin{array}{ccc}x &=& \arctan\left[\pm(1 \pm\sqrt{2})\right] \\ \\[-3mm]<br />
y &=& \arctan\left[\pm(1\mp\sqrt{2})\right]\end{array}


    I'll let you sort out the answers and determine if any are extraneous.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    534
    Hell,: red dog and Soroban ;Thanks for solution.
    I'have solution when can calculate x and y.
    I'have this identity :


    For : :



    I'have this equation :

    The solution is :

    coclusion I :
    same solution for
    For :
    I'have this equation :
    The solution is :
    Conclusion the solution of system is :

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigo value
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 11th 2010, 07:10 PM
  2. New trigo-system
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 21st 2010, 02:06 PM
  3. Trigo-system
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: May 19th 2010, 11:21 PM
  4. Trigo Eqn
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: December 2nd 2009, 02:53 AM
  5. Transfer Funtion for a cascade system and sub system
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 7th 2009, 05:13 AM

Search Tags


/mathhelpforum @mathhelpforum