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Some help needed

  1. Sep 10, 2006 #1
    Can you please help me for making 'z' the subject of formula;

    (z-x)/g = tan(0.5(arctan((z-y)/g)+arctan((y-x)/g)))

    I'll be thankfull...
     
  2. jcsd
  3. Sep 10, 2006 #2

    quasar987

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    Sometimes it's not possible to make 'z' the subject of a formula.
     
  4. Sep 11, 2006 #3

    ok let's begin by writing that more mathematically.

    [tex]\frac{z-x} {g} = tan{\frac{tan^{-1}(\frac{z-y} {g})+tan^{-1}(\frac{y-x} {g})} {2}[/tex]

    IS this what you mean? 6am...I might have done it wrong.

    tip: maybe trying to get both sides to look messy might help...i have a way to solve this in mind and if it works that way, this is a beautiful problem. if it doesn't... i dont like this problem.
     
    Last edited: Sep 11, 2006
  5. Sep 11, 2006 #4
    Oh I hate you now, it awakened my curiousity and...50 minutes later I think I have it. One Z on the process decided to turn squared on me and instead of a nice common factor i had to use quadratic formula. indication that i did something awfully wrong but...i checked and it looks ok.

    EDIT: Ok I re-read the stickies, this isnt the homework section so i can post the work I did. Basically i got rid of that nasty 1/2 by...well this way: (sorry for the notation, i wrote this is a .txt file.

     
    Last edited: Sep 11, 2006
  6. Sep 11, 2006 #5
    thanks robokapp, I'll be thankfull if you do the whole thing (applying the formulas) what is z equal to????

    I don't really know which formulas you are talking about...

    thanks once again
     
  7. Sep 11, 2006 #6
    Okay...sure. After all I do am curious if I did any horrible mistakes (i might have)

    I disliked the 1/2 part. I knew there are formulas for different half-angle, duble angle, angle plus angle etc so i went to wikipedia and typed in 'trigonometry.' the formulas are on bottom.

    I knew the "tan" and "arctan" will cancel eachother out so I picked formulas so that I only work in terms of tan. This is my full work:

    (z-x)/g = tan(0.5(arctan((z-y)/g)+arctan((y-x)/g)))


    2arctan[(z-x)/g] = arctan((z-y)/g)+arctan((y-x)/g)


    tan{2arctan[(z-x)/g]} = tan{arctan((z-y)/g)+arctan((y-x)/g)}


    Using the double angle formula in first part and angle plus angle formula in second part...


    2tan{arctan[(z-x)/g]}
    ----------------------------------------------- =
    1 - tan{arctan[(z-x)/g]} * tan{arctan[(z-x)/g]}


    tan{arctan((z-y)/g)}+tan{arctan((y-x)/g)}
    = -------------------------------------------- ====>
    1+tan{arctan((z-y)/g)}*tan{arctan((y-x)/g)}


    2(z-x)/g.............(z-y)/g+(y-x)/g
    -------------- = ------------------- ====>
    1-(z-x)^2/g^2.....1+(z-y)/g * (y-x)/g


    2(z-x).................(z-y)+(y-x)
    -------------- = ---------------- =====>
    1-(z-x)^2/g.......1+(z-y) * (y-x)


    2gz-2gx ...............z-x
    ------------- = ---------------- ======> flip it.
    g-z^2-2zx+x^2......1+zy-zx-y^2+xy


    g-z^2-2zx+x^2....1+zy-zx-y^2+xy
    ------------- = -------------- =======>
    2gz-2gx...............z-x


    g-z^2-2zx+x^2...... 1+zy-zx-y^2+xy
    ------------- = -------------- =======> multiply second by 2g/2g
    2g(z-x).................. z-x


    g-z^2-2zx+x^2....2g(1+zy-zx-y^2+xy)
    ------------- = ------------------- =======>
    2g(z-x)..................2g(z-x)


    g-z^2-2zx+x^2 = 2g+2zgy-2zgx-2gy^2+2gxy


    -z^2-2zx-2zgy-2zgx = 2g+2gxy-2gy^2-g-x^2


    -z^2-2zx-2zgy-2zgx = g+2gxy-2gy^2-x^2


    -z^2+z(2x-2gy-2gx) - g-2gxy+2gy^2+x^2 = 0


    Calling

    a= -1
    b= 2x-2gy-2gx
    c= -g-2gxy+2gy^2+x^2


    z=0.5{-(2x-2gy-2gx) +/- sqrt[(2x-2gy-2gx)^2+4(-g-2gxy+2gy^2+x^2)]}


    -------------------------------------------

    Now...I dont think it should have 2 answers, I dont seem to have any way around that z^2 however.


    The ......... takes place of..balnk so numbers dont run into eachother.

    plz someone check my work...
     
    Last edited: Sep 11, 2006
  8. Sep 12, 2006 #7
    Thanks Robokapp, I had actually done a sily mistake in writing the question, It should had been:

    (z-x)/2g = tan(0.5(arctan((z-y)/g)+arctan((y-x)/g)))

    But I used the formulas you provided and got the following results:

    z = 2x - y +/- sqrt(3y^2-6yx+3x^2-4g^2)

    Thankyou you were a great help
     
  9. Sep 12, 2006 #8
    but chk it out, Did I do it correctly? becase I'm little puzzeled It's not working as I wanted...
     
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