Results 1 to 3 of 3

Math Help - Amount of Rotation Formula

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    6

    Exclamation Amount of Rotation Formula

    6x-6xy+14y=45

    write it in rotated xy-system (prime or whatever)

    please list it out in steps.
    i understand the first step, but i do not understand how to get Θ.
    cot2Θ=(A-C)/B
    so: cot2Θ=(6-14)/-6 =(-8)/-6 = 4/3
    so how can you do: cot2Θ=4/3 !!!!!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by needmathhelptoujours View Post
    6x-6xy+14y=45

    write it in rotated xy-system (prime or whatever)

    please list it out in steps.
    i understand the first step, but i do not understand how to get Θ.
    cot2Θ=(A-C)/B
    so: cot2Θ=(6-14)/-6 =(-8)/-6 = 4/3
    so how can you do: cot2Θ=4/3 !!!!!!!
    If \cot2\theta=4/3 then \tan2\theta=3/4. (You don't have to flip the thing upside down like that. It's just that I know the formula for \tan2\theta but not the formula for \cot2\theta.)

    Write t = \tan\theta. Then \tan2\theta = \frac{2t}{1-t^2} = \frac34. That is a quadratic equation for t, with solutions t=–3, t=1/3. Taking t=1/3, you get \sin\theta = 1/\sqrt{10} and \cos\theta = 3/\sqrt{10}, and you can plug those values into the formula for rotating axes. You could equally well take t=–3, which would have the effect of rotating the axes in the opposite direction. In either case, you should end up with an equation without an x'y' term.
    Last edited by Opalg; April 21st 2010 at 06:34 AM. Reason: Minus signs were missing
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,682
    Thanks
    614
    Hello, needmathhelptoujours!


    We have: . \cot2\theta \:=\:\frac{4}{3}\:=\:\frac{adj}{opp}


    2\theta is in a right triangle with: . opp = 3,\;adj = 4
    Code:
                      *
                    * |
                  *   |
                *     | 3
              *       |
            * 2θ      |
          * - - - - - *
                4
    Pythagorus says: . hyp = 5

    . . Hence: . \cos2\theta \:=\:\frac{4}{5}


    We have these two half-angle identities:

    . . \sin\theta \:=\:\sqrt{\frac{1-\cos2\theta}{2}} \qquad\cos\theta \:=\: \sqrt{\frac{1+\cos2\theta}{2}}


    Therefore: . \begin{Bmatrix}\;\sin\theta &=& \sqrt{\dfrac{1-\frac{4}{5}}{2}} &=& \sqrt{\dfrac{1}{10}} &=& \dfrac{1}{\sqrt{10}}\; \\ \\[-3mm]<br />
\;\cos\theta &=& \sqrt{\dfrac{1+\frac{4}{5}}{2}} &=& \sqrt{\dfrac{9}{10}} &=& \dfrac{3}{\sqrt{10}}\; \end{Bmatrix}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: January 14th 2012, 08:02 PM
  2. Rotation formula challange - Help
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: December 14th 2010, 04:43 PM
  3. Help w/ formula for variable loan amount
    Posted in the Business Math Forum
    Replies: 2
    Last Post: January 16th 2010, 11:48 AM
  4. Expansion of a rotation-formula
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 14th 2010, 12:45 PM
  5. amount
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 21st 2009, 02:12 PM

Search Tags


/mathhelpforum @mathhelpforum