Results 1 to 4 of 4

Math Help - [SOLVED] Rearrange I = f(A) to A = f(I)

  1. #1
    jamesparker_1
    Guest

    [SOLVED] Rearrange I = f(A) to A = f(I)

    Hi

    I need to rearrange the following formula from I = f(A) to A = f(I)

    The formula is :


    tan(I) = sin(A)/(M+cos(A))

    Can anyone tell me how I rearrange this to get A as a function of I.

    Thanks

    James
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,735
    Thanks
    642
    Hello, James!

    I have a way . . . but it's really ugly . . .


    The formula is: . \tan I \:= \:\frac{\sin A}{M + \cos A}

    Can anyone tell me how to get A as a function of I?

    We have: . \tan I(M + \cos A) \:=\:\sin A

    .From \sin^2\!\!A + \cos^2\!\!A \:=\:1, we have: . \sin A \:=\:\pm\sqrt{1 - \cos^2\!\!A}

    Substitute: . \tan I(M + \cos A)\:=\:\pm\sqrt{1 - \cos^2\!\!A}

    Square both sides: . \tan^2\!I\left(M^2 + 2M\cos A + \cos^2\!\!A\right) \:=\: 1 - \cos^2\!\!A

    This simplifies to the quadratic: . (\sec^2\!I)\cos^2\!\!A + (2M\tan^2\!I)\cos A + (M^2\tan^2\!I - 1)\:=\:0

    Quadratic Formula: . \cos A \;=\;\frac{-2M\tan^2\!I \pm \sqrt{(2M\tan^2\!I)^2 - 4(\sec^2\!I)(M^2\tan^2\!I - 1)}}{2(\sec^2\!I)}

    . . . Good luck!

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member AlvinCY's Avatar
    Joined
    Feb 2007
    From
    I live in Sydney, Australia.
    Posts
    69
    tan(I) = sin(A)/(M+cos(A))

    we know for sin(X)/cos(X)=sin(Y)/cos(Y), then Y = X is one solution.

    tan(I) = sin(A)/(M+cos(A)) is the same as
    sin(I)/cos(I) = sin(A)/(M+cos(A))

    so one solution by my guess would be.

    I = inv.cos (M + cos A)
    cos A + M = cos I
    A = inv. cos (cos I - M)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,920
    Thanks
    332
    Awards
    1
    Quote Originally Posted by AlvinCY View Post
    tan(I) = sin(A)/(M+cos(A))

    we know for sin(X)/cos(X)=sin(Y)/cos(Y), then Y = X is one solution.

    tan(I) = sin(A)/(M+cos(A)) is the same as
    sin(I)/cos(I) = sin(A)/(M+cos(A))

    so one solution by my guess would be.

    I = inv.cos (M + cos A)
    cos A + M = cos I
    A = inv. cos (cos I - M)
    This solution implies that there exists an I such that cos(I) = M + cos(A), which may not be true. (If |M| > 2 there isn't one.) It is valid as a test, but may not produce all (or even any) solutions.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to rearrange DE?
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 3rd 2010, 12:14 AM
  2. [SOLVED] Rearrange to solve
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 11th 2009, 02:32 PM
  3. Please Help rearrange
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 11th 2008, 06:51 AM
  4. Someone please rearrange this for me
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 29th 2008, 11:45 PM
  5. Rearrange
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: September 25th 2007, 10:46 PM

/mathhelpforum @mathhelpforum