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Math Help - Rearranging an equation

  1. #1
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    Rearranging an equation

    Hi,
    I currently have the equation:

    γ tan(α) = sin(α') / (cos(α') + β)

    I was using α as the subject however in order to do my calculations I need α' to be the subject.
    Does anyone know how I could rearrange this to that I have one case of α' on its own?
    Thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Topin View Post
    Hi,
    I currently have the equation:

    γ tan(α) = sin(α') / (cos(α') + β)

    I was using α as the subject however in order to do my calculations I need α' to be the subject.
    Does anyone know how I could rearrange this to that I have one case of α' on its own?
    Thanks
    Ummm....

    In terms of symbols, is this supposed to be
    \gamma ~ tan(\alpha) = \frac{sin(\alpha ^{\prime})}{cos(\alpha ^{\prime} ) + \beta}

    Anyway:
    \gamma ~ tan(\alpha) ~ ( cos(\alpha ^{\prime} ) + \beta ) = sin(\alpha ^{\prime})

    \gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ) = sin(\alpha ^{\prime})

    Because we don't know which quadrant \alpha ^{\prime} is in we need to do the following:
    (\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ))^2 = sin^2(\alpha ^{\prime})

    And sin^2( \alpha ^{\prime} ) = 1 - cos^2( \alpha ^{\prime} ), so
    (\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ))^2 = 1 - cos^2( \alpha ^{\prime} )

    When you simplify this, you will have a quadratic equation in cos( \alpha ^{\prime} ) which you can do the quadratic equation on.

    -Dan
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  3. #3
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    Thanks for your reply, Im still a little confused though
    (\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ))^2 = 1 - cos^2( \alpha ^{\prime} )
    I think there was a missing bracket but I tried to rearrange this but I couldn't get all of the cos(\alpha ^{\prime} ) together in a way that I could then simplify. The last line of my working before I seemed to double back on myself was:
     cos^2(\alpha ^{\prime} ) + 2 cos(\alpha ^{\prime} ) \beta + \beta^2 = \frac{1 - cos^2(\alpha ^{\prime} )} {\gamma^2 tan^2 (\alpha)}
    I was wondering where I could procede from here. Thanks for your help so far though.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    (\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta )^2 = 1 - cos^2( \alpha ^{\prime} )
    \gamma ^2 ~ tan^2(\alpha) ~ cos^2(\alpha ^{\prime} ) + 2 \gamma~tan(\alpha)~cos(\alpha ^{\prime} ) \cdot \gamma \beta + \gamma^2 \beta^2 = 1 - cos^2( \alpha ^{\prime} )

    (\gamma ^2 ~ tan^2(\alpha) + 1) cos^2(\alpha ^{\prime} ) + 2 \gamma^2 \beta~tan(\alpha)~cos(\alpha ^{\prime} ) + \gamma^2 \beta^2 - 1 = 0

    Set y = cos(\alpha ^{\prime} ).

    Then we have
    (\gamma ^2 ~ tan^2(\alpha) + 1) y^2 + 2 \gamma^2 \beta~tan(\alpha)~y + (\gamma^2 \beta^2 - 1) = 0
    which is a quadratic in y.

    -Dan
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  5. #5
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    Finally done it now. I found that in my example it came to
    (\gamma ^2 ~ tan^2(\alpha) + 1) y^2 + (2 \gamma^2 \beta~tan^2(\alpha))~y + (\gamma^2 \beta^2 ~ tan^2 - 1) = 0<br />
    Using that method I've got the answer that I was expecting, also got another answer come out that I thought I would have to do another calculation for
    Thankyou very much for your help - would never have managed this on my own!
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Topin View Post
    Finally done it now. I found that in my example it came to
    (\gamma ^2 ~ tan^2(\alpha) + 1) y^2 + (2 \gamma^2 \beta~tan^2(\alpha))~y + (\gamma^2 \beta^2 ~ tan^2 - 1) = 0<br />
    Using that method I've got the answer that I was expecting, also got another answer come out that I thought I would have to do another calculation for
    Thankyou very much for your help - would never have managed this on my own!
    It's a useful trick.

    -Dan
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