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Math Help - Sin x / (1-cos x)

  1. #1
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    Sin x / (1-cos x)

    I have been working on a problem I have sort of worked myself into, and don't really know how to solve it.

    I end up at

    k = sin x / (1- cos x)

    and I need to solve for x.

    I cant find any identities that will make this easier, and I don't really remember how to break it up any further (as I am an engineer, not a mathematician)

    I can plug it into a calculator and solve for a given value, and that works well enough for what I am doing. However, I would like to further solve it, to make using it easier in the future.

    Anyway you could aide me would be greatly appreciated.
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  2. #2
    Member SENTINEL4's Avatar
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    What was the original problem you had to solve?
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  3. #3
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    Talking Original Problem

    I would like a formula (or a couple of formulas) that allow me to take a given change in X and Y, treat it as a chord of a circle, and determine the angle of the chord and the radius of the circle.

    I probably just worked myself into a corner.

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

    Here is where I got:

    I know
    x = sin theta
    y= 1- cos theta

    so,
    x/y = sin theta / (1-cos theta)

    I can use my graphing calc to find theta for a given ratio x/y

    Then I plug x and theta back into the first equation and solve for r.

    This works... almost. when I build an arc (in a cad prog) based on my calculated r and theta, it gets close, but doesn't quite match up with the original x and y.
    Last edited by gwammy; September 30th 2009 at 09:32 AM. Reason: more information
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  4. #4
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    this identity may help
     \sin x = 2\sin(\frac {x}{2})\cos (\frac{x}{2})
    \ 1-\cos x= 2 \sin^2(\frac{x}{2})
    these identities reduce the equation to
    \ k=\frac{\sin x}{1-\cos x}
    \ k=\frac{2\sin(\frac {x}{2})\cos (\frac{x}{2})}{2 \sin^2(\frac{x}{2})}
    \ k= \frac{\cos(\frac{x}{2})}{\sin(\frac{x}{2})}
    \ k=\cot(\frac{x}{2})
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  5. #5
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    Thanks

    That gave me exactly what I needed.

    theta = 2 arctan(y/x)
    r = x / sin theta
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  6. #6
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    Hello, gwammy!

    This is a tricky problem, requiring special treatment.


    Solve for x\!:\;\;k \:=\:\frac{\sin x}{1- \cos x}
    We have: . k - k\cos x \:=\:\sin x \quad\Rightarrow\quad k\cos x + \sin x \:=\:k

    Divide by \sqrt{1+k^2}\!:\quad \cos x \cdot\frac{k}{\sqrt{1+k^2}} \;+\; \sin x\cdot\frac{1}{\sqrt{1+k^2}} \;\;=\;\;\frac{k}{\sqrt{1+k^2}} .[1]


    Consider angle \theta in a right triangle with: .  opp = 1,\:adj = k
    Code:
                            *
               ____      *  |
              √1+kČ   *     |
                   *        | 1
                *           |
             * θ            |
          * - - - - - - - - *
                    k
    The hypotenuse is \sqrt{1+k^2}

    And: . \cos\theta = \frac{k}{\sqrt{1+k^2}}, \quad \sin\theta = \frac{1}{\sqrt{1+k^2}} .[2]

    . . Note that: . \theta \:=\:\arccos\left(\frac{k}{\sqrt{1+k^2}}\right) .[3]


    Substitute [2] into [1]: . \cos x \cos\theta + \sin x\sin\theta \:=\:\frac{k}{\sqrt{1+k^2}}

    . . And we have: . \cos(x - \theta) \;=\;\frac{k}{\sqrt{1+k^2}}

    . . Then: . x-\theta \:=\:\arccos\left(\frac{k}{\sqrt{1+k^2}}\right) \quad\Rightarrow\quad x \;=\;\arccos\left(\frac{k}{\sqrt{1+k^2}}\right) + \theta


    Substitute [3]: . x \;=\;\arccos\left(\frac{k}{\sqrt{1+k^2}}\right) + \arccos\left(\frac{k}{\sqrt{1+k^2}}\right)

    . . Therefore: . x \;=\;2\arccos\left(\frac{k}{\sqrt{1+k^2}}\right)

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