# Sin x / (1-cos x)

• Sep 30th 2009, 08:10 AM
gwammy
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.
• Sep 30th 2009, 08:27 AM
SENTINEL4
What was the original problem you had to solve?
• Sep 30th 2009, 09:00 AM
gwammy
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.
• Sep 30th 2009, 09:39 AM
ramiee2010
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
$\ k=\frac{\sin x}{1-\cos x}$
$\ k=\frac{2\sin(\frac {x}{2})\cos (\frac{x}{2})}{2 \sin^2(\frac{x}{2})}$
$\ k= \frac{\cos(\frac{x}{2})}{\sin(\frac{x}{2})}$
$\ k=\cot(\frac{x}{2})$
• Sep 30th 2009, 10:03 AM
gwammy
Thanks
That gave me exactly what I needed.

$theta = 2 arctan(y/x)$
$r = x / sin theta$
• Sep 30th 2009, 10:07 AM
Soroban
Hello, gwammy!

This is a tricky problem, requiring special treatment.

Quote:

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