# Thread: Sin x / (1-cos x)

1. ## 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.

2. What was the original problem you had to solve?

3. ## 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.

4. 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
$\ k=\frac{\sin x}{1-\cos x}$
$\ k=\frac{2\sin(\frac {x}{2})\cos (\frac{x}{2})}{2 \sin^2(\frac{x}{2})}$
$\ k= \frac{\cos(\frac{x}{2})}{\sin(\frac{x}{2})}$
$\ k=\cot(\frac{x}{2})$

5. ## Thanks

That gave me exactly what I needed.

$theta = 2 arctan(y/x)$
$r = x / sin theta$

6. 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:
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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: . $x \;=\;2\arccos\left(\frac{k}{\sqrt{1+k^2}}\right)$