Please see attached image.
Knowing only theta and x, find theta_2 and theta_3
(the green curve is an ellipse inscribed within the reddish unit circle)
In general , we have
$\displaystyle \tan(\theta_n) = (\frac{1}{b})^n \tan(\theta_0)$
$\displaystyle x_n = (\frac{1}{b})^{n+1} \cdot \frac{ 1 - b }{ \sqrt{ 1 + (\frac{1}{b})^{2n} \tan^2(\theta_0) }} $
Once the value of $\displaystyle b $ is comfirmed ,
we can find out the angle . ( if we keep drawing the lines similarly , a sequence is formed ! )
These are very unexpected, interesting responses (to me), which is precisely what I hoped to get by posting this problem.
Opalg, I'm hesitant to say you're wrong, but your answer doesn't work when I graph it. Either it's wrong or I'm making some mistake when I graph it.
Simplependulum, I'm still trying to wrap my head around your response. I had never thought of expressing things in that way.
Here is my solution (obviously not the only one):
First, use the given info--$\displaystyle x, \sin{\theta}$--to find the semi-minor axis of the ellipse,$\displaystyle b$:
$\displaystyle x=\sin{\theta}-b\sin{\theta}$
so, $\displaystyle b=\frac{\sin{\theta}-x}{\sin{\theta}}$
Then, the equations for $\displaystyle \theta_2$ and $\displaystyle \theta_3$ are, respectively:
$\displaystyle \tan{\theta_2}=\sqrt{b^2(\sec^2{\theta}-1)}$
and
$\displaystyle \tan{\theta_3}=\arccos{\frac{b}{\sqrt{\tan^2{\thet a}+b^2}}}$
You derive these equations by fooling around with (x, y) intercepts and pythagorus' theorem. I won't go into the details unless you really want.
Happily, these equations turn out to be reducible to:
$\displaystyle \tan{\theta_2}=b\tan{\theta}$
and
$\displaystyle \tan{\theta_3}=\frac{\tan{\theta}}{b}$