1. Derive angle under spiral radius in terms of unit circle angle

Please see the attached thumbnails. I have here this lovely archimedean spiral inscribed within a unit circle and I want to derive angle $\theta_s$ in terms of angle $\theta$. $\theta$ is the angle under the pink line, $\theta_s$ is the angle under the spiral radius, green line.

Known: $\theta$ and the equation for the spiral radius $r_s$ which is just the polar equation for said spiral curve, and b, the y intercept of this curve:

$r_s=1+\frac{2\theta_s(b-1)}{\pi}$

$1\ge{b}>0$

Again, the spiral is inscribed within a unit circle. I want to express $\theta_s$ in terms of $\theta$.

(If $r_s$ can be solved for in terms of $\theta$ that would be fine too.)

So far I’ve taken the following approach:

1) I convert the $r_s$ curve from polar to Cartesian form:

$\sqrt{x^2+y^2}=1+\frac{2\arctan(\frac{y}{x})(b-1)}{\pi}$

2) I find the intercept between the spiral and the line $y=x\tan\theta_s$ by substituting in $x\tan\theta_s$ for y:

$\sqrt{x^2+x^2\tan^2\theta_s}=1+\frac{2\arctan(\fra c{x\tan\theta_s}{x})(b-1)}{\pi}$

Which simplifies to:

$x=\sqrt{\frac{\left(1+\frac{2\theta_s(b-1)}{\pi}\right)^2}{\tan^2\theta_s+1}}$

3) This, then, is equal to $\cos\theta$, so we can express $\theta$ in terms of $\theta_s$:

$\theta=\arccos\sqrt{\frac{\left(1+\frac{2\theta_s( b-1)}{\pi}\right)^2}{\tan^2\theta_s+1}}$

But, alas, the reverse is not so simple. I cannot find a way to solve this equation for $\theta_s$ in terms of $\theta$.

Perhaps a whole different approach is necessary. Your help is much appreciated.

2. Since that equation involve $\theta_s$ inside a transcendental function (tangent) as well as just squared, there probably is no simple formula for it.