Bound on angle implicitly defined via trigonometric equation Introduction / setup: In the article Spectral Gap for a Class of Random Billiards by Renato Feres and Hong-Kun Zhang (link: https://www.math.wustl.edu/~feres/spectralgapnew.pdf) a trigonometric equation is established (in proposition 8, equation (3.2)) between two angles $\theta$ and $\Theta = \Theta_{\theta,1}$, namely
\begin{equation}
K\sin\Theta = 1 - \cos\left( \frac{\theta + \Theta}{2} \right).
\end{equation}
Here $K \geq 2$ is the curvature / inverse of the radius of the circles $\Gamma_1$ and $\Gamma_2$, as seen in the following figure.

In proposition 9 (inequality (3.7), here you have to note that they define $\sigma(\theta) := \frac{\theta^2}{8K}$) they go on to say that from this equation we can deduce that if $0 \leq \theta \leq \sin^{-1}\left(\frac{1}{K}\right) := \theta_{0}$, then
$$\Theta \leq (1+\theta)\frac{\theta^2}{8K}.$$
Here it is also useful to note that we always have $0 \leq \Theta \leq \theta$.
My question: How can I prove that this bound indeed holds?
What I have tried / what may be useful to know: I do not believe that it is important how they established this equation and how the dynamics works (it comes from a billiard dynamics system). What is useful to know is the above and that they say one can derive this inequality by using estimates from Taylor approximations on the above trigonometric equation. You should for example think of: if $\theta \in \left[0,\frac{\pi}{2}\right]$, then $\sin(\theta) \leq \theta$ and $\sin(\theta) \geq \theta - \frac{\theta^3}{6}$, etc (see for example their proof of the lower bound in the inequality (3.7) from proposition 9). I have tried to do this along with trying some other trigonometric identities, but I just cannot get the right expression.
Many thanks in advance.
 A: Another approach
Let $\Theta=4 \tan ^{-1}(x)$, $c=\cos \left(\frac{\theta }{2}\right)$ and $s=\sin \left(\frac{\theta }{2}\right)$ to make the equation
$$\left(1+c\right)x^4+2\left(2K+ s\right)x^3+2x^2-2\left(2K- s\right)x+\left(1-c\right)=0\tag 1$$
which can be solved with (nasty) radicals.
For the given conditions, the discriminant
$\Delta$ is always negative and there are only two real roots in $x$.
Now, back to approximations : if, as it look, $\Theta$ is small, $x$ is small too.
Considering that $(1)$ is an expansion to $O(x^5)$, a series reversion gives
$$x=\frac{(1-c)}{2 (2 K-s)}+\frac{(1-c)^2}{4 (2 K-s)^3}+\frac{(1-c)^3 }{8 (2 K-s)^5}\left(4 K^2+2-s^2\right)+$$ $$\frac{(1-c)^4 }{32 (2 K-s)^7}\left(4 (c+11) K^2-4 (c+1) sK+\left((c-9)
   s^2+10\right)\right)$$
For $K=3$ and $\lambda=0.8$, this gives
$$\Theta=\color{red}{0.00314645047544}11$$ while the exact solution is
$$\Theta= \color{red}{0.0031464504754461}$$
For $K=4$ and $\lambda=0.6$, this gives
$$\Theta=\color{red}{0.00072481622655383}49$$ while the exact solution is
$$\Theta=\color{red}{0.0007248162265538377}$$
We still could improve considering that $(1)$ is an expansion to $O(x^{5+p})$ with $p>1$ and have $p$ extra terms. For example, the next term is
$$-\frac{(1-c)^5}{32 (2 K-s)^9}\left(48 K^4+K^2 \left(12 c-24 s^2+96\right)-12 K (c s+s)+\left(3 c s^2+3 s^4-18
   s^2+14\right)\right)$$
Expanded as a series for large values of $K$, we have
$$\Theta=\frac{1-c}{K}+\frac{(1-c) s}{2 K^2}+O\left(\frac{1}{K^3}\right)=\frac{2 \sin ^2\left(\frac{\theta }{4}\right)}{K}+\frac{\sin \left(\frac{\theta }{2}\right) \sin ^2\left(\frac{\theta
   }{4}\right)}{K^2}+O\left(\frac{1}{K^3}\right)$$
