Where $$f(\theta)=\sin \theta\left(1+\frac{\cos\theta}{\epsilon}\right)+\frac{\cos\theta}{\epsilon}\sqrt{\sin^2\theta+2\epsilon(\eta-\cos\theta)}~~,$$ with $0<\epsilon\le1$ and $\eta\ge1$. All I have for the moment is numerical evidence that this is true. It seems complicated for me to prove even special cases like $\eta=1$ or $\epsilon=1$.
I do not know if it is useful, but $f(\theta)$ is one of the roots of $$\frac{\epsilon}{2}\left(x\sec\theta+\tan\theta\right)^2+\tan\theta(x-\sin\theta)+(\eta-\cos\theta)\tag{1}$$
I think it should be sufficient to prove that the envelope curve of (1) for varying $\theta$ is a parabola. Check out the picture, looks nice. How can we show this?
We use Equation (1) and calculate the envelope. Let $$ F(\theta,x,y) = \frac{\epsilon}{2}\left(x\sec\theta+\tan\theta\right)^2+\tan\theta(x-\sin\theta)+(\eta-\cos\theta)-y~~.\tag{2} $$ Then, $F(\theta,x,y)=0$ defines the parabolas which, in turn, define the equation in theta for $y=0$. The other root of the parabola can be ignored because it will be contained inside the envelope.
Now, to find the envelope we find the curve such that $0=\partial_\theta F=\sec ^3(\theta ) (\sin (\theta )-x) (\epsilon (x \sin (\theta )-1)-\cos (\theta ))$. The $(\sin (\theta )-x)=0$ condition give the inner envelope (the half circle). The other is $(x \sin (\theta )-1)-\cos (\theta ))=0$ which means $x=\frac{\csc (\theta ) (\cos (\theta )+\epsilon )}{\epsilon }$. We include this in $F(\theta,x,y)=0$.
Then $$ \begin{align} -2\epsilon(y-1/2*(1/\epsilon + \epsilon) - \eta)&=\left(\epsilon ^2 \tan ^2(\theta )-2 \epsilon ^2 \sec ^2(\theta )+\epsilon ^2+2 \epsilon \cos (\theta )-4 \epsilon \sec (\theta )+2 \epsilon \sin (\theta ) \tan (\theta )+\csc ^2(\theta ) (\epsilon \sec (\theta )+1)^2-1\right)\\ &=\frac{\left(1-\mu ^2\right) \epsilon ^2}{\mu ^2}-\frac{2 \epsilon ^2}{\mu ^2}+\epsilon ^2+\frac{2 \left(1-\mu ^2\right) \epsilon }{\mu }+\frac{\left(\frac{\epsilon }{\mu }+1\right)^2}{1-\mu ^2}+2 \mu \epsilon -\frac{4 \epsilon }{\mu }-1 \end{align}$$ where $\mu=\cos\theta$. $$ \begin{align} &=\frac{1}{1-\mu ^2}+\frac{\epsilon ^2}{\mu ^2 \left(1-\mu ^2\right)}-\frac{\epsilon ^2}{\mu ^2}+\frac{2 \epsilon }{\mu \left(1-\mu ^2\right)}-\frac{2 \epsilon }{\mu }-1\\ &=-\frac{(\mu +\epsilon )^2}{\mu ^2-1} \end{align} $$ and the last expression is proportional to $x^2=\left(\frac{\csc (\theta ) (\cos (\theta )+\epsilon )}{\epsilon }\right)^2$. Therefore, the envelope of the curves is a parabola with axis coincident with the y-axis. The roots of the parabolas forming the family in Eq. (1), therefore, cover a symmetric interval around zero.