I would like to find a function $f(\theta)$ on the domain $\frac{-\pi}{2} \leqslant \theta \leqslant \frac{\pi}{2}$ for which $$ \tan \left( \frac{f(\theta) +f(-\theta))}{2}\right) = \cos \left( \frac{f(\theta) - f(-\theta))}{2}\right)$$ I know several things about the function:
- it is monotonically increasing and concave over this domain
- $f(\frac{-\pi}{2})=\frac{-\pi}{2}$ and $f(\frac{\pi}{2})=\frac{\pi}{2}$
- $f(0)=\frac{\pi}{4}$ and $f(\frac{-\pi}{4})=0$
- $f'(\frac{-\pi}{2})=\infty$ and $f'(\frac{\pi}{2})=0$
- more generally the function has diagonal symmetry (across $y=-\theta$)
The function is similar in form but not identical to a Lame curve $$ y = -\sqrt[n]{{\pi}^n - x^n}$$ for $n=\log_2{(1+\sqrt{5})}-1$ and $0 \leqslant x \leqslant \pi$
It is not clear to me that there should be a unique solution to your problem.
Let us introduce the symmetric function $S(\theta) = (f(\theta)+f(-\theta))/2$ and the anti-symmetric function $A(\theta) = (f(\theta)-f(-\theta))/2$. Then your equation states that $$tan(S(\theta)) = cos(A(\theta))$$
This is a not very restrictive relation between $S$ and $A$. It has solution $S(\theta) = tan^{-1}(cos(A(\theta))$. So we can choose a suitable $A(\theta)$, then evaluate the matching $S(\theta)$ and finally decompose into $f(\theta)$ and $f(-\theta)$.