Define the family of densities:
$$p(\phi;\theta) = \Big(f\big(\hspace{-1pt}\cos(\phi-\theta)\big) – f\big(\hspace{-1pt}\cos(\phi+\theta)\big)\Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)}, \quad 0 \le \phi,\theta\le \pi/2$$
where $f(x)=g(x^2)$ with $g$ non-negative, increasing, and convex or concave, on $[0,\infty)$. (Interestingly these densities indeed have the same measure for all $\theta$ whenever $g$ is concave or convex, which can be shown by an integral representation.)
Edit: As shown in the answer, these conditions are not sufficient. I believe it is true though if additionally $g^{(3)}(x) \ge 0$, for $x>0$.
Show that $p(\phi;\theta)$ has a monotone likelihood ratio (decreasing for concave $g$, increasing for convex $g$). I.e., for $0\le\theta_1 < \theta_2\le\pi/2$:
$$ h(\phi) = \frac{f\big(\hspace{-1pt}\cos(\phi-\theta_2)\big) – f\big(\hspace{-1pt}\cos(\phi+\theta_2)\big)}{f\big(\hspace{-1pt}\cos(\phi-\theta_1)\big) – f\big(\hspace{-1pt}\cos(\phi+\theta_1)\big)}$$
is monotonic on $[0,\pi/2]$.
Examples of functions are $f(x) = |x|^p$, $1\le p<2$, or for $p>2$. (For $p=2$, $p(\phi;\theta) = \sin^2(2\phi)$.) And the function $f(x) = \log( \cosh(x))$, which is concave in $x^2$ (i.e. $\log(\cosh(\sqrt{x}))$ is concave on $[0,\infty)$), and twice differentiable (unlike $|x|^p$, $p<2$).
This result is important to prove uniqueness of stable optima in blind source separation and deconvolution with strongly sub- and super-gaussian input densities, using the Karlin-Rubin theorem. The result can be proved for $f(x)=x^4$ by simplifying the derivative expression. This corresponds to using kurtosis as the the cost function, and the uniqueness result is already known in this case. For $f(x) = |x|$, the likelihood ratio is non-increasing, constant around $\phi=0$ and $\phi=\pi/2$.
Best Answer
$\newcommand{\ep}{\varepsilon}$This conjecture is not true in general.
Indeed, suppose the "convex" part of your conjecture is true. Then (letting $x:=\phi$, $t:=\theta_1$, and $\theta_2\downarrow\theta_1=t$) we see that for any strictly increasing convex smooth function $g$ and all $x$ and $t$ in $(0,\pi/2)$ we would have $h_2(g;x,t):=\partial_x\partial_t\,\ln(g(\cos^2(x-t))-g(\cos^2(x+t))\ge0$. (Note that for all $x$ and $t$ in $(0,\pi/2)$ we have $\cos^2(x-t)-\cos^2(x+t)=\sin2x\,\sin2t>0$, so that $h_2(g;x,t)$ is well defined.)
For real $\ep>0$ and real $u$, let $u_{+;\ep}:=\frac12(u+\sqrt{\ep^2+u^2})$, an "$\ep$-smoothed" version of $u_+:=\max(0,u)$. For $c$ and $c_*$ in $[0,\infty)$, let $g_{c_*,\ep}(c):=(c-c_*)_{+;\ep}$.
Then the function $g_{c_*,\ep}$ is strictly increasing, convex, and smooth on $[0,\infty)$. However, $h_2(g;x,t)=-44051.358\ldots\not\ge0$ if $g=g_{c_*,\ep}$, $c_*=\frac12$, $\ep=\frac1{1000}$, $x=\frac{39}{100}$, and $t=\frac{118}{100}$. So, the "convex" part of your conjecture is not true in general.
Suppose now the "concave" part of your conjecture is true. Then for any strictly increasing concave smooth function $g$ and all $x$ and $t$ in $(0,\pi/2)$ we would have $h_2(g;x,t)\le0$.
For $c$ and $c_*$ in $[0,\infty)$, let $G_{c_*,\ep}(c):=c-\sqrt{\ep^2+(c-c_*)^2}$.
Then the function $G_{c_*,\ep}$ is strictly increasing, concave, and smooth on $[0,\infty)$. However, $h_2(G;x,t)=32614.565\ldots\not\le0$ if $G=G_{c_*,\ep}$, $c_*=\frac12$, $\ep=\frac1{1000}$, and $x=\frac{39}{100}=t$. So, the "concave" part of your conjecture is not true in general either. $\quad\Box$