$\newcommand{\ep}{\varepsilon}$The "convex" part of this conjecture is not true in general.

Indeed, suppose it 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 $g$ with $g'''\ge0$ 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 $c$ and $c_*$ in $[0,\infty)$, let $g(c):=g_{c_*,\ep}(c):=(\sqrt{(c-c_*)^2+\ep^2}+c-c_*)^2$.

Then the function $g$ is strictly increasing, convex, and smooth on $\mathbb R$, and $g'''>0$. However, $h_2(g;x,t)=-461586.955\ldots\not\ge0$ if $c_*=\frac{526}{1000}$, $\ep=\frac1{1000}$, $x=\frac{812}{1000}$, and $t=\frac{157}{100}$. So, the "convex" part of your conjecture is not true in general. $\quad\Box$

$\newcommand{\ep}{\varepsilon}$The "convex" part of this conjecture is not true in general.

Indeed, suppose it 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 $g$ with $g'''\ge0$ 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 $c$ and $c_*$ in $[0,\infty)$ and real $\ep>0$, let $g(c):=g_{c_*,\ep}(c):=(\sqrt{(c-c_*)^2+\ep^2}+c-c_*)^2$.

Then the function $g$ is strictly increasing, convex, and smooth on $\mathbb R$, and $g'''>0$. However, $h_2(g;x,t)=-461586.955\ldots\not\ge0$ if $c_*=\frac{526}{1000}$, $\ep=\frac1{1000}$, $x=\frac{812}{1000}$, and $t=\frac{157}{100}$. So, the "convex" part of your conjecture is not true in general. $\quad\Box$