This is a continuation of the an earlier post where the geometric motivation was presented. Here I'd like to ask: is it possible to prove $\Delta > 0$ always?
$$\begin{align} \Delta &\equiv \sin(t) \sin\left(r+ (2 \pi -2 r - t)\frac{\epsilon}4 \right) \sin\left( \frac{2 - \epsilon}2 (\pi-r-t)\right) \\ &\quad {} - \frac{2 - \epsilon}2 \sin(r) \sin\left(t-\frac{\epsilon \; t}{4}\right) \sin(r+t) \end{align}$$ where $$0<r<\frac{\pi}{4} \qquad 0<t<\frac{\pi}{4} \qquad 0<\epsilon <1$$
Some relevant posts include this one that renders the final form of $\Delta$, which hasn't gotten satisfactory answers.
Given $$0<r<\frac{\pi}{4} \qquad 0<t<\frac{\pi}{4} \qquad 0<\varepsilon <1\tag1$$
Easily to see that $$t<\dfrac\pi2,\quad \dfrac{\varepsilon t}4 <\dfrac\pi{16}.$$
At the same time, sine increases in $\left(0,\dfrac\pi2\right).$
Therefore, $$\begin{align} &\sin t >\sin\left(t-\dfrac{\varepsilon t}4\right),\\[4pt] &\Delta > \sin t\left(\sin\left(r+(2 \pi -2 r - t)\frac{\varepsilon}4 \right) \sin\left(\frac{2 - \varepsilon}2 (\pi-r-t)\right) - \frac{2 - \varepsilon}2\sin r\sin(r+t)\right). \end{align}$$
On the other hand, $$\sin\left(\frac{2-\varepsilon}2(\pi-r-t)\right) = \sin\left((\pi-r-t)-(2\pi-2r-2t)\frac\varepsilon4\right)\\ = \sin\left(r+t+(2\pi-2r-2t)\frac\varepsilon4\right) = \sin\left(r+t+\dfrac{3\varepsilon t}4+(2\pi-2r-t)\frac\varepsilon4\right).$$ So it is sufficiently to prove inequality $\delta(\varepsilon) >0,$ where $$\delta(\varepsilon) = \sin(r+\varepsilon\varphi)\sin\left(r+t+\dfrac{3\varepsilon t}4+\varepsilon\varphi\right) - \frac{2 - \varepsilon}2 \sin(r)\sin(r+t),\tag2$$ $$\varphi = \dfrac{2\pi-2r-t}4 \in\left(\dfrac{5\pi}{16},\dfrac\pi2\right),\quad \dfrac{3\varepsilon t}4 <\dfrac{3\pi}{16},\tag3$$ under the conditions $(1).$
Really, $$\delta(\varepsilon) = \sin(r+\varepsilon\varphi)\sin\left(r+t+\dfrac{3\varepsilon t}4 +\varepsilon\varphi\right) - \frac{2 - \varepsilon}2 \sin(r)\sin(r+t)\\ = \frac12\left(\cos\left(t+\dfrac{3\varepsilon t}4\right) - \cos\left(2r+t+\dfrac{3\varepsilon t}4+2\varepsilon\varphi\right)- \cos(t) + \cos(2r+t) +\varepsilon\sin(r)\sin(r+t)\right)\\ = \frac12\left(\cos\left(t+\dfrac{3\varepsilon t}4\right) - \cos(t) - \cos\left(2r+t+\dfrac{3\varepsilon t}4+2\varepsilon\varphi\right) + \cos(2r+t) +\varepsilon\sin(r)\sin(r+t)\right)\\ = -\sin\dfrac{3\varepsilon t}8 \sin\left(t+\dfrac{3\varepsilon t}8\right) + \sin\left(\varepsilon\varphi+\dfrac{3\varepsilon t}8\right) \sin\left(2r+t+\varepsilon\varphi+\dfrac{3\varepsilon t}8\right) + \dfrac\varepsilon2\sin(r)\sin(r+t).$$ Taking in account that $$\varepsilon < 1 < 1+\dfrac{\pi-2r-2t}{\pi-r+t} = 2-\dfrac{r+3t}{\pi-r+t},\tag4$$ one can get $$t+\dfrac{3\varepsilon t}8 < t+\dfrac{3\varepsilon t}8+2r +\varepsilon\varphi, \tag5$$ $$ t+\dfrac{3\varepsilon t}8+2r+\varepsilon\varphi = \pi-t-\dfrac{3\varepsilon t}8-\pi+2t+\dfrac{3\varepsilon t}4+2r+\varepsilon\dfrac{2\pi-2r-t}4\\ = \pi-t-\dfrac{3\varepsilon t}8-\pi+2t+2r + \varepsilon\dfrac{\pi-r+t}2\\ < \pi-t-\dfrac{3\varepsilon t}8-\pi+2t+2r + \left(2-\dfrac{r+3t}{\pi-r+t}\right)\dfrac{\pi-r+t}2\\ = \pi-t-\dfrac{3\varepsilon t}8-\pi+2t+2r + \pi-r+t -r- 3t = \pi-t-\dfrac{3\varepsilon t}8,$$ $$ t+\dfrac{3\varepsilon t}8+2r+\varepsilon\varphi < \pi-t-\dfrac{3\varepsilon t}8.\tag6$$ From $(5)-(6)$ should $$\sin\left(t+\dfrac{\varepsilon t}8+2r+\varepsilon\varphi\right) > \sin\left(t+\dfrac{\varepsilon t}8\right),$$ so $$\delta(\varepsilon) = \sin\left(\dfrac{\varepsilon t}8+\varepsilon\varphi\right) \sin\left(t+\dfrac{\varepsilon t}8+2r+\varepsilon\varphi\right) - \sin\dfrac{\varepsilon t}8 \sin\left(t+\dfrac{\varepsilon t}8\right) + \dfrac\varepsilon2\sin(r)\sin(r+t) > 0.$$
$\mathbf{Proved.}$