Possible to prove that a particular trigonometric expression is always positive? 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.
 A: 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
$$\delta(\varepsilon) = \sin\left(\dfrac{\varepsilon t}8+\varepsilon\varphi\right) \sin\left(t+\dfrac{\varepsilon t}8+2r+\varepsilon\varphi\right) - \sin\dfrac{\varepsilon t}8 \sin\left(t+\dfrac{\varepsilon t}8\right) + \dfrac\varepsilon2\sin(r)\sin(r+t) > 0.$$
$\mathbf{Proved.}$
