monotonically decreasing for $ x\!\in\!(0,0.11545 \ldots)$ and monotonically increasing for $ x\!\in\!(0.11545 \ldots ,\infty)$. $Q_7(0)<0$, $Q_7(0.98)<0$ and function $Q_7(x)$ has the first positive root $ x=0.98609 \ldots\;$. That means that $Q_7(x)<0$ for $ x\!\in\!(0,0.98]$. We can conclude following: $$\label{Ineq_93}
\begin{array}{rccl}
&Q_7(x)<0 & \mbox {for} & x\!\in\!(0,0.98]                       \\[1.0 ex]
\Longrightarrow\; & Q_{13}(x)>0 & \mbox {for} & x\!\in\!(0,0.98] \\[1.0 ex]
\Longrightarrow\; & f(x)>0 & \mbox {for} & x\!\in\!(0,0.98].
\end{array}$$ Let us notice that $x^*=0.98609\ldots$ is also the first positive root of the approximation of the function $f(x)$, i.e. of the polynomial $Q_{13}(x)$, defined at (\[Ineq\_89\]).

[**(C/II)**]{} [$ x\!\in\!\left(\!0.98,\nfrac{\pi}{2}\right)$]{} Let us define the function $$\label{Ineq_94}
\begin{array}{rcl}
\hat{f}(x)\!\!&\!\!\!=\!\!\!&\!\!f\left(\nfrac{\pi}{2}- x \right)=1\!-\!8\left(\nfrac{\pi}{2}\!-\! x \right)^{\!2}\!+\!\hat{h}_1(x)\cos4 x\!+\!\hat{h}_2(x)\cos2 x\!+\!\hat{h}_3(x)\sin2 x \\[1.0 ex]
\!\!&\!\!\!=\!\!\!&\!\!1-8\left(\nfrac{\pi}{2}- x \right)^{\!2}-\cos4 x+8\left(\nfrac{\pi}{2}- x \right)^{\!2} \cos2 x                                                                     \\[1.0 ex]
\!\!&\!\!\! \!\!\!&\!\!+\left(4\left(\nfrac{\pi}{2}- x \right)-4\left(\nfrac{16}{\pi^4} +c\left(\nfrac{\pi}{2}- x \right)\right)\!\left(\nfrac{\pi}{2}- x \right)^{\!5} \right)\sin2 x,
\end{array}$$ where $ x\!\in\!(0,c_3)$ for $c_3=\nfrac{\pi}{2}-0.98=\nfrac{\pi}{2}-\nfrac{49}{50} (= 0.59079\ldots$) and $\hat{h}_1(x)=-1<0$, $\hat{h}_2(x)=8 \left(\nfrac{\pi}{2}- x\right)^{\!2}>0$, $\hat{h}_3(x)=4 \left(\nfrac{\pi}{2}- x \right)-4 \left(\nfrac{16}{\pi^4} +c \left(\nfrac{\pi}{2}- x \right) \right)  \left(\nfrac{\pi}{2}- x\right)^{\!5}$.

We are proving that the function $\hat{f}(x)>0$.

It is important to find sign of the polynomial $\hat{h}_3(x)$. As we see, that polynomial is the polynomial of $6^{th}$ degree $$\label{Ineq_95}
\begin{array}{rcl}
\hat{h}_3(x)\!\!&\!\!\!=\!\!\!&\!\!\hat{P}_6(x)=4\left(\nfrac{\pi}{2}- x \right)-4\left(\nfrac{16}{\pi^4} +c\left(\nfrac{\pi}{2}- x \right) \right)\!\left(\nfrac{\pi}{2}- x \right)^{\!5}  \\[1.5 ex]
