$\alpha= 2.15\ldots 10^5$, $ x_1=-0.976\ldots $, $ x_2=0.674\ldots $, $ x_3=1.505\ldots $, $ x_4=3.509\ldots $ The polynomial $\hat {Q}_8^{(iv)}(x)$ has exactly four simple real roots with a symbolic radical representation and the corresponding numerical values: $ x_1$, $ x_2$, $ x_3$, $ x_4$. Therefore, the polynomial $\hat {Q}_8^{(iv)}(x)$ has no roots for $ x\!\in\!(0,c_1)$. Since $\hat {Q}_8^{(iv)}(0)<0$, we can conclude that the polynomial $\hat {Q}_8^{(iv)}(x)$ is negative function for $ x\!\in\!(0,c_1)$, which follows that the function $\hat{Q}_8^{{''}'}(x)$ is monotonically decreasing for $ x\!\in\!(0,c_1)$. Doing the same procedure for all derivates up to $\hat{Q}_8^{'}(x)$ we have the following: $$\begin{array}{lccccccr}
\hat{Q}_8^{{'''}}(c_1)\!>\!0: \!&\!\hat{Q}_8^{{'''}}(x)\!>\!0 \!&\! \mbox{for} \!&\! x\!\in\!(0, c_1) \!&\! \Longrightarrow\; \!&\! \hat{Q}_8^{{''}}(x)\nearrow \!&\! \mbox{for} \!&\! x\!\in\!(0, c_1), \\ [2.0 ex]
\hat{Q}_8^{{''}}(c_1)\!<\!0: \!&\!\hat{Q}_8^{{''}}(x)\!<\!0 \!&\! \mbox{for} \!&\! x\!\in\!(0, c_1) \!&\! \Longrightarrow\; \!&\! \hat{Q}_8^'(x) \searrow \!&\! \mbox{for} \!&\! x\!\in\!(0, c_1),       \\ [2.0 ex]
\hat{Q}_8^{'}(c_1)\!>\!0: \!&\!\hat{Q }_8^{'}(x)\!>\!0 \!&\! \mbox{for} \!&\! x\!\in\!(0,c_1) \!&\! \Longrightarrow\; \!&\! \hat{Q}_8(x) \nearrow \!&\! \mbox{for} \!&\! x\!\in\!(0, c_1).
 \end{array}$$ After all, we conclude following: $$\label{Ineq_50}
\begin{array}{rccl}
&\hat {Q}_8(x)<0 & \mbox {for} & x\!\in\!(0, c_1) \\[1.0 ex]
\Longrightarrow\; & \hat {Q}_{10}(x)>0 & \mbox {for} & x\!\in\!(0,c_1)\\[1.0 ex]
\Longrightarrow\; & \hat {f}(x)=f\left(\nfrac{\pi}{2}- x \right)>0 & \mbox {for} & x\!\in\!(0, c_1)\\[1.0 ex]
\Longrightarrow\; & f(x)>0 & \mbox 