a symbolic radical representation and corresponding numerical values $s_1$, $s_2$, $s_3$ given at (\[Ineq\_26\]). Since $P_3(0)>0$ it follows that $P_3(s)>0$ for $s\!\in\! \left(s_1, s_2 \right)$, so we have following conclusions: $$\label{Ineq_27}
\begin{array}{rccl}
                & P_3(s)\!>\!0 & \mbox {for} & s\!\in\!(0,2.4649] \subset \left(s_1,s_2\right)       \\[1.0 ex]
\Longrightarrow & P_6(x)\!>\!0 & \mbox {for} & x\!\in\!(0,1.57] \subset \left(0,\sqrt{s_2}\right)    \\[1.0 ex]
\Longrightarrow & P_7(x)\!>\!0 & \mbox {for} & x\!\in\!(0,1.57] \subset \left(0 ,\sqrt{s_2}\right).
\end{array}$$ where $\sqrt{s_2}=1.589\ldots >1.57\,$.

According to the Lemmas 2.3. and 2.4. and description of the method based on (\[Ineq\_14\]) and (\[Ineq\_17\]), the following inequalities: $\cos y \!<\! \overline T_{k}^{\cos ,0}(y)\;(k\!=\!20)$, $\cos y \!<\! \overline T_{k}^{\cos ,0}(y)\;(k\!=\!16)$, $\sin y \!>\! \underline T_k^{\sin ,0}(y)\;(k\!=\!11)$ are true, for $y\!\in\!\left(0, \sqrt{(k+3)(k+4)}\right)$. For $x\!\in\!(0,1.57]$ it is valid: $$\label{Ineq_28}
f(x)
>
Q_{20}(x)
=
1-8 x^2\!-\!\overline{T}_{20}^{\cos ,0}(4 x)\!-\!8 x^2 \overline{T}_{16}^{\cos ,0}(2 x)
\!+\!P_7(x)\underline{T}_{11}^{\sin ,0}(2 x),$$ where $Q_{20}(x)$ is the polynomial $$\label{Ineq_29}
\begin{array}{rcl}
Q_{20}(x)\!\!&\!=\!&\!\! -\nfrac{16}{9280784638125} x^{10}{\Big(}262144 x^{10}-5203625 x^8+69322260 x^6     \\[2.5 ex]
\!\!&\! \!&\!\! -665557650 x^4+3412527300 x^2-5237832600{\Big )}                                              \\[0.5 ex]
\!\!&\!=\!&\!\! -\nfrac{16}{9280784638125} x^{10} Q_{10}(x),
\end{array}$$ for $ x\!\in\!(0,1.57]$. Then, we have to determine sign of the polynomial $$\label{Ineq_30}
\begin{array}{rcl}
Q_{10}(x)\!\!&\!=\!&\!\! 262144 x^{10}-5203625 x^8+69322260 x^6-665557650 x^4\\[1.0 ex]
\!\!&\! \!&\!\! +3412527300 x^2-5237832600,
\end{array}$$ for $ x\!\in\!(0,1.57]$, which is the polynomial of $10^{th}$ degree. By substitution 