description of the method based on (14) and (17), the following inequalities: $\cos y < \overline T_{k}^{\cos ,0}(y)(k=0)$ , $\cos y > \underline T_{k}^{\cos ,0}(y)(k=2)$, $\sin y > \underline T_k^{\sin ,0}(y)(k=3)$ are true, for $y\!\in\!\left(0, \sqrt{(k+3)(k+4)}\right)$. For $x\!\in\!(0,c_1)$ it is valid: $$\label{Ineq_45}
\begin{array}{rcl}
\hat{f}(x)\!\!&\!=\!&\!\!f\left(\nfrac{\pi}{2}- x \right)>\hat{Q}_{10}(x) = 1-8 \left(\nfrac{\pi}{2}- x \right)^{\!2}-\overline T_0^{\cos ,0}(4 x) \\[1.5 ex]
\!\!&\! \!&\!\!+8\left(\nfrac{\pi}{2}- x \right)^{\!2} \underline T_2^{\cos ,0}(2 x)+\hat{P}_7(x) \underline T_3^{\sin ,0}(2 x),\\[1.0 ex]
\end{array}$$ where $\hat{Q}_{10}(x)$ is the polynomial $$\label{Ineq_46}
\begin{array}{rcl}
\!\!\hat{Q}_{10}(x)
\!\!&\!=\!&\!\! \nfrac{128}{2835}x^{10} - \nfrac{64}{405}\pi x^9+ \left(\nfrac{32}{135}\pi^2-\nfrac{64}{63}\right) x^8+\left(-\nfrac{16}{81}\pi^3+\nfrac{352}{135}\pi\right) x^7                                                           \\[1.5 ex]
\!\!&\! \!&\!\!+\left(\nfrac{8}{81}\pi^4-\nfrac{368}{135}\pi^2+\nfrac{64}{45}\right) x^6+\left(-\nfrac{4}{135}\pi^5+\nfrac{40}{27}\pi^3-\nfrac{32}{9}\pi \right) x^5                                                                       \\[1.5 ex]
\!\!&\! \!&\!\!\!+\left(\nfrac{2}{405}\pi^6\!-\!\nfrac{4}{9}\pi^4\!+\!\nfrac{32}{9}\pi^2\!-\!\nfrac{32}{3}\right)x^4 \!+\!\left(\nfrac{\!-\!1}{2835}\pi^7\!+\!\!\nfrac{2}{27}\pi^5\!-\!\nfrac{16}{9}\pi^3\!+\!\nfrac{40}{3}\pi\!\right)x^3 \\[1.5 ex]
\!\!&\! \!&\!\!\!+\left(\!-\nfrac{1}{135}\pi^6\!+\!\nfrac{4}{9}\pi^4\!-\!4\pi^2\!-\!8\!\right)x^2+\left(\nfrac{1}{1890}\pi^7-\nfrac{2}{45}\pi^5+4\pi \right) x                                                                           \\[1.5 ex]
\!\!&\! \!&\!\! = -\nfrac{1}{5670} x\left(-2 x+\pi \right) \hat{Q}_8(x).
\end{array}$$ Then, we have to determine sign of the polynomial $$\label{Ineq_47}
\begin{array}{rcl}
\hat{Q}_8(x)\!\!&\!=\!&\!\!128x^8\!-\!384\pi x^7\!+\!\left(480\pi^2\!-\!2880\right)x^6\!+\!\left(\!-\!320\pi^3\!+\!5952\pi \right)x^5 \\ [1.0 ex]
\!\!&\! \!&\!\!\!+\!\left(120\pi^4\!-\!4752\pi^2\!+\!4032 \right)x^4\!+\!\left(\!-\!24\pi^5\!+\!1824\pi^3\!-\!8064\pi \right)x^3\\ [1.0 ex]
\!\!&\! \!&\!\!\!+\!\left(2\pi^6\!-\!348\pi^4\!+\!6048\pi^2\!-\!30240\right)x^2\!+\!\left(36\pi^5\!-\!2016\pi^3\!+\!22680\pi \right)x\\[1.0 ex]
\!\!&\!\!&\!\! -3\pi^6+252\pi^4-22680,
\end{array}$$ for $ x\!\in\!(0,c_1)$. The fourth derivate of the polynomial $\hat {Q}_8(x)$ is the polynomial of $4^{th}$ degree $$\label{Ineq_48}
\begin{array}{rcl}
\hat {Q}_8^{(iv)}(x)\!\!&\!=\!&\!\! 215040 x^4-322560\pi x^3+(172800\pi^2-1036800) x^2\\[1.0 ex]
\!\!&\! \!&\!\! +(-38400\pi^3+714240\pi) x+2880\pi^4-114048\pi^2+96768.
\end{array}$$ Using MATLAB software we can determine the real numerical factorization of the polynomial $$\label{Ineq_49}
\begin{split}
\hat {Q}_8^{(iv)}(x)=\alpha(x- x_1)(x- x_2)(x- x_3)(x- x_4),
\end{split}$$ with values 