Vowe]{}, [J. Anglesio]{} proved the following double inequality $$\label{Ineq_3}
2+\nfrac{16}{\pi^4} x^3 \tan x
<
\left(\nfrac{\sin x}{x} \right)^{\!2} + \nfrac{\tan x}{x}
<
2+\nfrac{8}{45} x^3 \tan x,$$ for $x \!\in\! \left(0, \pi/2\right)$. In the paper [@[1]], [C. Mortici]{} has proved the following two statements:

For every $x \!\in\! \left(0, 1\right)$ we have$:$ $$\label{Ineq_4}
2+\left(\nfrac{8}{45}-a\left(x \right)\right) x^3 \tan x
<
\left(\nfrac{\sin x}{x}\right)^{\!2} + \nfrac{\tan x}{x}
<
2+\left(\nfrac{8}{45}-b\left(x \right)\right) x^3 \tan x,$$ where $a\left(x \right)= \nfrac{8}{945} x^2$, $ b\left(x \right)= \nfrac{8}{945} x^2- \nfrac{16}{14175} x^4$.

For every $x \!\in\! \left(\nfrac{\pi}{2}-\nfrac{1}{2}, \nfrac{\pi}{2} \right)$ in the left-hand side and for every $x \!\in\! \left(\nfrac{\pi}{3}-\nfrac{1}{2}, \nfrac{\pi}{2} \right)$ in the right-hand side the following inequalities are true$:$ $$\label{Ineq_5}
2+\left(\nfrac{16}{\pi^4}+c\left(x \right)\right)x^3 \tan x
<
\left(\nfrac{\sin x}{x} \right)^{\!2}+\nfrac{\tan x}{x}
<
2+\left(\nfrac{16}{\pi^4}+d\left(x \right) \right)x^3 \tan x ,$$ where\

$$c\left(x\right)
\!=\!
\left(\nfrac{160}{\pi^5}\!-\!\nfrac{16}{\pi^3}\right)\!\left(\nfrac{\pi}{2}\!-\!x\right)
,
~
d\left(x\right)
\!=\!
\left(\nfrac{160}{\pi^5} \!-\!\nfrac{16}{\pi^3}\right)\!\left(\nfrac{\pi}{2}\!-\!x\right)\!+\!\left(\nfrac{960}{\pi^6}\!-\!\nfrac{96}{\pi^4}\right)\!\left(\nfrac{\pi}{2}\!-\!x\right)^{\!2}.$$

Theorem 1.1. and Theorem 1.2. describe a subtly analysis of Wilker inequality by [C. Mortici]{}. The method of proving inequalites in this paper was given in the paper [@[7]] and it is based on use of appropriate approximations of some mixed trigonometric polynomials with finite Taylor series. The method presents continuation of method of [C. Mortici]{} presented in [@[6]]. The method from the paper [@[7]] 