For concrete indices $(i_0, i_1, i_2, i_3, i_4) = (1, 2, 1, 2, 2)$, we obtain

$$ 
\begin{aligned}
P_{1,2,1}^{+}(x) &= 190 \cdot x \cdot \underline{T}_{4 \cdot 1+2}^{\cos,0}(x) + 20 \cdot x \cdot \underline{T}_{4 \cdot 2+2}^{\cos,0}(2x) + 30 \cdot x^2 \cdot \underline{T}_{4 \cdot 1+3}^{\sin,0}(x) + 3 \cdot x^5, \\
P_{1,2,1}^{+}(x) &= -\frac{16}{2835} \cdot x^{11} + \frac{61}{504} \cdot x^9 - \frac{43}{24} \cdot x^7 + \frac{77}{4} \cdot x^5 - 105 \cdot x^3 + 210 \cdot x
\end{aligned}
$$

and

$$ 
\begin{aligned}
P_{2,2}^{-}(x) &= -70 \cdot \overline{T}_{4 \cdot 2+1}^{\sin,0}(x) - 70 \cdot \overline{T}_{4 \cdot 2+1}^{\sin,0}(2x), \\
P_{2,2}^{-}(x) &= -\frac{19}{192} \cdot x^9 + \frac{43}{24} \cdot x^7 - \frac{77}{4} \cdot x^5 + 105 \cdot x^3 - 210 \cdot x.
\end{aligned}
$$

Finally, for the MTP function

$$ f(x) = f^{+}(x) + f^{-}(x) $$

we obtain the concrete downward polynomial approximation

$$ 
\begin{aligned}
P(x) &= P_{1,2,1}^{+}(x) + P_{2,2}^{-}(x) \\
&= -\frac{16}{2835} \cdot x^{11} + \frac{89}{4032} \cdot x^9
\end{aligned}
$$

over $(0, \frac{\pi}{2})$, i.e. it holds that

$$ f(x) > P(x) $$

over $(0, \frac{\pi}{2})$.

IV (The final part) Based on the Sturm theorem, the following inequality

$$ P(x) > 0 $$

is true over $(0, \frac{\pi}{2})$. The stated conclusion for the polynomial function $P$ is correct based on the following facts:

1. We can conclude, by Sturm theorem, that the polynomial function $P(x)$ has only one zero over the concrete extended segment $[-0.1, 1.58]$ of the initial interval $(0, \frac{\pi}{2})$.

2. Facts $P(-0.1) \neq 0$, $P(0) = 0$ and $P(1.58) \neq 0$ are correct.

3. The polynomial $P$ is positive at boundary point $\frac{\pi}{2}$ ($P(\frac{\pi}{2}) = 0.47438... > 0$). Therefore, the following inequality

$$ f(x) > 0 $$

is true over $(0, \frac{\pi}{2})$. $\square$

## APPENDIX A7

The initial MTP function is

$$ 
\begin{aligned}
f(x) = & 16 \cdot x^6 + 135 \cdot x \cdot \sin x \cdot \cos^2 x + (90 \cdot x^3 - 360 \cdot x) \cdot \sin x \\
& +360 \cdot \cos^3 x + (585 \cdot x^2 - 360) \cdot \cos x
\end{aligned}
$$

and the initial interval is $\mathbb{S} = (0, \frac{\pi}{2})$.

Automated proof that $f(x) > 0$ for $x \in (0, \frac{\pi}{2})$:

I (Recognition of possible case) Facts $f(0) = 0$ and $f(\frac{\pi}{2}) \neq 0$ are correct. The MTP function $f$ is positive at boundary point $\frac{\pi}{2}$ ($f(\frac{\pi}{2}) = 23.68123... > 0$). Therefore, it is possible that $f(x) > 0$ over $(0, \frac{\pi}{2})$.
For concrete indices $(i_0, i_1, i_2, i_3, i_4) = (1, 2, 1, 2, 2)$, we obtain

$$ 
\begin{aligned}
P_{1,2,1}^{+}(x) &= 190 \cdot x \cdot \underline{T}_{4 \cdot 1+2}^{\cos,0}(x) + 20 \cdot x \cdot \underline{T}_{4 \cdot 2+2}^{\cos,0}(2x) + 30 \cdot x^2 \cdot \underline{T}_{4 \cdot 1+3}^{\sin,0}(x) + 3 \cdot x^5, \\
P_{1,2,1}^{+}(x) &= -\frac{16}{2835} \cdot x^{11} + \frac{61}{504} \cdot x^9 - \frac{43}{24} \cdot x^7 + \frac{77}{4} \cdot x^5 - 105 \cdot x^3 + 210 \cdot x
\end{aligned}
$$

and

$$ 
\begin{aligned}
P_{2,2}^{-}(x) &= -70 \cdot \overline{T}_{4 \cdot 2+1}^{\sin,0}(x) - 70 \cdot \overline{T}_{4 \cdot 2+1}^{\sin,0}(2x), \\
P_{2,2}^{-}(x) &= -\frac{19}{192} \cdot x^9 + \frac{43}{24} \cdot x^7 - \frac{77}{4} \cdot x^5 + 105 \cdot x^3 - 210 \cdot x.
\end{aligned}
$$

Finally, for the MTP function

$$ f(x) = f^{+}(x) + f^{-}(x) $$

we obtain the concrete downward polynomial approximation

$$ 
\begin{aligned}
P(x) &= P_{1,2,1}^{+}(x) + P_{2,2}^{-}(x) \\
&= -\frac{16}{2835} \cdot x^{11} + \frac{89}{4032} \cdot x^9
\end{aligned}
$$

over $(0, \frac{\pi}{2})$, i.e. it holds that

$$ f(x) > P(x) $$

over $(0, \frac{\pi}{2})$.

IV (The final part) Based on the Sturm theorem, the following inequality

$$ P(x) > 0 $$

is true over $(0, \frac{\pi}{2})$. The stated conclusion for the polynomial function $P$ is correct based on the following facts:

1. We can conclude, by Sturm theorem, that the polynomial function $P(x)$ has only one zero over the concrete extended segment $[-0.1, 1.58]$ of the initial interval $(0, \frac{\pi}{2})$.

2. Facts $P(-0.1) \neq 0$, $P(0) = 0$ and $P(1.58) \neq 0$ are correct.

3. The polynomial $P$ is positive at boundary point $\frac{\pi}{2}$ ($P(\frac{\pi}{2}) = 0.47438... > 0$). Therefore, the following inequality

$$ f(x) > 0 $$

is true over $(0, \frac{\pi}{2})$. $\square$

## APPENDIX A7

The initial MTP function is

$$ 
\begin{aligned}
f(x) = & 16 \cdot x^6 + 135 \cdot x \cdot \sin x \cdot \cos^2 x + (90 \cdot x^3 - 360 \cdot x) \cdot \sin x \\
& +360 \cdot \cos^3 x + (585 \cdot x^2 - 360) \cdot \cos x
\end{aligned}
$$

and the initial interval is $\mathbb{S} = (0, \frac{\pi}{2})$.

Automated proof that $f(x) > 0$ for $x \in (0, \frac{\pi}{2})$:

I (Recognition of possible case) Facts $f(0) = 0$ and $f(\frac{\pi}{2}) \neq 0$ are correct. The MTP function $f$ is positive at boundary point $\frac{\pi}{2}$ ($f(\frac{\pi}{2}) = 23.68123... > 0$). Therefore, it is possible that $f(x) > 0$ over $(0, \frac{\pi}{2})$.
II (Transformation of angles) After the transformation of terms $\cos^m x \cdot \sin^n x$ ($m, n \in N_0$) into the sum of sine and cosine functions of multiple angles, in the MTP function $f$, we obtain

$$f(x) = 90 \cdot \cos 3x + (90 \cdot x^3 - \frac{1305}{4} \cdot x) \cdot \sin x + (585 \cdot x^2 - 90) \cdot \cos x + 16 \cdot x^6 + \frac{135}{4} \cdot x \cdot \sin 3x.$$

Then, we consider the previous expression as two separate expressions, the first with positive and the second with negative terms next to sine and cosine functions

$$f^{+}(x) = 90 \cdot \cos 3x + \frac{135}{4} \cdot x \cdot \sin 3x + 585 \cdot x^2 \cdot \cos x + 90 \cdot x^3 \cdot \sin x + 16 \cdot x^6,$$

$$f^{-}(x) = -90 \cdot \cos x - \frac{1305}{4} \cdot x \cdot \sin x.$$

III (Determination of downward rational polynomial approximation) After substitution of sine and cosine functions by appropriate (downward or upward) polynomial approximations, we obtain downward polynomial approximations of $f^+(x)$ and $f^-(x)$ respectively

$$P_{i_0, i_1, i_2, i_3}^{+}(x) = 90 \cdot \frac{T_{4 \cdot i_0 + 2}^{\cos, 0}}{4} (3x) + \frac{135}{4} \cdot x \cdot \frac{T_{4 \cdot i_1 + 3}^{\sin, 0}}{4} (3x) + 585 \cdot x^2 \cdot \frac{T_{4 \cdot i_2 + 2}^{\cos, 0}}{4} (x) + 90 \cdot x^3 \cdot \frac{T_{4 \cdot i_3 + 3}^{\sin, 0}}{4} (x) + 16 \cdot x^6,$$

$$P_{i_4, i_5}^{-}(x) = -90 \cdot \frac{T_{4 \cdot i_4}^{\cos, 0}}{4} (x) - \frac{1305}{4} \cdot x \cdot \frac{T_{4 \cdot i_5 + 1}^{\sin, 0}}{4} (x).$$

For concrete indices $(i_0, i_1, i_2, i_3, i_4, i_5) = (2, 3, 1, 1, 2, 2)$, we obtain

$$P_{2,3,1,1}^{+}(x) = 90 \cdot \frac{T_{4 \cdot 2+2}^{\cos, 0}}{4} (3x) + \frac{135}{4} \cdot x \cdot \frac{T_{4 \cdot 3+3}^{\sin, 0}}{4} (3x) + 585 \cdot x^2 \cdot \frac{T_{4 \cdot 1+2}^{\cos, 0}}{4} (x) + 90 \cdot x^3 \cdot \frac{T_{4 \cdot 1+3}^{\sin, 0}}{4} (x) + 16 \cdot x^6,$$

$$P_{2,3,1,1}^{+}(x) = -\frac{531441}{1435033600} \cdot x^{16} + \frac{177147}{20500480} \cdot x^{14} - \frac{59049}{394240} \cdot x^{12} + \frac{6241}{17920} \cdot x^{10} - \frac{1}{16} \cdot x^{8} + \frac{83}{32} \cdot x^{6} - \frac{405}{8} \cdot x^{4} + \frac{1125}{4} \cdot x^{2} + 90$$

and

$$P_{2,2}^{-}(x) = -90 \cdot \frac{T_{4 \cdot 2}^{\cos, 0}}{4} (x) - \frac{1305}{4} \cdot x \cdot \frac{T_{4 \cdot 2+1}^{\sin, 0}}{4} (x),$$

$$P_{2,2}^{-}(x) = -\frac{29}{32256} \cdot x^{10} + \frac{1}{16} \cdot x^{8} - \frac{83}{32} \cdot x^{6} + \frac{405}{8} \cdot x^{4} - \frac{1125}{4} \cdot x^{2} - 90.$$

Finally, for the MTP function

$$f(x) = f^{+}(x) + f^{-}(x)$$

we obtain the concrete downward polynomial approximation

$$P(x) = P_{2,3,1,1}^{+}(x) + P_{2,2}^{-}(x) = -\frac{531441}{1435033600} \cdot x^{16} + \frac{177147}{20500480} \cdot x^{14} - \frac{59049}{394240} \cdot x^{12} + \frac{7003}{20160} \cdot x^{10}$$

over $(0, \frac{\pi}{2})$, i.e. it holds that

$$f(x) > P(x)$$

over $(0, \frac{\pi}{2})$.
II (Transformation of angles) After the transformation of terms $\cos^m x \cdot \sin^n x$ ($m, n \in N_0$) into the sum of sine and cosine functions of multiple angles, in the MTP function $f$, we obtain

$$f(x) = 90 \cdot \cos 3x + (90 \cdot x^3 - \frac{1305}{4} \cdot x) \cdot \sin x + (585 \cdot x^2 - 90) \cdot \cos x + 16 \cdot x^6 + \frac{135}{4} \cdot x \cdot \sin 3x.$$

Then, we consider the previous expression as two separate expressions, the first with positive and the second with negative terms next to sine and cosine functions

$$f^{+}(x) = 90 \cdot \cos 3x + \frac{135}{4} \cdot x \cdot \sin 3x + 585 \cdot x^2 \cdot \cos x + 90 \cdot x^3 \cdot \sin x + 16 \cdot x^6,$$

$$f^{-}(x) = -90 \cdot \cos x - \frac{1305}{4} \cdot x \cdot \sin x.$$

III (Determination of downward rational polynomial approximation) After substitution of sine and cosine functions by appropriate (downward or upward) polynomial approximations, we obtain downward polynomial approximations of $f^+(x)$ and $f^-(x)$ respectively

$$P_{i_0, i_1, i_2, i_3}^{+}(x) = 90 \cdot \frac{T_{4 \cdot i_0 + 2}^{\cos, 0}}{4} (3x) + \frac{135}{4} \cdot x \cdot \frac{T_{4 \cdot i_1 + 3}^{\sin, 0}}{4} (3x) + 585 \cdot x^2 \cdot \frac{T_{4 \cdot i_2 + 2}^{\cos, 0}}{4} (x) + 90 \cdot x^3 \cdot \frac{T_{4 \cdot i_3 + 3}^{\sin, 0}}{4} (x) + 16 \cdot x^6,$$

$$P_{i_4, i_5}^{-}(x) = -90 \cdot \frac{T_{4 \cdot i_4}^{\cos, 0}}{4} (x) - \frac{1305}{4} \cdot x \cdot \frac{T_{4 \cdot i_5 + 1}^{\sin, 0}}{4} (x).$$

For concrete indices $(i_0, i_1, i_2, i_3, i_4, i_5) = (2, 3, 1, 1, 2, 2)$, we obtain

$$P_{2,3,1,1}^{+}(x) = 90 \cdot \frac{T_{4 \cdot 2+2}^{\cos, 0}}{4} (3x) + \frac{135}{4} \cdot x \cdot \frac{T_{4 \cdot 3+3}^{\sin, 0}}{4} (3x) + 585 \cdot x^2 \cdot \frac{T_{4 \cdot 1+2}^{\cos, 0}}{4} (x) + 90 \cdot x^3 \cdot \frac{T_{4 \cdot 1+3}^{\sin, 0}}{4} (x) + 16 \cdot x^6,$$

$$P_{2,3,1,1}^{+}(x) = -\frac{531441}{1435033600} \cdot x^{16} + \frac{177147}{20500480} \cdot x^{14} - \frac{59049}{394240} \cdot x^{12} + \frac{6241}{17920} \cdot x^{10} - \frac{1}{16} \cdot x^{8} + \frac{83}{32} \cdot x^{6} - \frac{405}{8} \cdot x^{4} + \frac{1125}{4} \cdot x^{2} + 90$$

and

$$P_{2,2}^{-}(x) = -90 \cdot \frac{T_{4 \cdot 2}^{\cos, 0}}{4} (x) - \frac{1305}{4} \cdot x \cdot \frac{T_{4 \cdot 2+1}^{\sin, 0}}{4} (x),$$

$$P_{2,2}^{-}(x) = -\frac{29}{32256} \cdot x^{10} + \frac{1}{16} \cdot x^{8} - \frac{83}{32} \cdot x^{6} + \frac{405}{8} \cdot x^{4} - \frac{1125}{4} \cdot x^{2} - 90.$$

Finally, for the MTP function

$$f(x) = f^{+}(x) + f^{-}(x)$$

we obtain the concrete downward polynomial approximation

$$P(x) = P_{2,3,1,1}^{+}(x) + P_{2,2}^{-}(x) = -\frac{531441}{1435033600} \cdot x^{16} + \frac{177147}{20500480} \cdot x^{14} - \frac{59049}{394240} \cdot x^{12} + \frac{7003}{20160} \cdot x^{10}$$

over $(0, \frac{\pi}{2})$, i.e. it holds that

$$f(x) > P(x)$$

over $(0, \frac{\pi}{2})$.
IV (The final part) Based on the Sturm theorem, the following inequality

$$P(x) > 0$$

is true over $(0, \frac{\pi}{2})$. The stated conclusion for the polynomial function $P$ is correct based on the following facts:

1. We can conclude, by Sturm theorem, that the polynomial function $P(x)$ has only one zero over the concrete extended segment $[-0.1, 1.58]$ of the initial interval $(0, \frac{\pi}{2})$.

2. Facts $P(-0.1) \neq 0$, $P(0) = 0$ and $P(1.58) \neq 0$ are correct.

3. The polynomial $P$ is positive at boundary point $\frac{\pi}{2}$ ($P(\frac{\pi}{2}) = 2.27261... > 0$). Therefore, the following inequality

$$f(x) > 0$$

is true over $(0, \frac{\pi}{2})$.

□

**Bojan Banjac**
Computer Graphics Chair,
Faculty of Technical Sciences, University of Novi Sad,
Trg Dositeja Obradovića 16, Novi Sad, Serbia,
E-mail: bojan.banjac@uns.ac.rs

(Received 08. 03. 2024.)

(Revised 19. 04. 2024.)

**Branko Malešević**
Department of Applied Mathematics,
School of Electrical Engineering, University of Belgrade,
Bulevar kralja Aleksandra 73, Belgrade, Serbia,
E-mail: branko.malesevic@etf.bg.ac.rs

**Miloš Mićović**
Department of Applied Mathematics,
School of Electrical Engineering, University of Belgrade,
Bulevar kralja Aleksandra 73, Belgrade, Serbia,
E-mail: milos.micovic@etf.bg.ac.rs

**Bojana Mihailović**
Department of Applied Mathematics,
School of Electrical Engineering, University of Belgrade,
Bulevar kralja Aleksandra 73, Belgrade, Serbia,
E-mail: mihailovicb@etf.bg.ac.rs

**Milica Savatović**
Department of Applied Mathematics,
School of Electrical Engineering, University of Belgrade,
Bulevar kralja Aleksandra 73, Belgrade, Serbia,
E-mail: milica.makragic@etf.bg.ac.rs
IV (The final part) Based on the Sturm theorem, the following inequality

$$P(x) > 0$$

is true over $(0, \frac{\pi}{2})$. The stated conclusion for the polynomial function $P$ is correct based on the following facts:

1. We can conclude, by Sturm theorem, that the polynomial function $P(x)$ has only one zero over the concrete extended segment $[-0.1, 1.58]$ of the initial interval $(0, \frac{\pi}{2})$.

2. Facts $P(-0.1) \neq 0$, $P(0) = 0$ and $P(1.58) \neq 0$ are correct.

3. The polynomial $P$ is positive at boundary point $\frac{\pi}{2}$ ($P(\frac{\pi}{2}) = 2.27261... > 0$). Therefore, the following inequality

$$f(x) > 0$$

is true over $(0, \frac{\pi}{2})$.

□

**Bojan Banjac**
Computer Graphics Chair,

Faculty of Technical Sciences, University of Novi Sad,
Trg Dositeja Obradovića 16, Novi Sad, Serbia,
E-mail: bojan.banjac@uns.ac.rs

**Branko Malešević**
Department of Applied Mathematics,

School of Electrical Engineering, University of Belgrade,
Bulevar kralja Aleksandra 73, Belgrade, Serbia,
E-mail: branko.malesevic@etf.bg.ac.rs

**Miloš Mićović**
Department of Applied Mathematics,

School of Electrical Engineering, University of Belgrade,
Bulevar kralja Aleksandra 73, Belgrade, Serbia,
E-mail: milos.micovic@etf.bg.ac.rs

**Bojana Mihailović**
Department of Applied Mathematics,

School of Electrical Engineering, University of Belgrade,
Bulevar kralja Aleksandra 73, Belgrade, Serbia,
E-mail: mihailovicb@etf.bg.ac.rs

**Milica Savatović**
Department of Applied Mathematics,

School of Electrical Engineering, University of Belgrade,
Bulevar kralja Aleksandra 73, Belgrade, Serbia,
E-mail: milica.makragic@etf.bg.ac.rs

(Received 08. 03. 2024.)

(Revised 19. 04. 2024.)
SUPPLEMENTARY MATERIAL

Figure 1 illustrates the stratified family of functions

$$ \varphi_p(x) = -\cos x + \left(\frac{\sin x}{x}\right)^3 - \frac{1}{15}x^4 + p x^6 \quad (\text{for } x \in \left(0, \frac{\pi}{2}\right)) $$

from Lemma 2. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 1.

<figure><img src="image_2.png" /><figcaption>Figure 1: Stratified family of functions from Lemma 2</figcaption></figure>

Figure 2 illustrates the stratified family of functions

$$ \varphi_p(x) = -\frac{\sin x}{x} + \frac{\cos x + 2}{3} - \frac{1}{180}x^4 + p x^6 \quad (\text{for } x \in \left(0, \frac{\pi}{2}\right)) $$

from Lemma 4. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 2.
SUPPLEMENTARY MATERIAL

Figure 1 illustrates the stratified family of functions

$$ \varphi_p(x) = -\cos x + \left(\frac{\sin x}{x}\right)^3 - \frac{1}{15}x^4 + p x^6 \quad (\text{for } x \in \left(0, \frac{\pi}{2}\right)) $$

from Lemma 2. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 1.

<figure><img src="image_4.png" /><figcaption>Figure 1: Stratified family of functions from Lemma 2</figcaption></figure>

Figure 2 illustrates the stratified family of functions

$$ \varphi_p(x) = -\frac{\sin x}{x} + \frac{\cos x + 2}{3} - \frac{1}{180}x^4 + p x^6 \quad (\text{for } x \in \left(0, \frac{\pi}{2}\right)) $$

from Lemma 4. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 2.
<figure><img src="image_3.png" /><figcaption>Figure 2: Stratified family of functions from Lemma 4</figcaption></figure>

Figure 3 illustrates the stratified family of functions

$$ \varphi_p(x) = 2 \frac{\sin x}{x} + \frac{\tan x}{x} - 3 - \frac{3}{20} x^4 \frac{1}{\cos x} + p x^6 \frac{1}{\cos x} \quad (\text{for } x \in (0, \frac{\pi}{2})) $$

from Lemma 6. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 3.

<figure><img src="image_4.png" /><figcaption>Figure 3: Stratified family of functions from Lemma 6</figcaption></figure>
<figure><img src="image_5.png" /><figcaption>Figure 2: Stratified family of functions from Lemma 4</figcaption></figure>

Figure 3 illustrates the stratified family of functions

$$ \varphi_p(x) = 2 \frac{\sin x}{x} + \frac{\tan x}{x} - 3 - \frac{3}{20} x^4 \frac{1}{\cos x} + p x^6 \frac{1}{\cos x} \quad (\text{for } x \in (0, \frac{\pi}{2})) $$

from Lemma 6. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 3.

<figure><img src="image_6.png" /><figcaption>Figure 3: Stratified family of functions from Lemma 6</figcaption></figure>
Figure 4 illustrates the stratified family of functions

$$ \varphi_p(x) = \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} - 2 - \frac{8}{45} x^4 \frac{1}{\cos x} + p x^6 \frac{1}{\cos x} \quad (\text{for } x \in (0, \frac{\pi}{2})) $$

from Lemma 8. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 4.

<figure><img src="image_5.png" /><figcaption>Figure 4: Stratified family of functions from Lemma 8</figcaption></figure>

Figure 5 illustrates the stratified family of functions

$$ \varphi_p(x) = \left(\frac{x}{\sin x}\right)^2 + \frac{x}{\tan x} - 2 - p x^4 \quad (\text{for } x \in (0, \frac{\pi}{2})) $$

from Lemma 10. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 5.
Figure 4 illustrates the stratified family of functions

$$ \varphi_p(x) = \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} - 2 - \frac{8}{45} x^4 \frac{1}{\cos x} + p x^6 \frac{1}{\cos x} \quad (\text{for } x \in (0, \frac{\pi}{2})) $$

from Lemma 8. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 4.

<figure><img src="image_7.png" /><figcaption>Figure 4: Stratified family of functions from Lemma 8</figcaption></figure>

Figure 5 illustrates the stratified family of functions

$$ \varphi_p(x) = \left(\frac{x}{\sin x}\right)^2 + \frac{x}{\tan x} - 2 - p x^4 \quad (\text{for } x \in (0, \frac{\pi}{2})) $$

from Lemma 10. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 5.
<figure><img src="image_6.png" /><figcaption>Figure 5: Stratified family of functions from Lemma 10</figcaption></figure>

Figure 6 illustrates the stratified family of functions

$$ \varphi_p(x) = 3 \frac{x}{\sin x} + \cos x - 4 - \frac{1}{10} x^4 - p x^6 \quad (\text{for } x \in (0, \frac{\pi}{2})) $$

from Lemma 11. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 6.

<figure><img src="image_7.png" /><figcaption>Figure 6: Stratified family of functions from Lemma 11</figcaption></figure>
<figure><img src="image_8.png" /><figcaption>Figure 5: Stratified family of functions from Lemma 10</figcaption></figure>

Figure 6 illustrates the stratified family of functions

$$ \varphi_p(x) = 3 \frac{x}{\sin x} + \cos x - 4 - \frac{1}{10} x^4 - p x^6 \quad (\text{for } x \in (0, \frac{\pi}{2})) $$

from Lemma 11. Cases for all values of the parameter $p \in R^+$ are shown, highlighting those with constants obtained in Statement 6.

<figure><img src="image_9.png" /><figcaption>Figure 6: Stratified family of functions from Lemma 11</figcaption></figure>