Without any software and approximations prove that $\sec(52^{\circ})-\cos(52^{\circ})>1$ Without any software and approximations prove that $$\sec(52^{\circ})-\cos(52^{\circ})>1$$
We can use some known trig values like $18^{\circ}$,$54^{\circ}$,etc
My try:
I considered the function:
$$f(x)=\sec(x)-\cos(x)-1,\: x\in \left (0, \frac{\pi}{3}\right)$$
We have the derivative as:
$$f'(x)=\sec x\tan x+\sin x >0$$ so $f$ is Monotone increasing.
So we have:
$$f(52^{\circ})>f(45^{\circ})=\frac{1}{\sqrt{2}}-1$$ but not able to proceed
 A: For angles less than 60 degrees, $\cos(3x)$ is a decreasing function of $\cos(x)$. So apply triple angle identity twice to $\cos(52)<(\sqrt{5}-1)/2$ to get the equivalent formulation $\cos(72) > (7033 -3145\sqrt{5})/2$. Since $\cos(72)=(\sqrt{5}-1)/4$, the inequality reduces to $521/233 < \sqrt{5}$, which is true, though just barely.
A: We have that for $\theta\in(0,\pi/2)$, $f(\theta)=\sec \theta -\cos \theta$ is an increasing function and
$$\sec \theta-\cos \theta =1 \implies \cos \theta = \frac{\sqrt 5-1}2 =\frac 1\varphi$$
that is $\theta$ is an angle of a Kepler_triangle, precisely $\theta=51.83°$ (almost equal to the slope of the Great Pyramid of Giza), therefore
$$\sec (52°)-\cos (52°) >1$$

(credit)

Assuming we don't know the value for Kepler angle, according to the following triangles

we have
$$\beta - \theta > \sqrt{(\sin \beta - \sin \theta)^2+(\cos \beta - \cos \theta)^2}>0.0379 \;\text{rad}>2° \implies \theta <52°$$
A: Thanks to "user" for giving me thought to complete the proof.
Here is the proof:
We are aiming to prove $\sec(52^{\circ})-\cos(52^{\circ})>1$. Let $\phi$ be the Golden ratio and $\psi$ be its reciprocal..
Consider $$f(x)=\sec x-\cos x,\:\:0<x<\frac{\pi}{2}$$
Its evident that $f$ is Monotone increasing.
Also $$\begin{aligned}
 & f(t)=1 \\
 \Rightarrow & \sec t-\frac{1}{\sec t}=1 \\
 \Rightarrow & \sec t=\phi \\
 \Rightarrow & t=\sec ^{-1}(\phi)
\end{aligned}$$
We know that by taylor's series:
$$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+\cdots,|x| \leqslant 1$$
Using the fact that:
$$\sec ^{-1}(v)=\cos ^{-1}\left(\frac{1}{v}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{v}\right)$$
So we have
$$t=\sec ^{-1}(\phi)=\frac{\pi}{2}-\left(\psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}+p\right)$$ Where $p>0$.
Also let $$t_0=\frac{\pi}{2}-\left(\psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}\right)$$
We know that $\psi$ satisfies $$\psi^2+\psi-1=0$$
We have the following results which can be easily derived:
$$\begin{aligned}
 \psi^2 &=1-\psi \\
 \psi^3 &=2 \psi-1 \\
 \psi^5 &=5 \psi-3 \\
 \Rightarrow & \psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}=\frac{205 \psi-47}{120} .
\end{aligned}$$
We have:
$$\begin{aligned}
 & t_0=\frac{\pi}{2}-\left(\psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}\right) \\
 \Rightarrow & t_0=\frac{\pi}{2}-\left(\frac{205 \psi-47}{120}\right) \\
 \Rightarrow & \frac{52 \pi}{180}-t_0=\frac{205 \psi-47}{120}-\frac{19 \pi}{90}=\frac{615 \psi-(141+76 \pi)}{360}>0
\end{aligned}$$
Hence we have :
$$\begin{aligned}
 & t_0<\frac{52 \pi}{180} \\
 \Rightarrow & t=t_0-p<\frac{52 \pi}{180}
\end{aligned}$$
Finally we have
$$\begin{aligned}
 1 &=f(t)<f\left(\frac{52 \pi}{180}\right) \\
 & \Rightarrow \sec (52^{\circ})-\cos (52^{\circ})>1 .
\end{aligned}$$
A: Hint .
Can you show the inequality for $10\leq x\leq 65$:
$$f\left(x\right)=x\left(\sec\left(x\cdot\frac{\pi}{180}\right)-\cos\left(\frac{\pi}{180}\cdot x\right)\right)> \frac{2(x-9)^{3}-2(x-9)^{2}+15(x-9)+75}{3000}$$
?
Some other hint :
Define :
$$h(x)=\left(\sec\left(x\cdot\frac{\pi}{180}\right)-\cos\left(\frac{\pi}{180}\cdot x\right)\right),p(x)=\frac{2(x-9)^{3}-2(x-9)^{2}+15(x-9)+75}{3000}$$
Then we have for $10<x<65$ :
$$h''(x)>0,p''(x)>0$$
So we have using strong convexity for $x\in[45,52]$ :
$$xh(x)\geq x\left(h'\left(45\right)\left(x-45\right)+h\left(45\right)+\frac{h''\left(45\right)}{2}\left(x-45\right)^{2}\right)>p(x)$$



If $0<a<1$ and $0\leq x\leq 2a\pi$ then :
$$1-\cos\left(x\right)-\left(\frac{\sin\left(a\pi\right)}{a\pi}\right)^{2}\cdot\frac{x^{2}}{2}\geq 0$$
See [1] for a reference .
Now a lemma using the concavity of $\sin(x)$ on $(0,\pi/2)$ :
We have :
$$0<\frac{\sin\left(\frac{\pi}{6.25}\right)-\sin\left(\frac{\pi}{6}\right)}{\frac{\pi}{6.25}-\frac{\pi}{6}}-\frac{\sqrt{3}}{2}$$
A second lemma :
$$\pi<\frac{185}{100}\sqrt{3}$$
Using lemma 1 and 2 with $a=1/6.25$ and $x=52\cdot\frac{\pi}{180}$ for the first inequality we got :
$$\left(\frac{6.25}{\pi}\left(\frac{\sqrt{3}}{2}\left(\frac{1.85\sqrt{3}}{6.25}-\frac{1.85\sqrt{3}}{6}\right)+\frac{1}{2}\right)\right)^{2}\cdot\frac{\left(52\cdot\frac{\pi}{180}\right)^{2}}{2}=1934881/512000<1-\cos\left(52\cdot\frac{\pi}{180}\right)$$
Wich is sufficient to show the claim proposed in the comment @MartinR.
Reference :
[1] Becker,Michael and Lawrence E.Stark An extremal inequality for the Fourier coefficients of positive cosine polynomials ,Univ. Beograd .Publ. Elektrotehn Fak. Ser . Mat ., No.577-No.588 (1977)57-58.
A: I propose another way using the idea of golden ratio with triple-angle formula.
We have that for $\theta\in(0,\pi/2)$, $f(\theta)=\sec \theta -\cos \theta$ is an increasing function and
$$\sec \theta-\cos \theta =1 \implies \cos \theta = \frac{\sqrt 5-1}2 =\frac 1\varphi$$
Since $\theta < \beta =54°$ we have that
$$\theta < \alpha =52° \iff \cos \theta<\cos \alpha \iff \cos (3\theta)> \cos (3\alpha)$$
with
$$\cos(3\theta)= 4\cos^3 \theta-3\cos \theta = \frac4{\varphi^3}-\frac3{\varphi}=5\varphi-9$$
and using golden gnomon properties

$$\cos(3\alpha)=\cos( 4\beta-60°)=\cos(4\beta)\cos(60°)+\sin(4\beta)\sin(60°)=$$
$$=-\cos(36°)\cos(60°)-\sin(36°)\sin(60°)=-\frac{\varphi}4-\frac{\sqrt 3}2\sqrt{\frac 3 4 -\frac \varphi 4}$$
which leads to
$$\cos (3\theta)> \cos (3\alpha)\iff \frac{\sqrt 3}2\sqrt{\frac 3 4 -\frac \varphi 4}>9-\frac{21}4 \varphi \iff \varphi >\frac{144}{89}$$
which is true, indeed $F_{11}=89$ and $F_{12}=144$ are two consecutive Fibonacci numbers and their ratios approximate $\varphi$ such that
$$\frac{F_{n+1}}{F_{n}}<\varphi $$
for $n$ odd, therefore $\theta <52°$ and the given inequality holds.
