<s>[INST] 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
 [/INST] 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
$$\begin{aligned}
 1 &=f(t)<f\left(\frac{52 \pi}{180}\right) \\
 & \Rightarrow \sec (52^{\circ})-\cos (52^{\circ})>1 .
\end{aligned}$$
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