\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Since \(f''(t) > 0\) for \(t \in (0, x_0)\), we must have that for all \(x \in (0, x_0)\), we have \(f''(t) > 0\) for \(t \in (0, x)\), and hence \[ \int _{0}^{x} f''(t) \Diff t = f'(x) - f'(0) > 0. \]
But since \(f'(0) = 0\), this implies that \(f'(x) > 0\) for \(x \in (0, x_0)\).
Repeating this exact step gives that \(f(x) > 0\) for \(x \in (0, x_0)\) as desired.
We would like to show \(f(x) = 1 - \cos x \cosh x > 0\) for \(x \in \left (0, \frac {1}{2}\pi \right )\). Notice that \(f(0) = 1 - 1 \cdot 1 = 0\), and \[ f'(x) = \sin x \cosh x - \cos x \sinh x, \] which means \[ f'(0) = 0 \cdot 1 - 1 \cdot 0 = 0. \]
Further differentiation gives \[ f''(x) = \cos x \cosh x + \sin x \sinh x + \sin x \sinh x - \cos x \cosh x = 2 \sin x \sinh x. \]
If \(x \in \left (0, \frac {\pi }{2}\right )\), we have \(\sin x > 0\) and \(\sinh x > 0\), which gives \(f''(x) > 0\).
From the lemma we proved we have \(f(x) > 0\) for \(x \in \left (0, \frac {\pi }{2}\right )\), which is exactly \(\cos x \cosh x < 1\) as desired.
What is desired is to show \(\sin x \cosh x - x > 0\) and \(x^2 - \sin x \sinh x > 0\) for \(x \in \left (0, \frac {\pi }{2}\right )\).
Let \(g(x) = \sin x \cosh x - x\) and \(h(x) = x^2 - \sin x \sinh x\). \(g(0) = 0 \cdot 1 - 0 = 0\) and \(h(0) = 0^2 - 0 \cdot 0 = 0\).
Differentiating gives \[ g'(x) = \cos x \cosh x + \sin x \sinh x - 1, \] and \[ h'(x) = 2x - \cos x \sinh x - \sin x \cosh x. \]
Hence, \[ g'(0) = 1 \cdot 1 + 0 \cdot 0 - 1 = 0, \] and \[ h'(0) = 2 \cdot 0 - 1 \cdot 0 - 0 \cdot 1 = 0. \]
Differentiating this again gives \[ g''(x) = - \sin x \cosh x + \cos x \sinh x + \cos x \sinh x + \sin x \cosh x = 2 \cos x \sinh x, \] and \[ h''(x) = 2 + \sin x \sinh x - \cos x \cosh x - \cos x \cosh x - \sin x \sinh x = 2 - 2 \cos x \cosh x. \]
For \(x \in (0, \frac {\pi }{2})\), we notice that \(\cos x > 0\) and \(\sinh x > 0\), and so \(g''(x) > 0\). Also, notice that \(h''(x) = 2 f(x)\) so \(h''(x) > 0\).
Hence, \(g(x) > 0, h(x) > 0\) when \(x \in (0, \frac {\pi }{2})\) which proves the result as desired.