\(% 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}}\)

2016.3.6 Question 6

• In the case where \(B > A > 0\) or \(-B < -A < 0\), notice that \[ R \cosh (x + \gamma ) = R \cosh x \cosh \gamma + R \sinh x \sinh \gamma . \]

  Therefore, we would like \(R \sinh \gamma = A\) and \(R \cosh \gamma = B\).

  Since \(\cosh \gamma ^2 - \sinh \gamma ^2 = 1\), we have \(R^2 = B^2 - A^2\).

  We also have \(\tanh \gamma = \frac {A}{B}\), and therefore \(\gamma = \artanh \frac {A}{B}\).

  Notice that \(\cosh \gamma > 0\), so \(R\) must have the same sign as \(B\).

  • If \(B > A > 0\), \(R = \sqrt {B^2 - A^2}\).
  • If \(B < -A < 0\), \(R = - \sqrt {B^2 - A^2}\).

• In the case where \(-A < B < A\), notice that \[ R \sinh (x + \gamma ) = R \sinh \gamma \cosh x + R \cosh \gamma \sinh x. \]

  Therefore, we would like \(R \cosh \gamma = A\) and \(R \sinh \gamma = B\).

  Since \(\cosh \gamma ^2 - \sinh \gamma ^2 = 1\), we have \(R^2 = B^2 - A^2\).

  We also have \(\tanh \gamma = \frac {B}{A}\), and therefore \(\gamma = \artanh \frac {B}{A}\).

  Notice that \(\cosh \gamma > 0\), so \(R\) will have the same sign as \(A\), and hence \(R = \sqrt {A^2 - B^2}\).

• When \(B = A\), we have \begin {align*} A \sinh x + B \cosh x & = A \frac {e^x - e^{-x}}{2} + A \frac {e^x + e^{-x}}{2} \\ & = A e^x. \end {align*}
• When \(B = -A\), we have \begin {align*} A \sinh x + B \cosh x & = A \frac {e^x - e^{-x}}{2} - A \frac {e^x + e^{-x}}{2} \\ & = A e^{-x}. \end {align*}

Therefore, in conclusion, \[ A \sinh x + B \cosh x = \begin {cases} \sqrt {B^2 - A^2} \cosh \left (x + \artanh \frac {A}{B}\right ), & 0 < A < B, \\ Ae^x, & 0 < B = A, \\ \sqrt {A^2 - B^2} \sinh \left (x + \artanh \frac {B}{A}\right ), & -A < B < A, \\ -Ae^{-x}, & B = -A < 0, \\ -\sqrt {B^2 - A^2} \cosh \left (x + \artanh \frac {A}{B}\right ), & -B < -A < 0. \\ \end {cases} \]

1. We have \(\sech x = a \tanh x + b\), and hence \(1 = a \sinh x + b \cosh x\). If \(b > a > 0\), we have \[ \sqrt {b^2 - a^2} \cosh \left (x + \artanh \frac {a}{b}\right ) = 1. \]

   Therefore, \begin {align*} \cosh \left (x + \artanh \frac {a}{b}\right ) & = \frac {1}{\sqrt {b^2 - a^2}} \\ x + \artanh \frac {a}{b} & = \pm \arcosh \frac {1}{\sqrt {b^2 - a^2}} \\ x & = \pm \arcosh \frac {1}{\sqrt {b^2 - a^2}} - \artanh \frac {a}{b}, \end {align*}

   as desired.

2. When \(a > b > 0\), \[ \sqrt {a^2 - b^2} \sinh \left (x + \artanh \frac {b}{a}\right ) = 1. \]

   Therefore, \begin {align*} \sinh \left (x + \artanh \frac {b}{a}\right ) & = \frac {1}{\sqrt {a^2 - b^2}} \\ x + \artanh \frac {b}{a} & = \arsinh \frac {1}{\sqrt {a^2 - b^2}} \\ x & = \arsinh \frac {1}{\sqrt {a^2 - b^2}} - \artanh \frac {b}{a}. \end {align*}

3. We would like to have two solutions to the equation \(1 = a \sinh x + b \cosh x\).

   • \(0 < a < b\), this gives \[ x = \pm \arcosh \frac {1}{\sqrt {b^2 - a^2}} - \artanh \frac {a}{b}, \]

     For this to make sense, we must have \(\frac {1}{\sqrt {b^2 - a^2}} \geq 1\), and therefore \(0 < \sqrt {b^2 - a^2} \leq 1\), which is \(0 < b^2 - a^2 \leq 1\).

     For this to have two distinct points, we would like to have \(\arcosh \frac {1}{\sqrt {b^2 - a^2}} \neq 0\) as well. This means \(b^2 - a^2 \neq 1\).

     Therefore, in this case, this means that \(a < b < \sqrt {a^2 + 1}\).

   • \(b = a\), this gives \(a e^x = 1\), which gives a unique solution \(x = - \ln a\).
   • \(-a < b < a\), this gives \[ \sqrt {A^2 - B^2} \sinh \left (x + \artanh \frac {B}{A}\right ) = 1, \] which can only give the solution \(x = \arsinh \frac {1}{\sqrt {A^2 - B^2}} - \artanh \frac {B}{A}\).
   • \(b = -a\), this gives \(-a e^{-x} = 1\), which does not have a solution.
   • \(-b < -a < 0\), this gives \[ -\sqrt {b^2 - a^2} \cosh \left (x + \artanh \frac {a}{b}\right ) = 1, \] but this is impossible, since both square root and \(\cosh \) are always positive.

   Therefore, the only possibility is when \(a < b < \sqrt {a^2 + 1}\).

4. When they touch at a point, this will mean at this value, the number of solutions will change on both sides. This is only possible when \(b = \sqrt {a^2 + 1}\).

   Therefore, \[ x = - \artanh \frac {a}{\sqrt {a^2 + 1}}. \]

   Hence, \begin {align*} y & = a \tanh x + b \\ & = - a \cdot \frac {a}{\sqrt {a^2 + 1}} + \sqrt {a^2 + 1} \\ & = \frac {-a^2 + a^2 + 1}{\sqrt {a^2 + 1}} \\ & = \frac {1}{\sqrt {a^2 + 1}}. \end {align*}