Namespace TrigHyperbolic

Method:

We know, \( \cosh y = \frac{e^y + e^{-y}}{2} \). Solving for \( y \) and substituting \( \cosh y = x \), \[ \cosh^{-1} x = \ln \left( x + \sqrt{x^2-1} \right) \]

Method:

We know, \( \sinh y = \frac{e^y - e^{-y}}{2} \). Solving for \( y \) and substituting \( \sinh y = x \), \[ \sinh^{-1} x = \ln \left( x + \sqrt{x^2+1} \right) \]

Method:

We know, \( \tanh y = \frac{e^{2y} - 1}{e^{2y} + 1} \). Solving for \( y \) and substituting \( \tanh y = x \), \[ \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \]

Method:

Use the standard series definition of \( \cosh \).

\[ \cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \]

Method:

Use the standard series definition of \( \sinh \).

\[ \sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \]

Method:

\[ \tanh x = \frac{\sinh x}{\cosh x} \]