Hyperbolic Functions

y=\frac{e^{x}}{2}|dotted
y=-\frac{e^{-x}}{2}|dotted
y=\frac{e^{-x}}{2}|dotted
\sinh(x)
\cosh(x)
\tanh(x)

to prove identities
- convert to Exponential Functions, simplify
Normal trig functions make a circle ( $\sin^{2} x + \cos^{2} x = 1$ ), but hyperbolic functions make a hyperbola ( $\cosh^{2} x - \sinh^{2} x = 1$ )
- Any point on this curve has coordinates $(\cosh x, \sinh x)$

Osborn's Rule

Take a Trigonometric Identities
If there is a product of 2 sines, change its sign

Sinh

$\frac{e^{x} - e^{- x}}{2}$
odd

Cosh

$\frac{e^{x} + e^{- x}}{2}$
even
Shape of a free hanging chain
- Literally the same anywhere (if you graph $\frac{x}{a}$ and $\frac{y}{a}$ , they are both dimensionless so they don't depend on anything)

Tanh

$\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$
Looks similar to a squished $\arctan$

Inverse

Find $\sinh^{- 1} (x)$ :

\begin{aligned} y & = \sinh^{- 1} x \\ x & = \sinh y \\ = \frac{1}{2} (e^{y} - e^{- y}) \\ 2 x e^{y} & = e^{2 y} - 1 \\ e^{2 y} - 2 x e^{y} - 1 & = 0 \\ e^{y} & = \frac{2 x \pm \sqrt{(- 2 x)^{2} - 4 (- 1)}}{2} \\ = x \pm \frac{1}{2} \sqrt{4 x^{2} + 4} \\ = x \pm \sqrt{x^{2} + 1} \\ = x + \sqrt{x^{2} + 1} ∵ e^{y} > 0 \\ y & = \ln (x + \sqrt{x^{2} + 1}) for all x \in R \end{aligned}

Find $\tanh^{- 1} x$ :

\begin{aligned} y & = \tanh^{- 1} x \\ x & = \tanh y \\ = \frac{e^{y} - e^{- y}}{e^{y} + e^{- y}} \\ x (e^{y} + e^{- y}) & = e^{y} - e^{- y} \\ (x - 1) e^{y} & = - (1 + x) e^{- y} \\ (x - 1) e^{2 y} & = - (1 + x) \\ e^{2 y} & = \frac{1 + x}{1 - x} \\ y & = \frac{1}{2} \ln (\frac{1 + x}{1 - x}) for - 1 < x < 1 \end{aligned}