Hyperbolic functions

We have used some hyperbolic functions relations in some of our chapters.

This is a summary of these functions to enable us to use them in future chapters.

We’ll go through their definitions and we’ll use the similarities with the regular trigonometric functions to see most of the formulas.

$No.$	Equation
$1.$	$\sinh x=\frac{e^{x}-e^{-x}}{2}$
$2.$	$\cosh x=\frac{e^{x}+e^{-x}}{2}$
$3.$	$\cosh^{2} x-\sinh^{2} x=1$
$4.$	$\sinh 2x=2\sinh 2x \cosh 2x$
$5.$	$\cosh 2x=\cosh^{2} x+\sinh^{2} x$
$6.$	$\sinh ^{-1}x=\ln(x+ \sqrt{x^{2}+1})$
$7.$	$\cosh ^{-1}x=\ln(x+ \sqrt{x^{2}-1})$
$8.$	$\tanh ^{-1}x=\frac{1}{2}\ln(\frac{1+x}{1-x})$
$9.$	$\cosh x+ \sinh x=e^{x}$
$10.$	$\cosh x- \sinh x=e^{-x}$

Hyperbolic cosine function

The $Hyperbolic\;Cosine\; Function$ noted $\cosh$ is defined as follows:

$\cosh x=\frac{1}{2}(e^{x}+e^{-x})$

This straight forward definition will be used to generate other formulas.

Hyperbolic sine function

The $Hyperbolic\;Sine\; Function$ noted $\sinh$ is defined as follows:

$\sinh x=\frac{1}{2}(e^{x}-e^{-x})$

This straight forward definition will be used to generate other formulas.

Other hyperbolic functions
Without any surprise we can write:

$\tanh x=\frac{\sinh x}{\cosh x}$

We get

$\tanh x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$

Please note:

$\coth x=\frac{1}{\tanh x}$

When we note:
$\begin{cases}x=a \cdot \cosh t\\y=b \cdot \sinh t \end{cases}$

When $t$ changes in the real numbers, the points $(x,y)$ follow one branch of the hyperbola.

Hyperbolic Trigonometry

$\cosh x + \sinh x=e^{x}$

$\cosh x - \sinh x=e^{-x}$

$\cosh^{2} x - \sinh^{2} x=1$ This shows that $\cosh x \geq 1$

We can also show that:

$1-\tanh^{2} x=\frac{1}{\cosh^{2} x}$

Sum and Difference formulas

Using the above relations we can get:

$\cosh (a+b)=\cosh a \cosh b+ \sinh a \sinh b$

$\cosh (a+b)=\cosh a \cosh b- \sinh a \sinh b$

$\sinh (a+b)=\sinh a \cosh b+ \cosh a \sinh b$

$\sinh (a+b)=\sinh a \cosh b- \cosh a \sinh b$

Double angle

$\cosh 2a=\cosh^{2} a + \sinh^{2} a=2\cosh^{2} a -1=1+2\sinh^{2} a$

$\sinh 2a=2\sinh a \cosh a$

For the tangents:

$\tanh (a+b)=\frac{\tanh a +\tanh b}{1+ \tanh a \tanh b}$

$\tanh (a-b)=\frac{\tanh a -\tanh b}{1- \tanh a \tanh b}$

$\tanh (2a)=\frac{2\tanh a}{1+ \tanh^{2} a}$

Transformation formulas

From the formulas shown above, we can get:

$\cosh a \cosh b=\frac{1}{2}[\cosh (a+b)+ \cosh (a-b)]$

$\sinh a \cosh b=\frac{1}{2}[\sinh (a+b)+ \sinh (a-b)]$

$\sinh a \sinh b=\frac{1}{2}[\cosh (a+b)- \cosh (a-b)]$

$\cosh^{2} a=\frac{\cosh (2a)+1}{2}$

$\sinh^{2} a=\frac{\cosh (2a)-1}{2}$

$\cosh p+ \cosh q=2\cosh \frac{p+q}{2}\cosh \frac{p-q}{2}$

$\cosh p- \cosh q=2\sinh \frac{p+q}{2}\sinh \frac{p-q}{2}$

$\sinh p+ \sinh q=2\sinh \frac{p+q}{2}\cosh \frac{p-q}{2}$

$\sinh p- \sinh q=2\sinh \frac{p-q}{2}\cosh \frac{p+q}{2}$

If we have $t=\tanh \frac{x}{2}$

From above we get:

$\tanh x=\frac{2t}{1+t^{2}}$

$\cosh x=\frac{1+t^{2}}{1-t^{2}}$

$\sinh x=\frac{2t}{1-t^{2}}$

Inverse hyperbolic functions:

Let $y=\cosh x$

We can write:

$x=\cosh^{-1} y$ with $x \in [0, +\infty)$

But we know:

$e^{x}=\cosh x + \sinh x=\cosh x+ \sqrt{\cosh^{2} x-1}$

We plugin:

$e^{x}=y+\sqrt{y^{2}-1}$

Taking the $LOG$ of both sides:

$x=\ln (y+\sqrt{y^{2}-1})$

Let $y=\sinh x$

We can write:

$x=\sinh^{-1} y$ with $x \in [0, +\infty)$

But we know:

$e^{x}=\cosh x + \sinh x=\sinh x+ \sqrt{\sinh^{2} x+1}$

We plugin:

$e^{x}=y+\sqrt{y^{2}+1}$

Taking the $LOG$ of both sides:

$x=\ln (y+\sqrt{y^{2}+1})$

Problem 1:

Prove the sum formula below:

$\cosh (a+b)=\cosh a \cosh b+ \sinh a \sinh b$

Solution

To solve this problem we use the following formulas:

$\cosh x=\frac{1}{2}(e^{x}+e^{-x})$

$\sinh x=\frac{1}{2}(e^{x}-e^{-x})$

$\cosh x+\sinh x=\frac{1}{2}(e^{x}+e^{-x})+\frac{1}{2}(e^{x}-e^{-x})$

$\cosh x+\sinh x=\frac{1}{2}(e^{x}+e^{-x}+e^{x}-e^{-x})$

$\cosh x+\sinh x=\frac{1}{2}(e^{x}+e^{x})$

$\cosh x+\sinh x=\frac{1}{2}(2e^{x})$

$\cosh x+\sinh x=e^{x}$

$\cosh (-x)=\frac{1}{2}(e^{-x}+e^{+x})\Rightarrow \cosh (-x)=\cosh x$

$\cosh (-x)=\cosh (x)$

The same way we can prove that

$\sinh (-x)=-\sinh (x)$

Back to the formula

(1) $\begin{equation*} \begin{split} \cosh (x+y)&=\frac{1}{2}(e^{x+y}+e^{-(x+y)})\\ &= \frac{1}{2}(e^{x}e^{y}+e^{-x}e^{-y})\\ &=\frac{1}{2}((\cosh x+ \sinh x)(\cosh y+ \sinh y)+(\cosh x- \sinh x)(\cosh y- \sinh y)) \\ &=\frac{1}{2}(\cosh x \cosh y+ \cosh x \sinh y+ \sinh x \cosh y+ \sinh x \sinh y+\\ &\cosh x \cosh y- coh x \sinh y-\sinh x \cosh y+\sinh x \sinh y)\\ &=\frac{1}{2}(2(\cosh x \cosh y+\sinh x \sinh y))\\&=\cosh x \cosh y+\sinh x \sinh y \end{split} \end{equation*}$