# Hyperbolic Function identities

1. Jul 17, 2010

### Xyius

I have always been curious as to where the definition of cosh(x) and sinh(x) come from and how they are related to the natural exponential. I know it has something to do with Euler's formula but I don't know the details of the derivation. Could anyone shed some light on this? I haven't yet been formally taught hyperbolic functions.

Math level:Up to Differential Equations

Thanks!!!

2. Jul 18, 2010

### HallsofIvy

Staff Emeritus
There are a number of reasons why the hyperbolic functions are called that and are called sinh and cosh.

1) We can define the trig functions (also called "circular functions") in the following way: construct the unit circle on a coordinate system- the graph of $x^2+ y^2= 1$. Start from the point (1, 0) and measure around the circle a distanct t. The coordinates of the end point are (cos(t), sin(t)). We can define the hyperbolic functions by drawing the graph of $x^2- y^2= 1$, a hyperbola. Start from the point (1, 0) and measure along the hyperbola a distance t. The coordinates of the endpoint are (cosh(t), sinh(t)).

2) Given any function, f(x), we can define the "even part" of f to be fe(x)= (f(x)+ f(-x))/2 and the "odd part" to be fo= (f(x)- f(-x))/2. They are, of course, even and odd and their sum is f(x). sin(x) and cos(x) are odd and even themselves. The odd and even parts of ex are sinh(x) and cosh(x).

3) Just as $cosh(x)= (e^x+ e^{-x})/2$ and $sinh(x)= (e^x- e^{-x})/2$ so $cos(x)= (e^{ix}+ e^{-ix})/2$ and $sin(x)= (e^{ix}- e^{-ix})/(2i)$.

4) Sine and cosine are the "fundamental solutions" to the differential equation y"= -y: sin(x) satisfys the equation and the initial values y(0)= 0, y'(0)= 1 while cos(x) satisfies the same equation with initial values y(0)= 1, y'(0)= 0 so that if y(x) satisfies that equation with y(0)= A, y'(0)= B, then y(x)= A cos(x)+ B sin(x). Similarly, sinh(x) and cosh(x) are the "fundamental solutions" to y"= y. sinh(x) satisfies the differential equation with initial values y(0)= 0, y'(0)= 1 and cosh(x) satisfies the differential equation with initial values y(0)= 1, y'(0)= 0 so that if y(x) satisifies y"= y with initial values y(0)= A, y'(0)= B, y(x)= A cosh(x)+ B sinh(x).