# Hyperbolic trigonometric functions in terms of e ?

1. Mar 9, 2010

### morrobay

cosh x= e^x+e^-x/2
sinh x= e^x-e^-x/2

Can someone explain why the hyperbolic trigonometric functions are defined in terms of the natural exponential function, e^x ?

2. Mar 9, 2010

### g_edgar

Well, tell us the definition of cosh you want to use. Then we can work on why it is equal to (exp(x)+exp(-x))/2 ...

3. Mar 9, 2010

### Redbelly98

Staff Emeritus
It probably has to do with analogous formulas for the standard trig functions:

cos x = (eix + e-ix)/2
sin x = (eix - e-ix)/2​

Simply remove the i's and they become the hyperbolic functions.

4. Mar 10, 2010

### morrobay

That cosh (a) ,( a = area of hyperbolic sector of unit hyperbola )
is equal to the x coordinate.

Last edited by a moderator: May 4, 2017
5. Mar 11, 2010

### morrobay

Whats up Doc ?

Last edited by a moderator: May 4, 2017
6. Mar 11, 2010

### HallsofIvy

It is also worth noting that the "fundamental solutions at x= 0" to the differential equation y"+ y= 0, the solutions such that y1(0)= 1, y1'(0)= 0 and y2(0)= 0, y2'(0)= 0, are y1(x)= cos(x) and y2(x)= sin(x).

They are so called because is y is any function of x satisfying that differential equation, then y(x)= y(0)cos(x)+ y'(0) sin(x). That is, the coefficients in the linear combination are just y(0) and y'(0).

The fundamental solutions for y"- y= 0, at x= 0, are, similarly, cosh(x) and sinh(x).

Of course, ex and e-x are also independent solutions of that equation so they can be written in terms of each other.

7. Mar 11, 2010

### elibj123

Well, they parametrize a very basic hyperbola:

$$cosh^{2}(x)-sinh^{2}(x)=1$$

while the trigonometric cosines and sines describe a very basic circle:

$$cos^{2}(x)+sin^{2}(x)=1$$