# Hyperbolic trigonometric functions in terms of e ?

Gold Member

## Main Question or Discussion Point

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 ?

## Answers and Replies

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 ...

Redbelly98
Staff Emeritus
Homework Helper
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.

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

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Gold Member
That cosh (a) ,( a = area of hyperbolic sector of unit hyperbola )
is equal to the x coordinate.
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 ...
Whats up Doc ?

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HallsofIvy
Homework Helper
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.

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:

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