Hyperbolic trigonometric functions in terms of e ?

[Point Conception]

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 ?

 

[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 ...

 

[Redbelly98]

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

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

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

 

[Point Conception]

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.
see my thread starter: (https://www.physicsforums.com/showthread.php?t=336897 )
also linked below.

 

[Point Conception]

That cosh (a) ,( a = area of hyperbolic sector of unit hyperbola )
is equal to the x coordinate.
see my thread starter: (https://www.physicsforums.com/showthread.php?t=336897 )
also linked below.

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 ?

 

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

 

[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:

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