Hyperbolic trigonometric functions in terms of e ?

  • Thread starter morrobay
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  • #1
morrobay
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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 ?
 

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  • #2
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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
Redbelly98
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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
morrobay
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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 [Broken])
also linked below.
 
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  • #5
morrobay
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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 [Broken])
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 ?
 
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  • #6
HallsofIvy
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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
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Well, they parametrize a very basic hyperbola:

[tex]cosh^{2}(x)-sinh^{2}(x)=1[/tex]

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

[tex]cos^{2}(x)+sin^{2}(x)=1[/tex]
 

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