Hyperbolic trigonometric functions in terms of e ?

In summary, the hyperbolic trigonometric functions, cosh x and sinh x, are defined in terms of the natural exponential function, e^x. This can be seen by comparing their definitions to the analogous formulas for the standard trigonometric functions, cos x and sin x, which are also defined in terms of e^x. The fundamental solutions for the differential equation y"- y= 0 at x= 0 are cosh x and sinh x, similarly to how the fundamental solutions of y"+ y= 0 at x= 0 are cos x and sin x. These hyperbolic functions can also be seen as parametrizing a basic hyperbola, while the standard trigonometric functions parametrize
  • #1
morrobay
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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
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
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
g_edgar said:
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.
 
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  • #5
morrobay said:
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.

g_edgar said:
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
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
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]
 

1. What are hyperbolic trigonometric functions in terms of e?

Hyperbolic trigonometric functions are mathematical functions that relate the sides and angles of a hyperbola, a type of curve, to the exponential function e. These functions are sinh (hyperbolic sine), cosh (hyperbolic cosine), tanh (hyperbolic tangent), coth (hyperbolic cotangent), sech (hyperbolic secant), and csch (hyperbolic cosecant).

2. How are hyperbolic trigonometric functions calculated in terms of e?

To calculate hyperbolic trigonometric functions in terms of e, you can use the following formulas:

sinh(x) = (e^x - e^-x) / 2

cosh(x) = (e^x + e^-x) / 2

tanh(x) = sinh(x) / cosh(x)

coth(x) = cosh(x) / sinh(x)

sech(x) = 1 / cosh(x)

csch(x) = 1 / sinh(x)

3. What is the relationship between hyperbolic trigonometric functions and e?

The relationship between hyperbolic trigonometric functions and e is that they are calculated using the exponential function e. The value of e is approximately 2.71828, and it is used as the base for calculating these functions.

4. What are some real-life applications of hyperbolic trigonometric functions in terms of e?

Hyperbolic trigonometric functions in terms of e have various applications in fields such as physics, engineering, and finance. For example, they can be used to model the shape of a hanging cable or a suspension bridge, calculate the trajectory of a projectile, or analyze the behavior of electrical circuits. They are also used in financial mathematics to model interest rates and stock prices.

5. How do hyperbolic trigonometric functions in terms of e differ from regular trigonometric functions?

Hyperbolic trigonometric functions in terms of e differ from regular trigonometric functions in several ways. One difference is that they are defined using the exponential function e instead of the unit circle. Another difference is that their values can be complex numbers, while regular trigonometric functions only have real values. Additionally, the graphs of hyperbolic trigonometric functions are shaped like hyperbolas, while the graphs of regular trigonometric functions are shaped like circles or waves.

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