Hyperbolic Function

  • Thread starter basty
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  • #1
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Could someone tell me what is a hyperbolic trigonometric function?

What is the difference between regular trigonometry and a hyperbolic trigonometry?

Also, why and how to derive and get

##\sinh x = \frac{e^x - e^{-x}}{2}##

?????
 

Answers and Replies

  • #2
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Could someone tell me what is a hyperbolic trigonometric function?
Try a web search, using "hyperbolic trigonometric function" as your search string. You should be able to find answers to all of your questions.
basty said:
What is the difference between regular trigonometry and a hyperbolic trigonometry?

Also, why and how to derive and get

##\sinh x = \frac{e^x - e^{-x}}{2}##

?????
 
  • #3
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I wasn't able to find over the internet and in my calculus book of why

##\sinh x = \frac{e^x - e^{-x}}{2}##

Please someone help me.
 
  • #4
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What definition of sinh are you using? I'm asking because that's usually taken to be the definition, so unless you're using an alternative one, there are not really any more "whys" involved.
 
  • #5
robphy
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From googling "hyperbolic trigonometry", the third link gave
http://www.sosmath.com/trig/hyper/hyper01/hyper01.html
the fourth gave a Khan academy video
https://www.khanacademy.org/math/precalculus/hyperbolic_trig_topic [Broken][probably okay... but I haven't watched it]
 
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  • #6
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From googling "hyperbolic trigonometry", the third link gave
http://www.sosmath.com/trig/hyper/hyper01/hyper01.html
the fourth gave a Khan academy video
https://www.khanacademy.org/math/precalculus/hyperbolic_trig_topic [Broken][probably okay... but I haven't watched it]
All above links shows the identities of hyperbolic functions, not shows why ##\sinh x = \frac{e^x - e^{-x}}{2}.##
 
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  • #7
robphy
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Do you know the analogous definition of [itex]\sin x=\frac{e^x-e^{-x}}{2i} [/itex]?

The starting point for these is the exponential function [itex]e^x[/itex].
In the hyperbolic case, x is real.
In the circular case, x is pure-imaginary [itex]x=i\theta[/itex].

Write the exponential function as the sum of its "even part" and its "odd part"
[itex]C(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)[/itex]
[itex]S(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)[/itex]
so [itex]e^{x}=C(x)+S(x)[/itex]
In the hyperbolic case, these are cosh and sinh.
[itex]e^{x}=\cosh{x}+\sinh{x}[/itex]

In the circular case, where [itex]x=i\theta[/itex],
we have
[itex]C(i\theta)=\frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right)[/itex]
[itex]S(i\theta)=\frac{1}{2}\left(e^{i\theta}-e^{-i\theta}\right)[/itex]
It turns out that ##C(i\theta)## is a real-valued function of ##\theta##, called ##\cos(\theta)##.
However, ##S(i\theta)## is a pure-imaginary function of ##\theta##. By defining the real-valued function of ##\theta## called ##\sin\theta\equiv \frac{S(i\theta)}{i}=\frac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right)##,
we can write
[itex]e^{\theta}=\cos{\theta}+i\sin{\theta}[/itex]
 
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  • #8
HallsofIvy
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The regular trig functions are also called "circular functions" because if you let [itex]x= cos(t)[/itex] and [itex]y= sin(t)[/itex], [itex]x^2+ y^2= cos^2(t)+ sin^2(t)= 1[/itex]. That is, they give the x and y components of a point on the unit circle.

The name for "hyperbolic functions" comes from the fact that [itex]cosh^2(t)- sinh^2(t)= \frac{\left(e^{x}+ e^{-x}\right)^2}{4}- \frac{\left(e^{x}- e^{-x}\right)^2}{4}= \frac{e^{2x}+ 2+ e^{-2x}- e^{2x}+ 2- e^{-2x}}{4}= 1[/itex] so that setting x= cosh(t), y= sinh(t), [itex]x^2- y^2= 1[/itex]. They give the x and y components of a point on the unit hyperbola.
 

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