- #1

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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}##

?????

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- Thread starter basty
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- #1

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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}##

?????

- #2

Mark44

Mentor

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Try a web search, using "hyperbolic trigonometric function" as your search string. You should be able to find answers to all of your questions.Could someone tell me what is a hyperbolic trigonometric function?

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

Please someone help me.

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- #5

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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]

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

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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]

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 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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