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I've been studying hyperbolic functions and was wondering where the following two identities were derived from:

$$\sinh(x) = \frac{e^{x}-e^{-x}}{2}$$

$$\cosh(x) = \frac{e^{x}+e^{-x}}{2}$$

I understand how to use these to prove other identities and I understand how to use Euler's formula to find the identities for $\sin(x)$ and $\cos(x)$ but I am unable to find any proof for these two. Perhaps I am just unsure what to search for, so if there is a proof somewhere already I would love some directions or links.

Thank you.

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closed as unclear what you're asking by Nosrati, Yves Daoust, Carsten S, Lord Shark the Unknown, Cesareo Oct 2 '18 at 16:58

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    $\begingroup$ How do you define $\sinh$ and $\cosh$? $\endgroup$ – José Carlos Santos Oct 2 '18 at 9:47
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    $\begingroup$ Possible duplicate of What's the intuition behind the identities $\cos(z)= \cosh(iz)$ and $\sin(z)=-i\sinh(iz)$? $\endgroup$ – Nosrati Oct 2 '18 at 10:48
  • $\begingroup$ The proposed "duplicate" question asked for intuition, not proof. My interpretation of this question is it is asking for proof. There is some ambiguity in the question because it does not give a sufficient hint about what facts we might derive the formulas from, but nevertheless it is asking for a derviation. $\endgroup$ – David K Oct 2 '18 at 12:52
  • $\begingroup$ If you don't give a definition of these functions, no proof of anything can be given and the question should be closed. $\endgroup$ – Yves Daoust Oct 2 '18 at 13:55
  • $\begingroup$ see how hyperbolic function of an imaginary angle $ \cosh ix = \cos x$ is derived as a real quantity. $\endgroup$ – Narasimham Oct 2 '18 at 16:48
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You can't answer this until you have defined the hyperbolic functions !

One way is via the complex numbers,

$$i\sinh x:=\sin ix, \\\cosh x:=\cos ix,$$ giving

$$\sin ix=\frac{e^{i^2x}-e^{-i^2x}}{2i}, \\\cos ix=\frac{e^{i^2x}+e^{-i^2x}}2,$$ or $$\sin ix=\frac{e^{-x}-e^{x}}{2i}, \\\cos ix=\frac{e^{-x}+e^x}2.$$


The qualifier hyperbolic comes from the relation

$$\cosh^2x-\sinh^2x=1$$ or $$u^2-v^2=1,$$ i.e. an equilateral hyperbola. (Compare to $u^2+v^2=1$.)

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The hyperbolic functions are defined as the even and odd parts of $\exp x$ so $\exp\pm x=\cosh x\pm\sinh x$, in analogy with $\exp\pm ix=\cos x\pm i\sin x$. Rearranging gives the desired results.

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Another way to define $\cosh(x)$ and $\sinh(x)$ is to use the Taylor series expansions for $e^x$ and and $e^{-x}$, that is $$e^x = \sum_{n=0}^{\infty} \dfrac {x^n}{n!}$$ $$e^{-x} = \sum_{n=0}^{\infty} \dfrac {(-1)^nx^n}{n!}$$

We add the first two series to get $$e^x + e^{-x} = 2 \sum_{n=0}^{\infty} \dfrac {x^{2n}}{2n!}$$ or $$\dfrac {e^x + e^{-x}}{2} = \sum_{n=0}^{\infty} \dfrac {x^{2n}}{2n!}$$ As the sum of the right side is similar to $\cos (x)$, we can define $\cosh(x)$ as $$\cosh (x) \Leftrightarrow \dfrac {e^x + e^{-x}}{2}.$$

Now, subtracting the two series yields $$e^x - e^{-x} = 2 \sum_{n=0}^{\infty} \dfrac {(-1)^{n}x^{2n+1}}{(2n+1)!}$$ or $$\dfrac {e^x - e^{-x}}{2} = \sum_{n=0}^{\infty} \dfrac {(-1)^{n}x^{2n+1}}{(2n+1)!}$$ As the sum of the right side is similar to the series for $\sin(x)$, we can define $\sinh(x)$ as $$\sinh (x) \Leftrightarrow \dfrac {e^x - e^{-x}}{2}$$.

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  • $\begingroup$ "As the right side is the series for $\cos(x)$" -- Did you mean $\cosh(x)$? $\endgroup$ – David K Oct 2 '18 at 12:45
  • $\begingroup$ Yes, sir...I was thinking one thing and doing another. I've corrected the post. $\endgroup$ – bjcolby15 Oct 2 '18 at 12:57
  • $\begingroup$ In what way does this address the question ? $\endgroup$ – Yves Daoust Oct 2 '18 at 13:52
  • $\begingroup$ The original poster was asking how the hyperbolic functions were derived. In addition to the other ways other posters have illustrated, we can also derive $\cosh x$ and $\sinh x$ through the Taylor series of $e^x$ and $e^{−x}$. $\endgroup$ – bjcolby15 Oct 5 '18 at 1:28
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I think nobody has quite said that the most common definition of $\sinh,\cosh$ are in fact by the expressions you just wrote down. From these and the properties of $\exp$ we can derive all their other properties. Certainly the mathematicians who originally defined these did not know about complex numbers, much less envision the close relation between $\sinh,\cosh$ and $\sin,\cos$ via the complex exponential function. So if we define these four functions by the following for any complex $z$: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

$\cosh(z) = \lfrac12(\exp(z)+\exp(-z))$.

$\sinh(z) = \lfrac12(\exp(z)-\exp(-z))$.

$\cos(z) = \lfrac12(\exp(iz)+\exp(-iz))$.

$\sin(z) = \lfrac1{2i}(\exp(iz)-\exp(-iz))$.

Then we get the following easy results:

$\cosh(z) = \cos(iz)$.

$\cos(z) = \cosh(iz)$.

$\sinh(z) = -i\sin(iz)$.

$\sin(z) = -i\sinh(iz)$.

And hence every trigonometric identity can be easily transformed into a hyperbolic identity and vice versa.

Once you prove that $\exp' = \exp$, you can recover all the basic properties of $\exp$ and hence $\cosh,\sinh,\cos,\sin$, including:

$\cosh' = \sinh$.

$\sinh' = \cosh$.

$\cos' = -\sin$.

$\sin' = \cos$.

$\cos(x+y) = \cos(x)\cos(y)-\sin(x)\sin(y)$.

$\sin(x+y) = \cos(x)\sin(y)+\sin(x)\cos(y)$.

If you further define $π$ as the first positive real root of $\cos$, which you can prove exists by some simple real analysis, you can also get the well-known:

$\exp(iπ)+1=0$, and more generally that $2πi$ is the period of $\exp$.

See this post for more details.

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