Proof for hyperbolic trigonometric identities

Question

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.

Possible duplicate of What's the intuition behind the identities $\cos(z)= \cosh(iz)$ and $\sin(z)=-i\sinh(iz)$? — Nosrati, Oct 02 '18 at 10:48
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. — David K, Oct 02 '18 at 12:52
If you don't give a definition of these functions, no proof of anything can be given and the question should be closed. — , Oct 02 '18 at 13:55
see how hyperbolic function of an imaginary angle $ \cosh ix = \cos x$ is derived as a real quantity. — Narasimham, Oct 02 '18 at 16:48
See this answer, which develops the functions geometrically and derives the exponential forms with the help of an elementary integral. — Blue, Oct 02 '18 at 20:46

score 5 · Answer 1 · 2018-10-02T13:57:08.603

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 $u^2+v^2=1$.)

score 4 · Answer 2 · answered Oct 02 '18 at 10:07

4

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.

answered Oct 02 '18 at 10:07

J.G.

115,835

bjcolby15 · Answer 3 · 2018-10-04T11:04:37.633

1

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 $$\sinh (x) \Leftrightarrow \dfrac {e^x - e^{-x}}{2}$$.

edited Oct 04 '18 at 11:04

answered Oct 02 '18 at 10:29

bjcolby15

3,599

"As the right side is the series for $\cos(x)$" -- Did you mean $\cosh(x)$? – David K Oct 02 '18 at 12:45
Yes, sir...I was thinking one thing and doing another. I've corrected the post. – bjcolby15 Oct 02 '18 at 12:57
In what way does this address the question ? – Oct 02 '18 at 13:52
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 $e^{−x}$. – bjcolby15 Oct 05 '18 at 01:28

score 0 · Answer 4 · edited Jun 12 '20 at 10:38

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.

Here is an interesting related MAA article: The Early Drama of the Hyperbolic Functions. — user21820, Oct 02 '18 at 13:20

Proof for hyperbolic trigonometric identities

4 Answers4