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.

Answer 1

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.

Answer 2

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

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

Answer 3

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

Answer 4

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.