Ordinary trigonometric functions are defined independently of exponential function, and then shown to be related to it by Euler's formula.

Can one define hyperbolic cosine so that the formula $$\cosh{x}=\dfrac{e^x+e^{-x}}{2}$$ becomes something to be proven?

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    $\begingroup$ This is a definition - definitions can't be proved. $\endgroup$
    – user122283
    Apr 16, 2014 at 23:53
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    $\begingroup$ @SanathDevalapurkar: One can define $\cosh u$ and $\sinh u$ geometrically as hyperbolic analogues of $\cos\theta$ and $\sin\theta$, taking $(\cosh u, \sinh u)$ to be points on the "unit hyperbola", $x^2 - y^2 = 1$. In that case, the relation between these values and exponentials does require proof. (I may have posted one on MSE at some point.) $\endgroup$
    – Blue
    Apr 16, 2014 at 23:58
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    $\begingroup$ How exactly have you defined $\cosh x$, if not through this very formula? $\endgroup$
    – user61527
    Apr 17, 2014 at 0:04
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    $\begingroup$ I don't understand why this question is causing so much confusion...OP is merely asking if there's another equivalent definition one can work with. $\cosh x$ can be characterized as the function $f:\mathbb{R} \to \mathbb{R}$ satisfying $f'' = f$, $f'(0) = 0$ and $f(0) = 1$. Then after proving existence/uniqueness it's easy to verify that the formula you have works. $\endgroup$ Jul 18, 2016 at 8:00

2 Answers 2


The more-geometrically-minded of us take $\cosh u$ and $\sinh u$ to be defined via the "unit hyperbola", $x^2 - y^2 = 1$, in a manner directly analogous to $\cos\theta$ and $\sin\theta$. Specifically, given $P$ a point on the hyperbola with vertex $V$, and defining $u$ as twice(?!) the area of the hyperbolic sector $OVP$, then $\cosh u$ and $\sinh u$ are, respectively the $x$- and $y$-coordinates of $P$.

enter image description here

Just as in circular trig, we can assign measures $u$ (in "hyperbolic radians") to angles ---from the flat angle (when $u=0$) to half a right angle (when $u=\infty$)--- and associate those measures with the lengths of the corresponding $\cosh$ and $\sinh$ segments. And, just as in circular trig (prior to the advent of imaginary numbers), we might be forgiven for suspecting that the correspondences $u \leftrightarrow \cosh u$ and $u \leftrightarrow \sinh u$ are "non-arithmetical", which is to say: that no arithmetical formula converts angle measures to their associated trig values.

However, it turns out that the correspondences are not non-arithmetical; to find the appropriate arithmetical conversion formula, all we need is a bit of calculus ...

Edit. (Two years later!) Check the edit history for an inelegant argument that I now streamline with the help of this trigonograph, in which lengths from the unit hyperbola have been scaled by $\sqrt{2}$ (and, thus, areas by $2$):

enter image description here

Because the hyperbola is rectangular, we have that $|OX|\cdot|XY|$ is a constant (here, $1$), which guarantees that the regions labeled $v$ have the same area (namely, $1/2$), and therefore that the regions labeled $u$ have the same area (namely, $u$). Now, the bit of calculus I promised, to evaluate $u$ as the area under the reciprocal curve: $$u = \int_1^{|OX|}\frac{1}{t}dt = \ln|OX| \quad\to\quad |OX| = e^{u} \quad\to\quad |XY| = \frac{1}{e^u}$$ With that, we clearly have $$2\,\sinh u \;=\; e^{u}- e^{-u} \qquad\qquad 2\,\cosh u \;=\; e^{u} + e^{-u}$$ as desired. Easy-peasy!

End of edit.

That hyperbolic radians are defined via doubling the area of a hyperbolic sector may seem at odds with the common definition of circular radians in terms of arc-length, but it's hard to argue with success, given the elegance of the formulas above. Even so, the hyperbolic twice-the-sector-area definition can be seen as directly analogous to the circular case, since circular radians are also definable as "twice-the-sector-area": In the unit circle, the sector with angle measure $\pi/2$ radians has area $\pi/4$ (it's a quarter-circle), the sector with angle measure $\pi$ radians has area $\pi/2$ (it's a half-circle), and the "sector" with angle measure $2\pi$ radians has area $\pi$ (it's the full circle); in these, and all other, cases, the angle measure is twice the sector area.

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    $\begingroup$ @solstafir: You can define $\cosh$ and $\sinh$ based on an arc-length parameter (your $z$); however, hyperbolic arc-length cannot be expressed in terms of elementary functions. (Lengths of curves are almost-always trickier to calculate than the areas they bound; circles (& lines) are the primary exceptions.) The length of arc $V^\prime P^\prime$ involves $\int \sqrt{1+x^4}/x^2 dx$, which is quite non-trivial, so hyperbolic trig values would effectively be "non-arithmetical" functions of an arc-length-based angle measure. It's certainly not the case that arc-length is twice the sector area. $\endgroup$
    – Blue
    Apr 17, 2014 at 4:23
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    $\begingroup$ Okay, that makes sense, thanks for the clarification. $\endgroup$
    – solstafir
    Apr 17, 2014 at 4:34
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    $\begingroup$ This is great! Loved it. $\endgroup$
    – MycrofD
    Sep 22, 2016 at 9:29
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    $\begingroup$ @LeeDavidChungLin: This is the highest-resolution version I have: here. Enjoy! If you share it, please be sure to credit me ("Blue, the Trigonographer") and link to the entry on trigonography.com. $\endgroup$
    – Blue
    Apr 20, 2018 at 10:26
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    $\begingroup$ @Blue Thanks Blue. Using matrices, I found that the equation $x^2 - y^2 = \sqrt{2}^2$ is equivalent to $xy=1$ if we rotate the graph $45^\circ$ anticlockwise. (This makes sense because in both cases the closest point to the origin is $\sqrt{2}$ units away.) I love this answer because it explains how the hyperbolic functions are connected to the exponential: a common way to define the exponential function is by first defining the logarithm as $\log x = \int_{1}^{x}\frac{1}{t} \, dt$. And the graph of $y=1/t$ is a hyperbola! $\endgroup$
    – Joe
    Jan 28, 2021 at 21:25

Well, that is usually simply taken to be the definition, but given that

$$\cos x=\cosh ix$$

you may be asking for a proof that

$$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$

From Taylor's theorem, we know that



$$e^{ix}=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}+i\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=\cos x+i\sin x$$

Using $e^{ix}=\cos x+i\sin x$, express $e^{ix}+e^{-ix}$ in terms of $\cos x$, noting that the cosine function is even and the sine function is odd.


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