# Alternative definition of hyperbolic cosine without relying on exponential function

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?

• This is a definition - definitions can't be proved.
– user122283
Apr 16, 2014 at 23:53
• @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.)
– Blue
Apr 16, 2014 at 23:58
• How exactly have you defined $\cosh x$, if not through this very formula?
– user61527
Apr 17, 2014 at 0:04
• 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. Jul 18, 2016 at 8:00

## 2 Answers

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 $$P$$.

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 $$2$$):

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.

• @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.
– Blue
Apr 17, 2014 at 4:23
• Okay, that makes sense, thanks for the clarification. Apr 17, 2014 at 4:34
• This is great! Loved it. Sep 22, 2016 at 9:29
• @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.
– Blue
Apr 20, 2018 at 10:26
• @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!
– 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

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

So

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

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.