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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 '14 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 '14 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 '14 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. $\endgroup$ – MathematicsStudent1122 Jul 18 '16 at 8:00
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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 $|\overline{OX}|\cdot|\overline{XY}|$ 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: $$u = \int_1^{|\overline{OX}|}\frac{1}{t}dt = \ln|\overline{OX}| \quad\to\quad |\overline{OX}| = e^{u} \quad\to\quad |\overline{XY}| = \frac{1}{e^u}$$ 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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  • $\begingroup$ This was fascinating! Would I be correct in assuming that, like with the circular trig functions, if $z$ gives the arc length from the vertex to the point $(x,y)$ on the hyperbola $x^2-y^2=r^2$, with a sign of positive or negative according to whether $y$ is positive or negative, then $\cosh z$ could also be defined as the ratio $\frac{x}{r}$, and $\sinh z$ as $\frac{y}{r}$? And then in the unit hyperbola, these ratios simply reduce to coordinates and the arc length becomes half the sector area? This would be an even nicer analogy to circular trigonometry. $\endgroup$ – solstafir Apr 17 '14 at 4:03
  • $\begingroup$ @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 '14 at 4:23
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    $\begingroup$ Okay, that makes sense, thanks for the clarification. $\endgroup$ – solstafir Apr 17 '14 at 4:34
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    $\begingroup$ This is great! Loved it. $\endgroup$ – MycrofD Sep 22 '16 at 9:29
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    $\begingroup$ Thank you very much. Surely I do my best to bring people's attention to your work whenever I can. $\endgroup$ – Lee David Chung Lin Apr 20 '18 at 10:31
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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.

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This is often taken as the definition of $\cosh$ so it can't really be proved.

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    $\begingroup$ From what ? This is often taken as the definition of cosh $\endgroup$ – mahmoud afefey Apr 16 '14 at 23:51
  • $\begingroup$ I don't understand what you're trying to ask. $\endgroup$ – Cameron Williams Apr 16 '14 at 23:51
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    $\begingroup$ why the downvote? $\endgroup$ – robjohn Apr 16 '14 at 23:53
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    $\begingroup$ @robjohn It's sitting at 3 downvotes now lol. I think people do not realize how poorly conceived this post was initially. $\endgroup$ – Cameron Williams Jul 13 '16 at 0:55
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    $\begingroup$ At the time of my comment the question simply read like this. $\endgroup$ – robjohn Sep 22 '16 at 16:50

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