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Trigonometric functions $\cos$ and $\sin$ are related to points on the unit circle: they are the $x$ and $y$ coordinates.

Is there such a relation for $\cosh$ and $\sinh$, hyperbolic functions? What would play the role of the unit circle here, and how would the function be obtained from it?

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closed as unclear what you're asking by Chappers, N. F. Taussig, drhab, Tim Raczkowski, 6005 Sep 22 '15 at 18:59

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    $\begingroup$ Trig functions are to circles as hyperbolic functions are to hyperbolas. $\endgroup$ – got it--thanks Sep 21 '15 at 23:14
  • $\begingroup$ I was told that they were related, so that is wrong then? $\endgroup$ – Carlos V Sep 21 '15 at 23:16
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    $\begingroup$ This answer to a related question may be helpful. $\endgroup$ – Blue Sep 21 '15 at 23:17
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    $\begingroup$ The two are indeed related by Euler's formula. Speaking in the abstract, a hyperbola can be imagined as a circle of imaginary radius. $\endgroup$ – Lucian Sep 22 '15 at 5:24
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Normally we think of an angle in terms of arc length. But, an alternative to define an angle as twice the area (within the unit circle) swept by the terminal side as it moves away from the initial side. Note that a unit circle has area and circumference of $\pi\cdot 1^2=\pi$ and $2\pi\cdot 1=2\pi$, respectively, so the arc length and area of any sector area linearly related (i.e. the arc length is always twice that of the area).

enter image description here

We can define a hyperbolic "angle" similarly, as twice the area swept from the initial side to the terminal side on the hyperbola $x^2-y^2=1$. enter image description here

You really need integral calculus to find $\sinh t$ and $\cosh t$ in terms of $t$, so I'll leave that as a fun project for you. :) Oh, and don't forget to derive the formulas for arsinh and arcos too; they're even more fun.

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    $\begingroup$ Since we're talking about area, shouldn't it be $\operatorname{arsinh}$ and $\operatorname{arcosh}$? After all, we're looking for areas, not arc lengths. $\endgroup$ – mrf Sep 22 '15 at 14:05
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    $\begingroup$ Not to spoil the "fun project" (which really is worth doing), this answer walks through the process. $\endgroup$ – Blue Sep 22 '15 at 14:11
  • $\begingroup$ +1 You're very right, mrf. I made the change as you suggested. +1 ty blue $\endgroup$ – John Joy Sep 22 '15 at 14:39

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