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I've been reading up on hyperbolic functions and was wondering if there was a geometric definition for the hyperbolic angle and hyperbolic function.

In particular I was reading this: Alternative definition of hyperbolic cosine without relying on exponential function and this: Proofs of Hyperbolic Functions

My question is this - when we consider trigonometric functions, we treat them as points on a unit circle where (conventionally) the angle is formed between the x-axis and a ray drawn to some other point. I believe that for hyperbolic functions, we take the ray dividing the hyperbola in half, and another ray, and the angle between them is the hyperbolic angle.

For the hyperbola $x^2 - y^2 = 1$, somehow, the area of the sector is half the angle. Is there a proof for this?

Also, for the hyperbola $y = 1/x$, the area of the sector is equal to the angle. Is there a proof here? Why are the two cases different?

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Rohil Verma
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  • "For the hyperbola $x^2-y^2=1$, [...] the area of the sector is half the [hyperbolic measure of] the angle." The answer (by me!) in your first link explains that the exponential formulas for sinh and cosh work out nicely when the exponent is twice the area of the sector; so, that relation is taken as the definition of hyperbolic angle measure. "[F]or the hyperbola $y=1/x$, the area of the sector is equal to the angle. [..] Why are the two cases different." The cases are different because the semi-transverse axis of the second hyperbola is $\sqrt{2}$, so corresponding areas are scaled by $2$. – Blue Apr 09 '15 at 21:49
  • OK, that's exactly why I was wondering. Thanks a lot. But all of this works because we have some idea of two hyperbolic functions sinh ($(e^x-e^(-x))/2$) and cosh ($(e^x+e^(-x))/2$). Is there anyway to arrive at this idea geometrically? – Rohil Verma Apr 10 '15 at 07:06
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    your answer has been answered now, by some edits made in the first link you gave! I too had the same question as you did. And now the answer by @Blue helped so much. – MycrofD Sep 22 '16 at 09:32

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