What does it mean that the hyperbolic functions are "analygous" to the circular functions?

Question

Earlier I asked for a valid proof for
sinh x = [ e^x − e^−x ] ⁄ 2
cosh x = [ e^x + e^−x ] ⁄ 2

I was told, simply, that these are defined as such... that it has to do with the hyperbolic functions being "analogous" to the normal trigonometric functions.

I accept this as truth now. But can someone explain it to me? Demonstrate this "analogy". I am still looking for some sort of proof to this analogy or this relationship.

How does the mere definition of the hyperbolic sine function, as the opposite (y-coordinate) on the unit hyperbola, and the definition of sine, allow us to extrapolate this relationship? Exactly how do we go from sin x = [ e^(ix) − e^(−ix) ] ⁄ 2i to sinh x = [ e^x − e^−x ] ⁄ 2

And dont give me the sin (ix) = i sinh x as a proof... because its not a valid proof

Many so-called "proofs" are not valid because they are not fundamental truths that form the basis of more complex relationships. Its merely a consequence of the relationship that can be illustrated

Thanks Cogito

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Rivet Gun... I appreciate your insight... but how do we get sinh x = [ e^x − e^−x ] ⁄ 2 from this definition? Based on that of sin x?

Answer 1

Johann Heinrich Lambert was the first to propose the analogous relation. He juxtaposed the functions of an angle whose reference is the unit circle ( circular functions ) to analogous functions of an angle whose reference is the equilateral hyperbola ( hyperbolic functions ) .
His motivation was the need to know the sine of an arc whose measure is an imaginary number, and in general the trigonometric functions of such arcs. He used hyperbolic trigonometry to make these calculations.

In one case use the cirle X^2 + Y^2 = 1

and in the other use the hyperbola X^2 - Y^2 = 1

That is where the analogy lies

Answer 2

The height of a right circular cylinder is twice the radius. Express the volume as a function of the radius. Then, express the radius as a function of the volume. H = 2R given V = pi R^2 H = pi R^2 (2R) = 2 pi R^3 whence R^3 = (V / 2pi) Radius = (V / 2pi)^(1/3) ANSWER

Answer 3

Consider the unit circle (x^2 +y^2 = 1)
From origin we draw a straight line to a point on the circle which we call a radius. Shade the region bounded by our radius, the circle and the x axis and call this region "sector". The angle (in radians) the radius makes with the x axis = the arc lengh = twice the area of the sector. Since these are equivalent definions, we could define "angle in radians" = "twice area of sector". Call this area (or angle) a
sin a and cos a then have their familiar definitions as the x and y coordinates of our point.

Now consider the unit hyperbola (x^2 - y^2 = 1)
We draw a straight line from the origin to a point on the hyperbola and shade the region bounded by our line, the hyperbola and the x axis. Call the shaded regioin "hyperbolic sector"

Define "hyperbolic angle" = "twice area of hyperbolic sector" (analogous to our definition of "angle in radians" for the circle.) Call this hyperbolic angle (or area) b

Now sinh b and cosh b are the x and y coordinates respectivly of our point on the hyperbola.

Answer 4

I don't know but I want to say that you can use the words analogous, analogy without placing them in quotes.
"Analygous" does not exist.
There is some info on wikipedia but I guess you need more arguments.

Cogito: Stop insulting members("stupid"), including me. Stop making broad generalizations("they dont care one iota about understanding anything like you or I do").. I am here to have fun and help people not to answer to rants.
I gave her an useful link to wikipedia, where there is some useful reliable info. I also corrected her word "analygous".
Sometimes you can give a partial answer that includes something useful, like I did, instead of full answers. Maybe I was too harsh in correcting the language and I will try to do better next time. I apologize if I offended the asker.
Incidentally, I am not very interested in sinh and conh but I am interested in understanding deeply other domains of mathematics. You can't be interested in everything.
I appreciated the fact that this a good question and this is why I was the only one that gave it a star.
I appreciate these kind of questions, that are not routine homework.

Answer 5

I think that you are over analysing this matter.
Accepting that:-
cosh x = (1/2) [ e^x + e^(-x) ]
cos x = (1/2) [ e^(ix) + e^(-ix) ]

Similar expressions for sin x and sinh x.

The word "analogous" simply means that these two expressions are alike.

Don`t become bogged down unnecessarily.

Good luck!