Why is sinh called "sinus hyperbolicus" despite being just a regular e function?

Asked Jan 16 '19 at 16:37

Active Jan 16 '19 at 16:37

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What does the sinus hyperbolicus have to do with the sinus? Their graphs do not look alike at all.

The only similarity I can find is that their exponential representation looks similar.

$sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$

$sinh(x) = \frac{1}{2}(e^{x}-e^{-x})$

asked Jan 16 '19 at 16:37

Sebastian Walker

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Also note the relation $$ i\sin(x) = \sinh(ix). $$ – MSDG Jan 16 '19 at 16:39
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See Hyperbolic function. – Mauro ALLEGRANZA Jan 16 '19 at 16:43
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See Vincenzo Riccati, Opuscula physico-mathematica, Vol.1 (1757) page 70 : "ita in hyperbola sumemus sinus et cosinus logarithmorum, quos sinus et cosinus hyperbolicos appellabimus." – Mauro ALLEGRANZA Jan 16 '19 at 16:54
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Well, I don’t know whether this helps, but the standard formulas for spherical trigonometry are mirrored by formulas in hyperbolic trigonometry by the replacement of the sine and cosine by the corresponding hyperbolic functions, when it’s a matter of side lengths. For instance, the spherical Pythagorean formula is $\cos c=\cos a\cos b$, while the corresponding hyperbolic formula is $\cosh c=\cosh a\cosh b$. – Lubin Jan 16 '19 at 20:22
See "Geometric construction of hyperbolic trigonometric functions". In particular, my answer. – Blue Jan 16 '19 at 23:05

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