Hyperbolic Functions Explorer

Hyperbolic vs circular functions, identity checks, and geometric intuition.

Worked example — Hyperbolic Functions Explorer

cosh(1)	1.543081
cosh^2(t)-sinh^2(t)	1

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Hyperbolic Functions Explorer

The hyperbolic functions are named after a curve, and the name is the whole idea. Where (cos t, sin t) traces the circle x² + y² = 1, the pair (cosh t, sinh t) traces the hyperbola x² − y² = 1 — one sign changed, one curve swapped. The harder question, and the one this lesson is really about, is what the shared parameter t means once you leave the circle behind.

What each symbol means

t: the parameter. On the circle you were taught to call it an angle. That reading does not survive the move to the hyperbola, and replacing it correctly is the point of steps 3 and 4.
cosh t: (eᵗ + e⁻ᵗ)/2, the x-coordinate of the point. Never smaller than 1, and never oscillating — two things the word "cosine" actively suggests and which are both false here.
sinh t: (eᵗ − e⁻ᵗ)/2, the y-coordinate. Unbounded in both directions, and zero only at t = 0.
x² − y² = 1: the unit hyperbola. This page prints cosh²t − sinh²t in its readout and always shows 1 — that number is this equation, which is the reason for the name.

Where the formula comes from

Start where the intuition was formed. The point (cos t, sin t) sits on the unit circle, and on a circle the parameter t has two readings that happen to give the same number: it is the arc length travelled from (1, 0), and it is twice the area of the pie-slice swept out from the centre. Because the two agree, nobody is ever forced to pick one, and everybody leaves school calling t an angle.
Now change one sign and ask for a point on x² − y² = 1. Take cosh t = (eᵗ + e⁻ᵗ)/2 and sinh t = (eᵗ − e⁻ᵗ)/2, square them and subtract: the cross terms cancel and you are left with (e²ᵗ + 2 + e⁻²ᵗ)/4 − (e²ᵗ − 2 + e⁻²ᵗ)/4 = 1. So (cosh t, sinh t) lies on that hyperbola for every real t, by construction. This is the identity the page prints, and it is the entire justification for the word "hyperbolic".
So which of the circle's two readings of t comes across? Not arc length. The distance travelled along the hyperbola from (1, 0) to (cosh t, sinh t) is not t, and it has no elementary formula at all — it is an elliptic integral, which is why no textbook ever quotes it. The reading that survives is the other one: area.
Here is that claim earned rather than asserted. The hyperbolic sector — bounded by the x-axis, the ray out to (cosh t, sinh t), and the curve itself — is the triangle under the ray minus the region under the hyperbola. The triangle is ½·cosh t·sinh t. For the other piece, ∫√(x² − 1) dx = ½(x√(x² − 1) − arcosh x), and evaluating from 1 to cosh t gives ½(cosh t·sinh t − t), since √(cosh²t − 1) = sinh t and arcosh(cosh t) = t. Subtract: ½·cosh t·sinh t − ½(cosh t·sinh t − t) = t/2. The sector area is exactly t/2 — the same rule as the circle.
So the honest statement is that t is twice a swept area, in both geometries, and that "angle" was a coincidence of the circle that does not travel. This is why the geometry panel here marks two points and no angle: as you drag the slider the circular dot sweeps at a steady angular rate while the hyperbolic one visibly does not, and the quantity the two share is the area behind them, not the direction they lie in.

How to read what you see

Two panels and two numbers. The left plot is the chosen function against t; the right one is the geometry, and it is the more interesting of the pair — a unit circle carrying (cos t, sin t) next to the hyperbola carrying (cosh t, sinh t), both driven by the one slider. The "compare with circular" box superimposes the circular partner so the divergence is visible directly. Of the readouts, the second is the one to watch: cosh²(t) − sinh²(t) reads 1.000000 at every setting, because it is not a computed result at all but the equation of the curve, restated.

Assumes: Real t, and the right-hand branch of the hyperbola — cosh t ≥ 1 always, so this parameterisation never reaches the left branch, and no real t ever will. The unit hyperbola specifically, with both semi-axes equal to 1; a general x²/a² − y²/b² = 1 needs (a·cosh t, b·sinh t) and the neat area result picks up a factor of ab. And area counted as a signed sector measured from the positive x-axis, so negative t sweeps below it.
Breaks when: The naming misleads in two directions at once, and both are worth holding on to. Read t as an angle and every hyperbolic plot becomes quietly wrong: there is no angle in the picture that equals t, and the arc length that would be the other natural candidate is not elementary. Read "cosine" as a bounded oscillation and the opposite error follows — cosh t is never less than 1 and never comes back down, which is exactly why it is the shape of a hanging chain rather than of a wave. The functions were named for the curve they parameterise, not for any behaviour they share with their circular namesakes.

Galileo took the hanging chain for a parabola 🖖

A chain hanging under its own weight looks so much like a parabola that Galileo wrote it down as one in 1638. It is not: the true curve is y = a cosh(x/a), the catenary, and the difference is real rather than a rounding detail. Joachim Jungius showed the parabola was wrong, and in 1691 Huygens, Leibniz and Johann Bernoulli independently produced the correct curve in answer to a public challenge from Jakob Bernoulli. The name comes from catena, Latin for chain. Turn the shape upside down and you get the one arch that carries its own weight in pure compression, with no bending anywhere — which is why the Gateway Arch in St. Louis is a catenary and not a parabola.

Just exponentials in disguise 🖖

Strip away the fancy names and sinh and cosh are simply built from the exponential functions e^t and e^-t: cosh(t) = (e^t + e^-t)/2 and sinh(t) = (e^t - e^-t)/2. So cosh is the average of a growing and a shrinking exponential, while sinh is half their difference. That is why cosh never dips below 1 and sinh passes cleanly through the origin.

The bridge that needs no i 🖖

You can convert between hyperbolic and circular functions without any imaginary numbers, using the Gudermannian gd(t): sin(gd t) = tanh(t) and tan(gd t) = sinh(t). Its inverse, ln(tan(π/4 + φ/2)), is exactly the vertical coordinate of the Mercator map projection. So every classic nautical sea chart is quietly drawn using the geometry of sinh and cosh.

Problem solved in full

Cosh 1 from the exponentials and the physical quantity 1.543081 6 steps
Compute cosh 1 from the exponentials, prove the identity that gives these functions their name — and then find out what physical quantity 1.543081 is.
1. Both functions are built from the same two exponentials, differing only in a sign. Everything else follows from that.
  
  $cosh t = \frac{e ^{t} + e ^{- t}}{2}, sinh t = \frac{e ^{t} - e ^{- t}}{2}$
2. Evaluate at t = 1 with e = 2.718282. This is the number the tool prints.
  
  $cosh 1 = \frac{2.718282 + 0.367879}{2} = \frac{3.086161}{2} = 1.543081$
3. Square both and subtract. The cross terms cancel and what survives is 4/4 — an identity with no approximation anywhere in it, and the reason the parameter traces a hyperbola rather than a circle.
  
  $cosh^{2} t - sinh^{2} t = \frac{( e ^{t} + e ^{- t} ) ^{2} - ( e ^{t} - e ^{- t} ) ^{2}}{4} = \frac{4}{4} = 1$
4. Add them instead and the exponential reappears intact. The ratio is tanh, which is bounded by ±1 no matter how large the argument grows — a ceiling that will matter in a moment.
  
  $cosh t + sinh t = e^{t}; tanh 1 = \frac{1.175201}{1.543081} = 0.761594$
5. Take that bounded ratio as a velocity in units of c and compute the Lorentz factor. It returns the number from step 2, exactly.
  
  $γ = \frac{1}{1 - β ^{2}} = \frac{1}{1 - 0.76159 4 ^{2}} = \frac{1}{0.419974} = 1.543081 = cosh 1$
6. Now add two rapidities the ordinary way and convert back. The result agrees with relativistic velocity addition, which is what makes rapidity the natural coordinate.
  
  $tanh (1 + 1) = 0.964028 = \frac{0.761594 + 0.761594}{1 + 0.76159 4 ^{2}}$
Answer

1.543081, which is the Lorentz factor of something moving at 0.7616 c. That is not a coincidence dressed up as one. If you define rapidity by β = tanh φ, then γ = cosh φ falls out of the identity proved in step 3, and the whole of one-dimensional special relativity becomes hyperbolic trigonometry. The payoff is that rapidities simply add: two boosts of rapidity 1 give tanh 2 = 0.9640 c, which is precisely what Einstein's velocity-addition formula returns for 0.7616 c combined with itself. Velocities do not add; the angles they are cosines of do.

$cosh 1 = 1.543081 = γ at v = 0.7616 c$

References (2)

The lesson — the definitions, the identity and the sector-area interpretation of the argument: F. W. J. Olver, D. W. Lozier, R. F. Boisvert and C. W. Clark (eds.), NIST Handbook of Mathematical Functions, ch. 4. Cambridge University Press, 2010. ISBN 978-0-521-19225-5.
Where the functions and their name come from: V. Riccati, Opusculorum ad res physicas et mathematicas pertinentium, vol. I. Bologna, 1757 — where the hyperbolic sine and cosine are introduced. J. H. Lambert, "Mémoire sur quelques propriétés remarquables des quantités transcendentes circulaires et logarithmiques", Mémoires de l'Académie royale des sciences de Berlin 17, 1768, which brought them into general use.

By the 42Math team · Updated 2026-07-22

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

identity check - Hyperbolic identity check: cosh^2(1) - sinh^2(1) = 1.
tanh asymptote - tanh approaches +/-1 as |t| increases, unlike tan which diverges.
catenary shape - cosh curve is the catenary profile used by hanging chains and cables.
derivative relation - Derivative relation: d/dx sinh = cosh and d/dx cosh = sinh.