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Hyperbolic Functions — Definition, Formula & Examples

Hyperbolic functions are a family of functions — sinh, cosh, tanh, and their reciprocals — defined using exponential functions. They share many structural similarities with trigonometric functions but describe hyperbolas rather than circles.

For any real number xx, the hyperbolic sine and hyperbolic cosine are defined as sinh⁡x=ex−e−x2\sinh x = \frac{e^x - e^{-x}}{2} and cosh⁡x=ex+e−x2\cosh x = \frac{e^x + e^{-x}}{2}, respectively. The remaining hyperbolic functions are derived from these: tanh⁡x=sinh⁡xcosh⁡x\tanh x = \frac{\sinh x}{\cosh x}, coth⁡x=cosh⁡xsinh⁡x\operatorname{coth} x = \frac{\cosh x}{\sinh x}, sech⁡x=1cosh⁡x\operatorname{sech} x = \frac{1}{\cosh x}, and csch⁡x=1sinh⁡x\operatorname{csch} x = \frac{1}{\sinh x}. The point (cosh⁡t,sinh⁡t)(\cosh t, \sinh t) traces the right branch of the unit hyperbola x2−y2=1x^2 - y^2 = 1, analogous to how (cos⁡t,sin⁡t)(\cos t, \sin t) traces the unit circle.

Key Formula

sinh⁡x=ex−e−x2,cosh⁡x=ex+e−x2,tanh⁡x=ex−e−xex+e−x\sinh x = \frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}, \qquad \tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}
Where:
  • xx = Any real number (the argument of the function)
  • ee = Euler's number, approximately 2.71828

How It Works

You evaluate hyperbolic functions by substituting into their exponential definitions. For instance, to find sinh⁡2\sinh 2, compute e2−e−22\frac{e^2 - e^{-2}}{2}. Hyperbolic functions satisfy identities that mirror trigonometric ones, with key sign changes: the fundamental identity is cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1 (note the minus sign, compared to cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1). Derivatives are also closely parallel: ddxsinh⁡x=cosh⁡x\frac{d}{dx}\sinh x = \cosh x and ddxcosh⁡x=sinh⁡x\frac{d}{dx}\cosh x = \sinh x — no alternating signs as with sine and cosine. These properties make hyperbolic functions appear naturally in solutions to certain differential equations, such as those governing a hanging cable (catenary).

Worked Example

Problem: Evaluate sinh(ln 3) exactly.
Step 1: Write the definition of sinh with the given argument.
sinh⁡(ln⁡3)=eln⁡3−e−ln⁡32\sinh(\ln 3) = \frac{e^{\ln 3} - e^{-\ln 3}}{2}
Step 2: Simplify each exponential. Since eln⁡3=3e^{\ln 3} = 3 and e−ln⁡3=13e^{-\ln 3} = \frac{1}{3}:
sinh⁡(ln⁡3)=3−132\sinh(\ln 3) = \frac{3 - \frac{1}{3}}{2}
Step 3: Combine the fractions in the numerator and divide.
=9−132=86=43= \frac{\frac{9-1}{3}}{2} = \frac{8}{6} = \frac{4}{3}
Answer: sinh⁡(ln⁡3)=43\sinh(\ln 3) = \dfrac{4}{3}

Another Example

Problem: Verify the identity cosh²(x) − sinh²(x) = 1 using the definitions.
Step 1: Square both definitions.
cosh⁡2x=(ex+e−x2)2=e2x+2+e−2x4\cosh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4}
Step 2: Similarly for sinh²:
sinh⁡2x=e2x−2+e−2x4\sinh^2 x = \frac{e^{2x} - 2 + e^{-2x}}{4}
Step 3: Subtract sinh² from cosh².
cosh⁡2x−sinh⁡2x=(e2x+2+e−2x)−(e2x−2+e−2x)4=44=1\cosh^2 x - \sinh^2 x = \frac{(e^{2x}+2+e^{-2x}) - (e^{2x}-2+e^{-2x})}{4} = \frac{4}{4} = 1
Answer: The identity cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1 holds for all real xx.

Visualization

Why It Matters

Hyperbolic functions appear throughout Calculus II and Differential Equations. Engineers use cosh⁡x\cosh x to model the catenary curve of suspended cables and power lines. In physics, tanh⁡\tanh describes relativistic velocity addition, and hyperbolic functions arise in solutions to Laplace's equation in rectangular coordinates.

Common Mistakes

Mistake: Using the wrong sign in the fundamental identity, writing cosh⁡2x+sinh⁡2x=1\cosh^2 x + \sinh^2 x = 1 by analogy with the Pythagorean identity.
Correction: The correct identity is cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1. The minus sign reflects the hyperbola x2−y2=1x^2 - y^2 = 1, not the circle.
Mistake: Assuming cosh⁡x\cosh x can be negative, just as cos⁡x\cos x can.
Correction: cosh⁡x=ex+e−x2\cosh x = \frac{e^x + e^{-x}}{2} is a sum of positive quantities divided by 2, so cosh⁡x≥1\cosh x \geq 1 for all real xx. It is always positive and has a minimum value of 1 at x=0x = 0.

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