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Hyperbolic Cosine — Definition, Formula & Examples

Hyperbolic cosine, written cosh(x), is a function defined as the average of exe^x and e−xe^{-x}. It produces the shape of a hanging chain or cable, called a catenary, and is always greater than or equal to 1.

For any real number xx, the hyperbolic cosine function is defined by cosh⁡(x)=ex+e−x2\cosh(x) = \frac{e^{x} + e^{-x}}{2}. It is an even function satisfying cosh⁡(−x)=cosh⁡(x)\cosh(-x) = \cosh(x), with a global minimum of 1 at x=0x = 0. It is the real part of the identity cosh⁡(ix)=cos⁡(x)\cosh(ix) = \cos(x), linking it to the ordinary cosine through Euler's formula.

Key Formula

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

How It Works

To evaluate cosh⁡(x)\cosh(x), compute exe^x and e−xe^{-x}, add them, and divide by 2. The function starts at cosh⁡(0)=1\cosh(0) = 1 and grows exponentially in both directions, producing a smooth U-shaped curve. Its derivative is sinh⁡(x)\sinh(x), and it satisfies the fundamental identity cosh⁡2(x)−sinh⁡2(x)=1\cosh^2(x) - \sinh^2(x) = 1, which mirrors the Pythagorean identity for ordinary trig functions. You will encounter cosh⁡\cosh when solving certain differential equations, computing arc lengths, and modeling physical systems like hanging cables.

Worked Example

Problem: Evaluate cosh(2) to four decimal places.
Step 1: Compute e2e^2.
e2≈7.3891e^{2} \approx 7.3891
Step 2: Compute e−2e^{-2}.
e−2≈0.1353e^{-2} \approx 0.1353
Step 3: Add the two results and divide by 2.
cosh⁡(2)=7.3891+0.13532=7.52442=3.7622\cosh(2) = \frac{7.3891 + 0.1353}{2} = \frac{7.5244}{2} = 3.7622
Answer: cosh⁡(2)≈3.7622\cosh(2) \approx 3.7622

Another Example

Problem: Verify the identity cosh⁡2(1)−sinh⁡2(1)=1\cosh^2(1) - \sinh^2(1) = 1.
Step 1: Compute cosh⁡(1)\cosh(1) using the definition.
cosh⁡(1)=e+e−12≈2.7183+0.36792=1.5431\cosh(1) = \frac{e + e^{-1}}{2} \approx \frac{2.7183 + 0.3679}{2} = 1.5431
Step 2: Compute sinh⁡(1)\sinh(1) using its definition.
sinh⁡(1)=e−e−12≈2.7183−0.36792=1.1752\sinh(1) = \frac{e - e^{-1}}{2} \approx \frac{2.7183 - 0.3679}{2} = 1.1752
Step 3: Square each value and subtract.
1.54312−1.17522≈2.3812−1.3811=1.0001≈11.5431^2 - 1.1752^2 \approx 2.3812 - 1.3811 = 1.0001 \approx 1
Answer: The identity holds: cosh⁡2(1)−sinh⁡2(1)=1\cosh^2(1) - \sinh^2(1) = 1 (the tiny rounding error vanishes with exact values).

Visualization

Why It Matters

In Calculus II and differential equations courses, cosh⁡\cosh appears in catenary curve problems, integration techniques, and solutions to the equation y′′=yy'' = y. Engineers use it to model the shape of suspension cables and power lines. Physicists rely on it in special relativity, where rapidity—a measure of relativistic velocity—is expressed using hyperbolic functions.

Common Mistakes

Mistake: Confusing the sign in the definition: writing ex−e−x2\frac{e^x - e^{-x}}{2} instead of ex+e−x2\frac{e^x + e^{-x}}{2}.
Correction: The plus sign gives cosh⁡(x)\cosh(x); the minus sign gives sinh⁡(x)\sinh(x). Remember that cosh⁡\cosh uses addition, which is why cosh⁡(0)=1\cosh(0) = 1 (the two exponentials add up) rather than 0.
Mistake: Assuming cosh⁡(x)\cosh(x) behaves like cos⁡(x)\cos(x) and oscillates or has a range of [−1,1][-1, 1].
Correction: cosh⁡(x)\cosh(x) never oscillates. It has a minimum value of 1 at x=0x = 0 and increases without bound as ∣x∣|x| grows. Its range is [1,∞)[1, \infty).

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