sinh, cosh and tanh

Two halves of the exponential

The hyperbolic cosine and the hyperbolic sine are made from $e^x$eˣ and $e^{-x}$e⁻ˣ. Their average is $\cosh x=\frac{e^x+e^{-x}}{2}$cosh x = (eˣ + e⁻ˣ)/2, and half their difference is $\sinh x=\frac{e^x-e^{-x}}{2}$sinh x = (eˣ − e⁻ˣ)/2.

At x = 1, e = 2.718282 and $e^{-1}=0.367879$e⁻¹ = 0.367879, so $\cosh 1=\frac{2.718282+0.367879}{2}=1.543081$cosh 1 = (2.718282 + 0.367879)/2 = 1.543081 and $\sinh 1=\frac{2.718282-0.367879}{2}=1.175201$sinh 1 = (2.718282 − 0.367879)/2 = 1.175201.

Replacing x by −x swaps $e^x$eˣ and $e^{-x}$e⁻ˣ. The sum does not change, so cosh(−x) = cosh x: cosh is an even function, and its graph is symmetric about the y-axis. The difference changes sign, so sinh(−x) = −sinh x: sinh is an odd function.

The two plain curves are $y=e^x$y = eˣ, rising to the right, and $y=e^{-x}$y = e⁻ˣ, its reflection in the y-axis. The gold curve, cosh x, is exactly halfway between them at every x: at x = 1 it is 1.543081, halfway between 2.718282 and 0.367879. All three pass through (0, 1).

Adding them back together

Add the two definitions: $\frac{e^x+e^{-x}}{2}+\frac{e^x-e^{-x}}{2}$(eˣ + e⁻ˣ)/2 + (eˣ − e⁻ˣ)/2. The $e^{-x}$e⁻ˣ terms cancel and the two halves of $e^x$eˣ make a whole, so $\cosh x+\sinh x=e^x$cosh x + sinh x = eˣ. Subtracting instead gives $\cosh x-\sinh x=e^{-x}$cosh x − sinh x = e⁻ˣ. At x = 1, 1.543081 + 1.175201 = 2.718282 = e.

So $e^x$eˣ splits into an even part, cosh x, and an odd part, sinh x. The series shows the same split. $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$eˣ = 1 + x + x²/2! + x³/3! + …; its even powers make $\cosh x=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\dots$cosh x = 1 + x²/2! + x⁴/4! + …, and its odd powers make $\sinh x=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots$sinh x = x + x³/3! + x⁵/5! + …. At x = 1 the first four terms of cosh give 1 + 0.5 + 0.041667 + 0.001389 = 1.543056, close to 1.543081.

These are the series for cos x and sin x with every minus sign changed to a plus.

The graphs of cosh and sinh

cosh x is never less than 1. Here is why: $\cosh x-1=\frac{e^x-2+e^{-x}}{2}$cosh x − 1 = (eˣ − 2 + e⁻ˣ)/2, and $e^x-2+e^{-x}$eˣ − 2 + e⁻ˣ is the square ${\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right)}^2$(e^(x/2) − e^(−x/2))², since $e^{\frac{x}{2}}\times e^{-\frac{x}{2}}=1$e^(x/2) × e^(−x/2) = 1. A square is never negative, so $\cosh x-1\ge 0$cosh x − 1 ≥ 0. It is 0 only when $e^{\frac{x}{2}}=e^{-\frac{x}{2}}$e^(x/2) = e^(−x/2), which is at x = 0. So the lowest point of cosh is (0, 1), and the curve rises on both sides of it.

sinh x increases everywhere: $e^x$eˣ increases and $e^{-x}$e⁻ˣ decreases, so their difference increases. It passes through the origin, since sinh 0 = 0, and it takes every value, negative and positive.

Far to the right, $e^{-x}$e⁻ˣ is tiny, so both functions are close to $\frac{e^x}{2}$eˣ/2. At x = 3, cosh 3 = 10.067662, sinh 3 = 10.017875 and $\frac{e^3}{2}=10.042768$e³/2 = 10.042768. Far to the left, cosh x is close to $\frac{e^{-x}}{2}$e⁻ˣ/2 and sinh x is close to $-\frac{e^{-x}}{2}$−e⁻ˣ/2.

The gold curve is y = cosh x, with its lowest point at (0, 1). The plain curve is y = sinh x, rising through the origin. At x = 2 they are 3.762196 and 3.626860, already close together.

tanh

The hyperbolic tangent is $\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}$tanh x = (sinh x)/(cosh x) = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ). Like sinh, it is odd.

Divide the top and the bottom by $e^x$eˣ: $\tanh x=\frac{1-e^{-2x}}{1+e^{-2x}}$tanh x = (1 − e^(−2x))/(1 + e^(−2x)). For x > 0, $e^{-2x}$e^(−2x) is between 0 and 1, so the top is positive and smaller than the bottom, and 0 < tanh x < 1. As x grows, $e^{-2x}$e^(−2x) shrinks toward 0 and tanh x climbs toward 1 without reaching it. Because tanh is odd, it lies between −1 and 0 for x < 0, approaching −1 on the far left.

tanh 1 = 0.761594, tanh 2 = 0.964028 and tanh 3 = 0.995055. The lines y = 1 and y = −1 are horizontal asymptotes.

y = tanh x between its two asymptotes, y = 1 and y = −1. The marked points are (1, 0.761594) and (2, 0.964028): by x = 2 the curve is within 0.04 of 1.

The values at zero

At x = 0 both exponentials are $e^0=1$e⁰ = 1. So $\cosh 0=\frac{1+1}{2}=1$cosh 0 = (1 + 1)/2 = 1, $\sinh 0=\frac{1-1}{2}=0$sinh 0 = (1 − 1)/2 = 0 and $\tanh 0=\frac{0}{1}=0$tanh 0 = 0/1 = 0.

Another value worth knowing is x = ln 2, where $e^x=2$eˣ = 2 and $e^{-x}=\frac{1}{2}$e⁻ˣ = 1/2. Then $\cosh x=\frac{2+\frac{1}{2}}{2}=\frac{5}{4}$cosh x = (2 + 1/2)/2 = 5/4, $\sinh x=\frac{2-\frac{1}{2}}{2}=\frac{3}{4}$sinh x = (2 − 1/2)/2 = 3/4 and $\tanh x=\frac{3}{5}$tanh x = 3/5.

The usual mistakes

Leaving out the halving. $e^x+e^{-x}$eˣ + e⁻ˣ is 2 at x = 0; cosh 0 is half of that, 1.

Swapping the two definitions. cosh has the plus sign, and it is the even one; sinh has the minus sign, and it is the odd one.

Expecting cosh to take small values. cosh x = 0.5 has no solution, because cosh x is never below 1.

Using a calculator set to degrees. These are functions of a number, not of an angle.

A cable and a boat

A cable hanging between two supports takes the shape $y=a\cosh \left(\frac{x}{a}\right)$y = a cosh(x/a), called a catenary. In the first application below, the sag of a power line is the height of its cosh curve at a pylon minus the height at its lowest point.

In the second, a boat’s speed is a multiple of tanh, so the speed approaches that multiple and never reaches it.