The Identity cosh²x − sinh²x = 1

Square each definition

${\cosh }^{2}x$cosh²x means ${\left(\cosh x\right)}^2$(cosh x)², the square of cosh x. To square $\frac{e^x+e^{-x}}{2}$(eˣ + e⁻ˣ)/2, square the top and divide by 4. The top is ${\left(e^x+e^{-x}\right)}^2=e^{2x}+2e^x{e}^{-x}+e^{-2x}$(eˣ + e⁻ˣ)² = e²ˣ + 2eˣe⁻ˣ + e⁻²ˣ, and the middle term is simple, because $e^x\times e^{-x}=e^0=1$eˣ × e⁻ˣ = e⁰ = 1. So ${\cosh }^{2}x=\frac{e^{2x}+2+e^{-2x}}{4}$cosh²x = (e²ˣ + 2 + e⁻²ˣ)/4.

Squaring $\sinh x=\frac{e^x-e^{-x}}{2}$sinh x = (eˣ − e⁻ˣ)/2 the same way gives ${\sinh }^{2}x=\frac{e^{2x}-2+e^{-2x}}{4}$sinh²x = (e²ˣ − 2 + e⁻²ˣ)/4. The two squares have the same outer terms, $e^{2x}$e²ˣ and $e^{-2x}$e⁻²ˣ, and differ only in the sign of the middle term.

Subtract

In ${\cosh }^{2}x-{\sinh }^{2}x$cosh²x − sinh²x the outer terms cancel in pairs, and the middle terms leave $\frac{2-\left(-2\right)}{4}=\frac{4}{4}=1$(2 − (−2))/4 = 4/4 = 1. So ${\cosh }^{2}x-{\sinh }^{2}x=1$cosh²x − sinh²x = 1, for every value of x.

Check it at x = 1: ${\cosh }^{2}1=2.381098$cosh²1 = 2.381098 and ${\sinh }^{2}1=1.381098$sinh²1 = 1.381098, and the difference is 1. At x = 3 the squares are much larger, 101.357818 and 100.357818, and the difference is still 1. At x = ln 2, $\cosh x=\frac{5}{4}$cosh x = 5/4 and $\sinh x=\frac{3}{4}$sinh x = 3/4, and $\frac{25}{16}-\frac{9}{16}=\frac{16}{16}=1$25/16 − 9/16 = 16/16 = 1.

A second proof is shorter. ${\cosh }^{2}x-{\sinh }^{2}x$cosh²x − sinh²x is a difference of two squares, so it equals (cosh x − sinh x)(cosh x + sinh x). The second bracket is $e^x$eˣ and the first is $e^{-x}$e⁻ˣ, and $e^{-x}\times e^x=1$e⁻ˣ × eˣ = 1.

The sign matters. Adding the squares instead gives ${\cosh }^{2}x+{\sinh }^{2}x=\frac{2e^{2x}+2e^{-2x}}{4}=\cosh 2x$cosh²x + sinh²x = (2e²ˣ + 2e⁻²ˣ)/4 = cosh 2x, which is not a constant: at x = 1 it is 3.762196.

The circle

The trigonometric identity is ${\cos }^2\theta +{\sin }^2\theta =1$cos²θ + sin²θ = 1. Put $x=\cos \theta$x = cos θ and $y=\sin \theta$y = sin θ, and it says $x^2+y^2=1$x² + y² = 1: the point $\left(\cos \theta ,\sin \theta \right)$(cos θ, sin θ) lies on the circle of radius 1 centered at the origin, for every $\theta$θ.

The hyperbola

Do the same with the hyperbolic identity. Put x = cosh t and y = sinh t, and ${\cosh }^{2}t-{\sinh }^{2}t=1$cosh²t − sinh²t = 1 says $x^2-y^2=1$x² − y² = 1. That equation is a hyperbola, which is where the hyperbolic functions get their name.

The hyperbola $x^2-y^2=1$x² − y² = 1 has two branches, one with $x\ge 1$x ≥ 1 and one with $x\le -1$x ≤ −1. Since cosh t is never less than 1, the point (cosh t, sinh t) is always on the right-hand branch. And since sinh t increases and takes every value, the point runs up the whole of that branch as t increases, from the bottom to the top.

For large t, cosh t and sinh t are both close to $\frac{e^t}{2}$eᵗ/2, so y is close to x: the branch approaches the line y = x. In the same way it approaches y = −x below. These two lines are the asymptotes of the hyperbola.

The hyperbola $x^2-y^2=1$x² − y² = 1 with its asymptotes y = x and y = −x, and the dashed unit circle $x^2+y^2=1$x² + y² = 1. The two curves touch at (1, 0). The dots are (cosh t, sinh t) for t = −1, −0.5, 0, 0.5, 1 and 1.5, from (1.5431, −1.1752) at the bottom to (2.3524, 2.1293) at the top. Every one is on the right-hand branch.

On the left, the unit circle with the point (cos t, sin t); on the right, the hyperbola with the point (cosh t, sinh t). Each gold region has area $\frac{t}{2}$t/2: on the circle it is a sector with angle t, and on the hyperbola it is bounded by the axis, the curve and the line from the origin to the point. At t = 1.5 the point is (2.35, 2.13) and both areas are 0.75. Drag the point back to t = 1, where it is (1.54, 1.18) and both areas are 0.5.

An identity for tanh

Divide every term of ${\cosh }^{2}x-{\sinh }^{2}x=1$cosh²x − sinh²x = 1 by ${\cosh }^{2}x$cosh²x. The first term becomes 1, the second becomes ${\tanh }^{2}x$tanh²x, and the right-hand side becomes $\frac{1}{{\cosh }^{2}x}$1/cosh²x. So $1-{\tanh }^{2}x=\frac{1}{{\cosh }^{2}x}$1 − tanh²x = 1/cosh²x. The reciprocal of cosh x is written sech x, so the right-hand side is also written sech²x.

Check at x = ln 2, where $\tanh x=\frac{3}{5}$tanh x = 3/5 and $\cosh x=\frac{5}{4}$cosh x = 5/4: $1-\frac{9}{25}=\frac{16}{25}$1 − 9/25 = 16/25, and $\frac{1}{{\cosh }^{2}x}=\frac{16}{25}$1/cosh²x = 16/25.

The right-hand side is positive for every x, so ${\tanh }^{2}x$tanh²x is always less than 1. That is another way to see that tanh x stays between −1 and 1.

Using the identity

If $\sinh x=\frac{3}{4}$sinh x = 3/4, then ${\cosh }^{2}x=1+{\sinh }^{2}x=1+\frac{9}{16}=\frac{25}{16}$cosh²x = 1 + sinh²x = 1 + 9/16 = 25/16, so $\cosh x=\frac{5}{4}$cosh x = 5/4. Only the positive square root is possible, because cosh x is never negative.

Going the other way gives two answers. If $\cosh x=\frac{5}{4}$cosh x = 5/4, then ${\sinh }^{2}x=\frac{25}{16}-1=\frac{9}{16}$sinh²x = 25/16 − 1 = 9/16, so $\sinh x=\frac{3}{4}$sinh x = 3/4 or $\sinh x=-\frac{3}{4}$sinh x = −3/4. Both happen: cosh is even, so $\cosh x=\frac{5}{4}$cosh x = 5/4 at x = ln 2 and at x = −ln 2, where sinh x has opposite signs.

Other identities follow the trigonometric ones, with the sign changed wherever a product of two sines appears. This is Osborn’s rule. For example, $\cos 2x=1-2{\sin }^{2}x$cos 2x = 1 − 2 sin²x becomes $\cosh 2x=1+2{\sinh }^{2}x$cosh 2x = 1 + 2 sinh²x. At x = ln 2, $\cosh 2x=\frac{4+\frac{1}{4}}{2}=\frac{17}{8}$cosh 2x = (4 + 1/4)/2 = 17/8, and $1+2\times \frac{9}{16}=\frac{17}{8}$1 + 2 × 9/16 = 17/8.

The usual mistakes

Writing a plus sign. ${\cos }^2\theta +{\sin }^2\theta =1$cos²θ + sin²θ = 1 has a plus, but the hyperbolic identity has a minus: ${\cosh }^{2}x-{\sinh }^{2}x=1$cosh²x − sinh²x = 1.

Taking a negative value for cosh x. From ${\cosh }^{2}x=\frac{25}{16}$cosh²x = 25/16, cosh x is $\frac{5}{4}$5/4, never $-\frac{5}{4}$−5/4.

Forgetting the second sign for sinh x. From ${\sinh }^{2}x=\frac{9}{16}$sinh²x = 9/16, sinh x can be $\frac{3}{4}$3/4 or $-\frac{3}{4}$−3/4, and the question has to say which.

Putting (cosh t, sinh t) on the circle. The circle belongs to cos and sin, whose identity has the plus sign; the hyperbolic point is on the hyperbola.

A cable and a cooling tower

In the first application below, the pull in a hanging cable has a horizontal part and a vertical part, and Pythagoras’ theorem combines them. The identity turns $\sqrt{1+{\sinh }^{2}u}$√(1 + sinh²u) into a plain cosh u.

In the second, the outline of a cooling tower is a hyperbola, so every point on it can be written with cosh and sinh, and the identity moves between the radius and the height.