The Identity cosh²x − sinh²x = 1

Square each definition

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

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

Subtract

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

Check it at x = 1: $\cosh^{2}1 = 2.381098$ and $\sinh^{2}1 = 1.381098$, 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, $\cosh x = \frac{5}{4}$ and $\sinh x = \frac{3}{4}$, and $\frac{25}{16} - \frac{9}{16} = \frac{16}{16} = 1$.

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

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

The circle

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

The hyperbola

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

The hyperbola $x^{2} - y^{2} = 1$ has two branches, one with $x \ge 1$ and one with $x \le -1$. 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 $\frac{e^{t}}{2}$, 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 $x^{2} - y^{2} = 1$ with its asymptotes y = x and y = −x, and the dashed unit circle $x^{2} + y^{2} = 1$. 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 $\frac{t}{2}$: 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 $\cosh^{2}x - \sinh^{2}x = 1$ by $\cosh^{2}x$. The first term becomes 1, the second becomes $\tanh^{2}x$, and the right-hand side becomes $\frac{1}{\cosh^{2}x}$. So $1 - \tanh^{2}x = \frac{1}{\cosh^{2}x}$. The reciprocal of cosh x is written sech x, so the right-hand side is also written sech²x.

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

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

Using the identity

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

Going the other way gives two answers. If $\cosh x = \frac{5}{4}$, then $\sinh^{2}x = \frac{25}{16} - 1 = \frac{9}{16}$, so $\sinh x = \frac{3}{4}$ or $\sinh x = - \frac{3}{4}$. Both happen: cosh is even, so $\cosh x = \frac{5}{4}$ 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, $\cos 2x = 1 - 2 \sin^{2}x$ becomes $\cosh 2x = 1 + 2 \sinh^{2}x$. At x = ln 2, $\cosh 2x = \frac{{4 + \frac{1}{4}}}{2} = \frac{17}{8}$, and $1 + 2 \times \frac{9}{16} = \frac{17}{8}$.

The usual mistakes

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

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

Forgetting the second sign for sinh x. From $\sinh^{2}x = \frac{9}{16}$, sinh x can be $\frac{3}{4}$ or $- \frac{3}{4}$, 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 $\sqrt{1 + \sinh^{2}u}$ 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.