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A hyperbolic angle $t$ is twice the signed area $A$ enclosed by the two rays from the origin to $(1,0)$ and $(\cosh t,\sinh t)$ and the intervening arc of the unit hyperbola. Counterclockwise traversal is positive: $t>0$ gives an ordinary area $t/2$ above the horizontal axis, and $t<0$ gives signed area $t/2$ below it. This parallels the unit circle, whose sector has area $\theta /2$ for an angle $\theta$ measured in radians.

To check the area interpretation, write $x(u)=\cosh u$ and $y(u)=\sinh u$. Green's theorem expresses the signed area as a line integral; the straight radial edges contribute zero, leavingThus $t$ is an area parameter, rather than the ordinary angle of the ray. For $t>0$, that ordinary angle is $\arctan (\tanh t)$; it stays below $\pi /4$ even as the hyperbolic angle grows without bound.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.