Hyperbolic sine and cosine

The Euler e-function at a pure imaginary input splits into sine and cosine. At a real input that exponential is one real height. Which two functions split that height, and which identity do they satisfy together?

Split the exponential

For every complex $x$,

$$\sinh x:=\frac{e^x-e^{-x}}{2},\cosh x:=\frac{e^x+e^{-x}}{2}.$$

This picture keeps the input real and draws both functions on one pair of axes. Drag the input. The green height is $\sinh x$: it passes through the origin and changes sign with the input. The blue height is $\cosh x$: it equals $1$ when the input is $0$, and the height to the left of the axis matches the height to the right.

Substitute a pure imaginary argument

For every complex $x$,

$$\sinh (ix)=i\sin x,\cosh (ix)=\cos x.$$

A pure imaginary input to this pair returns sine and cosine. The difference of squares needs the substitution run the other way. Sine and cosine give $\cos x=(e^{ix}+e^{-ix})/2$ and $\sin x=(e^{ix}-e^{-ix})/(2i)$. Replace that input by $ix$. For every complex $x$,

$$\cos (ix)=\cosh x,\sin (ix)=i\sinh x.$$

The circle identity ${\cos }^{2}z+{\sin }^{2}z=1$, at $z=ix$, is then

$${\cosh }^{2}x-{\sinh }^{2}x=1.$$

Each later formula for this pair can be read from a sine or cosine formula by one of those two substitutions. The difference of squares is the first.

The right-hand branch

Hyperbolas in the plane already trace the right-hand branch, for positive $a$ and $b$, by $x=a\cosh t$ and $y=b\sinh t$. The left-hand branch on that page puts a minus in front of $x$. This picture freezes $a=b=1$, so the moving point is $(\cosh t,\sinh t)$. Drag the parameter. The point stays on $x^2-y^2=1$. That is the identity just written, read as a point on the right-hand branch.

Odd and even

The graphs already showed the symmetry for a real input. For every complex $x$,

$$\sinh (-x)=-\sinh x,\cosh (-x)=\cosh x.$$

The Euler e-function repeats after a shift of $2\pi i$. These two functions repeat with it:

$$\sinh (x+2\pi i)=\sinh x,\cosh (x+2\pi i)=\cosh x.$$

One step along the tangent

Differentiate the two definitions. For every complex $x$,

$$\frac{\mathrm{d}\sinh x}{\mathrm{d}x}=\cosh x,\frac{\mathrm{d}\cosh x}{\mathrm{d}x}=\sinh x.$$

This picture keeps the input real. Drag the point on $y=\sinh x$. The horizontal step has length $1$. The vertical step equals $\cosh x$ at that point, and the diagonal is the tangent. The height of $\sinh$ one unit farther to the right is a different number.

Sinh and cosh are the two combinations of $e^x$ and $e^{-x}$. For every complex number they satisfy ${\cosh }^{2}x-{\sinh }^{2}x=1$, and each derivative is the other function. On the real line the first changes sign through the origin, and the second stays positive, with value $1$ at $0$. A real parameter $t$ places $(\cosh t,\sinh t)$ on the right-hand branch of $x^2-y^2=1$. How two such parameters add is the hyperbolic addition formulas.