Hyperbolic addition formulas

Hyperbolic sine and cosine already place one real parameter on the right-hand branch of $x^2-y^2=1$. A second real parameter places another point on that same branch. Where does the point for their sum sit, and why do its coordinates factor?

The product is the height at the sum

The Euler e-function multiplies when the inputs add. Split that product into its odd and even parts, and the coordinates of the sum appear. For all complex $x$ and $y$,

$$\sinh (x\pm y)=\sinh x\cosh y\pm \cosh x\sinh y,$$

$$\cosh (x\pm y)=\cosh x\cosh y\pm \sinh x\sinh y.$$

Sine and cosine add with this same pattern for sine. The middle sign inside cosine is the one that flips. Here the middle sign inside cosh keeps the sign in front of $y$.

This picture keeps both inputs real. Drag either one. The gray marks sit on the exponential at those inputs. The green mark sits at their sum, and its height is the product of the two heights. The odd and even parts of that product are the two formulas.

Freeze the two parameters

Set $y=x$. The two parameters are now the same number. For every complex $x$,

$$\sinh 2x=2\sinh x\cosh x,\cosh 2x={\sinh }^{2}x+{\cosh }^{2}x.$$

The horizontal coordinate is the sum of the squares. The difference of those same squares is the identity that stays at $1$.

Drag the parameter. The gray point is that parameter on the right-hand branch. The green point is twice the parameter. The blue segment runs out to the horizontal coordinate, which is the sum of the squares.

The sign on the way back

Solving those double-angle formulas for the half parameter needs a real input. The square root is nonnegative, and $\sinh (x/2)$ is negative when $x$ is negative, so the sign has to be put back by hand. For real $x$,

$$\sinh \frac{x}{2}=\sqrt{\frac{1}{2}(\cosh x-1)}(x\ge 0),$$

$$\sinh \frac{x}{2}=-\sqrt{\frac{1}{2}(\cosh x-1)}(x<0),$$

$$\cosh \frac{x}{2}=\sqrt{\frac{1}{2}(\cosh x+1)}.$$

The second root stays nonnegative, which matches $\cosh$ on the real line, always at least $1$.

Drag the input across zero. The green point is the half parameter, so its height follows the sign. The blue point keeps both square roots nonnegative, so it stays above the axis. The two points share a place only while the input is positive. At zero both sit at $(1,0)$.

A power lengthens the parameter

The two combinations $e^x=\cosh x+\sinh x$ and $e^{-x}=\cosh x-\sinh x$ are already the definitions. Raising either one to a positive integer power replaces the parameter $x$ by $nx$. For every complex $x$ and every positive integer $n$,

$$(\cosh x\pm \sinh x{)}^n=\cosh (nx)\pm \sinh (nx).$$

This picture keeps the parameter real and positive, and keeps $n$ a positive integer, so both points stay on the right-hand branch. Drag either one. The gray point stays at the input. The green point is that many times the parameter, and it walks out along the branch as the integer grows.

Sums, briefly

A sum of values rewrites as a product of a half-sum and a half-difference:

$$\sinh x\pm \sinh y=2\sinh \frac{1}{2}(x\pm y)\cosh \frac{1}{2}(x\mp y),$$

$$\cosh x+\cosh y=2\cosh \frac{1}{2}(x+y)\cosh \frac{1}{2}(x-y),$$

$$\cosh x-\cosh y=2\sinh \frac{1}{2}(x+y)\sinh \frac{1}{2}(x-y).$$

Those three rows are the addition formulas read the other way.

Two real parameters add by factoring the product of the exponentials, and the sum stays on the right-hand branch. Freezing them doubles the parameter. On the way back, a real input lets the square root carry the sign of $\sinh (x/2)$. A positive integer power of $\cosh x\pm \sinh x$ only lengthens the parameter. The next quotients are tanh and coth.