# How do hyperbolic sine and cosine relate to the hyperbola?

Mar 24, 2016

Here's an attempt to explain in slightly touchy feely terms. If I can formulate it a bit better I may provide another answer later.

#### Explanation:

The trigonometric functions $\cos$, $\sin$, etc. relate to the unit circle.

The hyperbolic functions $\cosh$, $\sinh$, etc. relate to the hyperbola.

As conic sections, the unit circle and unit hyperbola are at right angles.

In fact the Real trigonometric and hyperbolic functions are just sections of the Complex $\cos$ and $\sin$ functions at right angles.

Consider the unit circle in the Complex plane and a point travelling around the circle with position $\cos t + i \sin t$ at time $t$. It's fairly easy to convince ourselves that the (tangential) velocity at time $t$ is $- \sin t + i \cos t$.

Hence we find $\frac{d}{\mathrm{dt}} \sin t = \cos t$ and $\frac{d}{\mathrm{dt}} \cos t = - \sin t$

From these derivatives and the values $\cos \left(0\right) = 1$, $\sin \left(0\right) = 0$, we can construct Maclaurin series for $\sin x$ and $\cos x$.

sin x = x/(1!)-x^3/(3!)+x^5/(5!)-x^7/(7!)+...

cos x = 1/(0!)-x^2/(2!)+x^4/(4!)-x^6/(6!)+...

These Maclaurin series are easy to prove convergent and very well behaved in general. We can use them to extend the definition of $\sin x$ and $\cos x$ to Complex values, defining:

sin z = z/(1!)-z^3/(3!)+z^5/(5!)-z^7/(7!)+...

cos z = 1/(0!)-z^2/(2!)+z^4/(4!)-z^6/(6!)+...

Then see what happens to these series when we substitute $i x$ for $z$:

sin ix = i (x/(1!)+x^3/(3!)+x^5/(5!)+x^7/(7!)+...)

cos ix = 1/(0!)+x^2/(2!)+x^4/(4!)+x^6/(6!)+...

So we find:

cos ix - i sin ix = 1/(0!)+x/(1!)+x^2/(2!)+x^3/(3!)+...

Define:

exp(z) = 1/(0!)+z/(1!)+z^2/(2!)+z^3/(3!)+... = cos iz - i sin iz

Then:

$\exp \left(i \theta\right) = \cos \left({i}^{2} \theta\right) - i \sin \left({i}^{2} \theta\right)$

$= \cos \left(- \theta\right) - i \sin \left(- \theta\right)$

$= \cos \theta + i \sin \theta$

If we define:

sinh z = 1/i sin iz = z/(1!)+z^3/(3!)+z^5/(5!)+z^7/(7!)+...

cosh z = cos iz = 1/(0!)+z^2/(2!)+z^4/(4!)+z^6/(6!)+...

then we are basically just rotating the axes of the coefficients through a right angle.

$\cosh x$ and $\sinh x$ are the even and odd parts of the function $\exp \left(x\right)$ and therefore satisfy:

$\cosh \left(x\right) = \frac{\exp \left(x\right) + \exp \left(- x\right)}{2}$

$\sinh \left(x\right) = \frac{\exp \left(x\right) - \exp \left(- x\right)}{2}$

These identities continue to work for Complex values too.

Apr 3, 2016

The parametric equation

$\left\{\begin{matrix}x = \cosh \left(t\right) \\ y = \sinh \left(t\right)\end{matrix}\right.$

defines a unit hyperbola.

This arises from the identity

${\cosh}^{2} \left(x\right) - {\sinh}^{2} \left(x\right) \equiv 1$