exp, trig, hyp

What can we do with Euler’s Formula?

\[ e^{i t} = \cos (t) + i \, \sin (t) \]

Substituting \(-t\) for \(t\) gives \(e^{i (-t)} = \cos (-t) + i \, \sin (-t) = \cos (t) - i \, \sin (t)\), because \(\cos (t)\) is an even function and \(\sin (t)\) is an odd function.

That gives us

\[ e^{i t} + e^{-i t} = 2 \cos (t) \]
\[ \frac {e^{i t} + e^{-i t}}{2} = \cos (t) \]

It also gives us

\[ e^{i t} - e^{-i t} = 2 i \sin (t) \]
\[ \frac {e^{i t} - e^{-i t}}{2 i} = \sin (t) \]

This tells us that the geometry of right triangles can be expressed through exponential functions.

Our right triangles come from the unit circle, which are the solutions to the equation \(x^2 + y^2 = 1\).

There is also a unit hyperbola. It consists of the points that satisfy the equation \(x^2 - y^2 = 1\).

We can define two functions from the two coordinates of points on the unit hyperbola. We call these hyperbolic sine (\(\sinh \)) and hyperbolic cosine (\(\cosh \)).

They can also be expressed using exponentials.

\[ \cosh (t) = \frac {e^t + e^{-t}}{2} \, \text { and } \, \sinh (t) = \frac {e^t - e^{-t}}{2} \]

The geometry of the circle and the geometry of the hyperbola are connected through complex numbers.

\[ \cos (t) = \frac {e^{i t} + e^{-i t}}{2} = \cosh (i \, t) \]
\[ \sin (t) = \frac {e^{i t} - e^{-i t}}{2 i} = -i \sinh (i \, t) \]

If we work with Complex numbers, then trigonometric and hyperbolic functions are expressible in terms of exponential functions. They are different ways of expressing the same structure.

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more examples can be found by following this link
More Examples of the Complex Bridge

2026-07-11 14:50:11