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

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

$$e^{i t} - e^{-i t} = 2 i \sin (t)$$

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

We have also seen that

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

This tells us that

$$\cos (t) = \frac {e^{i t} + e^{-i t}}{2} = \cosh (i t)$$

$$\sin (t) = \frac {e^{i t} - e^{-i t}}{2 i} = -i \sinh (i t)$$

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.