The reason you see $s - a$ in the numerator in the Laplace transform equation related to the cosine function comes from the shift theorem in the Laplace transform.

The general Laplace transform of $\cos(\omega t)$ is:

$\mathcal{L} \{ \cos(\omega t) \} = \frac{s}{s^2 + \omega^2}$

Now, if we consider a shifted cosine function like $e^{at} \cos(\omega t)$, the Laplace transform involves the shifting property:

$\mathcal{L} \{ e^{at} f(t) \} = F(s - a)$

Applying this to $e^{at} \cos(\omega t)$, the Laplace transform becomes:

$\mathcal{L} \{ e^{at} \cos(\omega t) \} = \frac{s - a}{(s - a)^2 + \omega^2}$

Thus, the $s - a$ in the numerator arises from the exponential term $e^{at}$ due to the shifting property in Laplace transforms.

Would you like further details, or do you have any questions on this?

Here are five related questions you might find helpful:

1. How does the shifting property apply to other trigonometric functions?
2. What is the Laplace transform of $\sin(\omega t)$ with an exponential shift?
3. How do you perform the inverse Laplace transform for shifted functions?
4. How do complex roots of characteristic equations relate to cosine terms in Laplace transforms?
5. Can you explain the connection between Laplace transforms and solving differential equations?

Tip: The shifting property of Laplace transforms is particularly useful when dealing with systems where damping or exponential growth occurs, such as in circuits or control systems!