So I'm just learning Laplace transform in college, and an exercise asked us to solve this:
$cos(at+b)$
Now, I found multiple answers saying that this evaluates to:
$\LARGE \frac{s * cos(b) - a * sin(b)}{s^2+a^2}$
using the well-known formula of $cos(a + b)$.
But here comes my question. Can we actually use the displacement theorem to solve this?
What I mean is rewrite $cos(at+b)$ as $cos(a(t+\frac{b}{a}))$ and then use the theorem as:
$L[f(t-c)] = e^{-cs}*L[f(t)] = e^{-cs} * F(s)$
Using this, I arrive at:
$\LARGE L[f(t)] = \frac{e^{\frac{bs}{a}}*s}{s^2+a^2}$
Which doesn't really look like the answer above. Am I doing something wrong?
In the second case you have the Heaviside function.
So that if you inverse the function you get: $$\mathcal {L^{-1}} \left \{e^{bs/a}\dfrac s {s^2+a^2} \right \}=U(t+\frac ba)\cos (a (t+\dfrac b a ))$$ $$\mathcal {L^{-1}} \left \{e^{bs/a}\dfrac s {s^2+a^2} \right \}=U(t+\frac ba)\cos (at +b))$$
For the first case you have the Laplace transform of: $$\mathcal {L} \left \{u(t)\cos (at+ b ) \right \}$$