Could someone please explain how to transform this to the Laplace domain?
I've tried to use the definition of Laplace (not sure this is the easiest way).
$$\int_{0}^{t}e^{-st}f(t)\,dt$$
$$\int_{0}^{t}e^{-st} \cos(4t+8)\,dt$$
But got stuck in a loop because, the integral of cosine is sine and the integral if sine is cosine. More or less same store for the $e^{-st}$…
I have a table with the standard laplace transformations. The closest I have is
$$\mathcal{L}\{\cos(a)\} = \frac{s}{s^2+a^2}$$
Best Answer
You can also use the addition formula $\cos(x+y)=\cos x\cos y-\sin x\sin y$, so \begin{align} \mathscr{L}\{\cos(4t+8)\}&=\mathscr{L}\{\cos 8\cdot\cos(4t)-\sin 8\cdot\sin (4t)\}\\ &=(\cos 8)\mathscr{L}\{\cos(4t)\}-(\sin 8)\mathscr{L}\{\sin (4t)\}\\ &=(\cos 8)\left(\frac{s}{s^2+16}\right)-(\sin 8)\left(\frac{4}{s^2+16}\right)\\ &=\frac{s\cos 8-4\sin 8}{s^2+16} \end{align}