Problem
I need to find the Laplace transform of $\cos(2t-(\frac\pi3))$
Attempt
I've tried to look up some relevant formulae in my book, but I can't find anything that looks useful. I suspect there is something there, that I'm just not seeing.
Wolfram Alpha suggests the answer is $\displaystyle\frac{s+2\sqrt3}{2s^2+8}$, but I can't "reverse engineer" it either.
Any help appreciated!
Best Answer
Case 1: $f(t)=\cos\left(2t-\frac{\pi}{3}\right) u\left(t\right)$
Observing that $\cos(\omega t -\phi)=\cos(\omega t)\cos\phi+\sin(\omega t)\sin\phi$ and that $\mathcal L\left\{\cos(\omega t)\right\}=\frac{s}{s^2+\omega^2}$ and $\mathcal L\left\{\sin(\omega t)\right\}=\frac{\omega}{s^2+\omega^2}$, we have \begin{align} \mathcal L\left\{\cos(\omega t -\phi)\right\}&=\mathcal L\left\{\cos(\omega t)\right\}\cos\phi+\mathcal L\left\{\sin(\omega t)\right\}\sin\phi\\ &=\frac{s}{s^2+\omega^2}\cos\phi+\frac{\omega}{s^2+\omega^2}\sin\phi\\ &=\frac{s\cos\phi+\omega\sin\phi}{s^2+\omega^2} \end{align} For $\omega=2$ and $\phi=\frac{\pi}{3}$, $\cos\left(\frac{\pi}{3}\right)=\frac 1 2 $ and $\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt 3}{2}$ we have $$ \mathcal L\left\{\cos\left(2t-\frac{\pi}{3}\right)u\left(t\right)\right\}= \frac{\frac{s}{2}+\sqrt{3}}{s^2+4}=\frac{s+2\sqrt{3}}{2s^2+8} $$
Case 2: $f(t)=\cos\left(2t-\frac{\pi}{3}\right) u\left(2t-\frac{\pi}{3}\right)$
Use the Time shifting property $$ \mathcal L\left\{f(t - a) \right\}=\mathrm e^{-as} F(s) $$ and that for $f(t)=\cos(at)$ $$ \mathcal L\left\{\cos(\omega t)\right\}=F(s)=\frac{s}{s^2+\omega^2} $$ So $$\mathcal L\left\{\cos(\omega(t-a))\right\}=\mathrm e^{-as}\frac{s}{s^2+\omega^2} $$ and for $\omega=2$ and $a=\frac{\pi}{6}$ we have $$ \mathcal L\left\{\cos\left(2t-\frac{\pi}{3}\right)u\left(t-\frac{\pi}{6}\right)\right\}=\mathrm e^{-\frac{\pi}{6}s}\frac{s}{s^2+4} $$