$\cos(t)$ as a function of (t-2)? As the title suggests,
I'm doing a Laplace transform problem using the $t$-shift theorem, and I've almost got it, I just can't work out how to transform $\cos(t)$ into a function of $(t-2)$, I saw a trig identity 
$$\cos(x)\cos(y) = \frac{1}{2} [\cos(x - y) + \cos(x + y)]$$
But I can't quite work out how to re-arrange it, 
any help much appreciated. 
Thanks. 

$\cos t = \cos((t-2)+2)$ is a function of $t-2$. If you prefer to express it as a linear combination of $\sin (t-2)$ and $\cos (t-2)$, you can indeed use a trig identity:
$\cos (x+y) = \cos x \cos y - \sin x \sin y$ will work (with $x=t-2$ and $y=2$).
