2
$\begingroup$

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.

  • 0
    Thanks but I'm looking to get f(t) = cos(t) but all terms that include t are in the form (t-2). eg something like; cos(t) = f(t) = A.sin(t-2) + B.cos(t-2)2011-05-21

1 Answers 1

5

$\cos t = \cos((t-2)+2)$ 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 $y=2$).

  • 0
    Aha, gotcha. Thank $y$ou very much.2011-05-21