# $\cos(t)$ 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.

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$\cos(t)=f(t-2)$ if $f(t)=\cos(t+2)$. Is that what you're looking for? –  Jonas Meyer May 21 '11 at 8:32
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) –  tomatosource May 21 '11 at 10:46
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## 1 Answer

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

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I think I'll need it as a linear combination of sin(t-2) and cos(t-2) , but am confused how to re arrange the above identity. Wouldnt that just become cos(t-2) = cos(t)cos(-2)-sin(t)sin(-2) giving cos(t)=cos(t-2)+sin(t)sin(-2) / cos(-2)? Unsure how to get to the magical cos(t) = Xcos(t-2) + Ysin(t-2). –  tomatosource May 21 '11 at 10:42
@tomatosource, I think the point is to use the identity for $\cos(x+y)$ with $x=t-2$ and $y=2$. –  Gerry Myerson May 21 '11 at 11:00
Aha, gotcha. Thank you very much. –  tomatosource May 21 '11 at 13:25
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