Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
$\cos(t)=f(t-2)$ 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
up vote 5 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.