# Finding a cosine curve from a sine curve?

• November 29th 2009, 03:34 AM
maybealways
Finding a cosine curve from a sine curve?
Hopefully I posted this in the right spot!

If I had an equation of a sine curve, and I wanted to find the cosine curve, would I just subtract $\frac\pi2$ from the horizontal translation value?

Or rather, if I had a set of data, which, when plotted showed sine curve characteristics, but I actually wanted to find the cosine curve of this data; how would I go about doing that? (hopefully that made sense)

Thanks.
• November 29th 2009, 03:40 AM
e^(i*pi)
Quote:

Originally Posted by maybealways
Hopefully I posted this in the right spot!

If I had an equation of a sine curve, and I wanted to find the cosine curve, would I just subtract $\frac\pi2$ from the horizontal translation value?

Or rather, if I had a set of data, which, when plotted showed sine curve characteristics, but I actually wanted to find the cosine curve of this data; how would I go about doing that? (hopefully that made sense)

Thanks.

Nay, you need to add $\frac{\pi}{2}$ to get a horizontal translation left.

See the spoiler for why this is

$sin(A \pm B)=sin(A)cos(B) \pm cos(A)sin(B)$

Spoiler:
$sin \left(x-\frac{\pi}{2}\right) = sin(x)cos\left(\frac{\pi}{2}\right) - sin\left(\frac{\pi}{2}\right)cos(x)
$

$cos\left(\frac{\pi}{2}\right)=0 \: \: , \: \: sin \left(\frac{\pi}{2}\right) = 1 \: \: \therefore \: sin \left(x-\frac{\pi}{2}\right)=-cos(x)$

==============================

$sin \left(x+\frac{\pi}{2}\right) = sin(x)cos\left(\frac{\pi}{2}\right) + sin\left(\frac{\pi}{2}\right)cos(x) = cos(x)
$