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Thread: Finding a cosine curve from a sine curve?

  1. November 29th 2009, 03:34 AM #1
    maybealways
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    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.
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  2. November 29th 2009, 03:40 AM #2
    e^(i*pi)
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    Quote Originally Posted by maybealways View Post
    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)<br />

    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)<br />
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