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Why is cos(-pi)=0?

  1. Oct 18, 2011 #1
    Sorry I made a mistake in the title, I was thinking of something else I meant "Why is cos(-pi)=-1?"

    cos(-pi)=-1, why is this so? I feel like maybe i'm missing the obvious, but I would think that it's equal to 1, considering cos(pi)=-1?
     
  2. jcsd
  3. Oct 18, 2011 #2
    The cosine function satisfies the identity

    cos (-x) = cos (x).

    Perhaps you are thinking of the sine function? It satisfies the identity

    sin (-x) = -sin (x).
     
  4. Oct 18, 2011 #3
    @Petek: Wow I feel slow, totally passed me that cosine is an even function lol. Thanks
     
  5. Oct 18, 2011 #4

    SammyS

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    Think about the unit circle !

    Where is the angle π relative to the angle -π ?
     
  6. Oct 18, 2011 #5
    As pointed out above, cosine is an even function.

    Another way to think about it is that the period of cosine is 2π. So if cos(m)=A
    Then cos(k2π+m) = A, where k is some integer. In your case, you say that cos(π) = -1, so it must be true that cos(π - 2π) = -1, ie: cos(-π) = -1
     
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