# Homework Help: Why is cos(-pi)=0?

1. Oct 18, 2011

### SMA_01

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. Oct 18, 2011

### Petek

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).

3. Oct 18, 2011

### SMA_01

@Petek: Wow I feel slow, totally passed me that cosine is an even function lol. Thanks

4. Oct 18, 2011

### SammyS

Staff Emeritus
Think about the unit circle !

Where is the angle π relative to the angle -π ?

5. Oct 18, 2011

### QuarkCharmer

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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