Uncovering the Mystery of cos(-pi)=-1: Insights and Explanations

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Homework Help Overview

The discussion revolves around understanding why cos(-π) equals -1, with participants exploring the properties of the cosine function and its behavior on the unit circle.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the even nature of the cosine function and its implications for negative angles. There are questions about the relationship between cos(π) and cos(-π), as well as considerations of the unit circle.

Discussion Status

Several participants have provided insights into the properties of the cosine function, including its evenness and periodicity. There seems to be a productive exploration of these concepts, although no consensus has been explicitly stated.

Contextual Notes

Some participants express confusion regarding the relationship between cosine values at positive and negative angles, indicating a potential gap in understanding the function's properties.

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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?
 
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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).
 
@Petek: Wow I feel slow, totally passed me that cosine is an even function lol. Thanks
 
Think about the unit circle !

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