Cosine of Integer Multiple of Pi

Theorem
Let $x \in \R$ be a real number.

Let $\cos x$ be the cosine of $x$.

Then:
 * $\forall n \in \Z: \cos n \pi = \left({-1}\right)^n$

or alternatively:

Proof
Recall the definition of the cosine function:


 * $\displaystyle \cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

From Cosine of Zero is One, we have that:
 * $\displaystyle \cos 0 = 1 - \frac {0^2} {2!} + \frac {0^4} {4!} - \cdots = 1$

This takes care of $n = 0$, and is sufficient for the derivation in Sine and Cosine are Periodic on Reals.

From Sine and Cosine are Periodic on Reals, we have that $\cos \left({x + 2 \pi}\right) = \cos x$, and thus:
 * $\forall m \in \Z: \cos \left({x + 2 m \pi}\right) = \cos x$.

That is:
 * $\cos \left({2 m \pi}\right) = 1$

Following on directly from the above and, from Sine and Cosine are Periodic on Reals, we have that:
 * $\cos \left({x + \pi}\right) = -\cos x$

So:
 * $\cos \left({2 m + 1}\right) \pi = -1$

We note that $\forall n \in \Z$:
 * If $n$ is even, it is in the form $n = 2m$ for some $m \in \Z$, and so $\cos n \pi = 1$
 * If $n$ is odd, it is in the form $n = 2m+1$ for some $m \in \Z$, and so $\cos n \pi = -1$

Therefore:
 * $\cos n \pi = \left({-1}\right)^n$