# Cosine of Multiple of Pi

From ProofWiki

## 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:

\(\displaystyle \) | \(\displaystyle \forall m \in \Z:\) | \(\displaystyle \) | \(\displaystyle \cos \left({2 m \pi}\right)\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \forall m \in \Z:\) | \(\displaystyle \) | \(\displaystyle \cos \left({\left({2 m + 1}\right) \pi}\right)\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle -1\) | \(\displaystyle \) | \(\displaystyle \) |

## Proof

Recall the definition of the cosine function:

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

From Cosine of Zero is One, we have that:

- $\displaystyle \cos 0 = 1$

This takes care of the case $n = 0$.

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

- $\forall m \in \Z: \cos \left({x + 2 m \pi}\right) = \cos x$.

The above observations combine to establish that:

- $\cos \left({2 m \pi}\right) = 1$

Again from Sine and Cosine are Periodic on Reals, we have that:

- $\cos \left({x + \pi}\right) = -\cos x$

Combining this with the above reasoning, it follows that:

- $\forall m \in \Z: \cos \left({2 m + 1}\right) \pi = -1$

Note that $\forall n \in \Z$:

- If $n$ is even, it is of the form $n = 2m$ for some $m \in \Z$, and so $\cos n \pi = 1$
- If $n$ is odd, it is of the form $n = 2m+1$ for some $m \in \Z$, and so $\cos n \pi = -1$

This leads to the desired conclusion:

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

$\blacksquare$