Cosine of Integer Multiple of Pi

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Theorem

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

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


Then:

$\forall n \in \Z: \cos n \pi = \paren {-1}^n$

or:

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


Proof

Recall the definition of the cosine function:

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


From Cosine of Zero is One, we have that:

$\cos 0 = 1$

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


From Sine and Cosine are Periodic on Reals:

$\cos \paren {x + 2 \pi}= \cos x$

and thus:

$\forall m \in \Z: \cos \paren {x + 2 m \pi} = \cos x$.

The above observations combine to establish that:

$\cos 2 m \pi = 1$


Again from Sine and Cosine are Periodic on Reals:

$\cos \paren {x + \pi} = -\cos x$

Combining this with the above reasoning, it follows that:

$\forall m \in \Z: \cos \paren {2 m + 1} \pi = -1$


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 $n = 2 m + 1$ for some $m \in \Z$, and so $\cos n \pi = -1$


This leads to the desired conclusion:

$\cos n \pi = \paren {-1}^n$

$\blacksquare$


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