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:
\(\ds \forall m \in \Z: \, \) | \(\ds \cos 2 m \pi\) | \(=\) | \(\ds 1\) | |||||||||||
\(\ds \forall m \in \Z: \, \) | \(\ds \cos \paren {2 m + 1} \pi\) | \(=\) | \(\ds -1\) |
Proof
Recall the definition of the cosine function:
- $\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$
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:
- $\cos n \pi = \paren {-1}^n$
$\blacksquare$