Properties of Real Cosine Function

Theorem
The cosine function satisfies the following properties:.


 * $(1): \quad \cos x$ is continuous on $\R$
 * $(2): \quad \cos x$ is absolutely convergent for all $x \in \R$
 * $(3): \quad \cos n\pi = (-1)^n$ for all $n \in \Z$
 * $(4): \quad \cos x$ is even, that is, $\cos \left({-x}\right) = \cos x$ for all $x \in \R$

Proof
Recall the definition:


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

$(1): \quad$ Continuity of $\cos x$:

$(2): \quad$ Absolute convergence of $\cos x$:

For:
 * $\displaystyle \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!}$

to be absolutely convergent, we want:
 * $\displaystyle \sum_{n=0}^\infty \left|{\left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!}}\right| = \sum_{n=0}^\infty \frac {\left|{x}\right|^{2n}}{\left({2n}\right)!}$

to be convergent.

But
 * $\displaystyle \sum_{n=0}^\infty \frac {\left|{x}\right|^{2n}}{\left({2n}\right)!}$

is just the terms of
 * $\displaystyle \sum_{n=0}^\infty \frac {\left|{x}\right|^n}{n!}$

for even $n$.

Thus:
 * $\displaystyle \sum_{n=0}^\infty \frac {\left|{x}\right|^{2n}}{\left({2n}\right)!} < \sum_{n=0}^\infty \frac {\left|{x}\right|^n}{n!}$

But:
 * $\displaystyle \sum_{n=0}^\infty \frac {\left|{x}\right|^n}{n!} = \exp \left|{x}\right|$

from the Taylor Series Expansion for Exponential Function of $\left|{x}\right|$, which converges for all $x \in \R$.

The result follows from the Squeeze Theorem.

$(3): \quad \cos n \pi = (-1)^n$:

Follows directly from the definition:
 * $\displaystyle \cos 0 = 1 - \frac {0^2} {2!} + \frac {0^4} {4!} - \cdots = 1$

This fact is sufficient for the derivation in Sine and Cosine are Periodic on Reals.

Therefore, by the corollary that $\cos(x+\pi) = -\cos x$ and induction we have $\cos n\pi = (-1)^n$.

$(4): \quad \cos \left({-x}\right) = \cos x$:

From Even Powers are Positive, we have that $\forall n \in \N: x^{2n} = \left({-x}\right)^{2n}$.

The result follows from the definition.