Properties of Real Cosine Function

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

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

Then:
 * $\cos x$ is continuous on $\R$.


 * $\cos x$ is absolutely convergent for all $x \in \R$.


 * $\cos 0 = 1$.


 * $\cos \left({-x}\right) = \cos x$.

Proof

 * Continuity of $\cos x$:


 * Absolute convergence of $\cos x$:

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

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.


 * $\cos 0 = 1$:

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


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

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

The result follows from the definition.