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

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

But $$\sum_{n=0}^\infty \frac {\left|{x}\right|^{2n}}{\left({2n}\right)!}$$ is just the terms of $$\sum_{n=0}^\infty \frac {\left|{x}\right|^n}{n!}$$ for even $$n$$.

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

But $$\sum_{n=0}^\infty \frac {\left|{x}\right|^n}{n!} = \exp \left|{x}\right|$$ 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: $$\cos 0 = 1 - \frac {0^2} {2!} + \frac {0^4} {4!} - \cdots = 1$$.


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

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

The result follows from the definition.