Properties of Real Sine Function

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

Let $$\sin x$$ be the sine of $x$.

Then:
 * $$\sin x$$ is absolutely convergent for all $$x \in \mathbb{R}$$.


 * $$\sin 0 = 0$$.


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

Proof

 * Absolute convergence of $$\sin x$$:

We have that $$\sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$$.

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

But $$\sum_{n=0}^\infty \frac {\left|{x}\right|^{2n+1}}{\left({2n+1}\right)!}$$ is just the terms of $$\sum_{n=0}^\infty \frac {\left|{x}\right|^n}{n!}$$ for odd $$n$$.

Thus $$\sum_{n=0}^\infty \frac {\left|{x}\right|^{2n+1}}{\left({2n+1}\right)!} < \sum_{n=0}^\infty \frac {\left|{x}\right|^n}{n!}$$.

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 $$x \in \mathbb{R}$$.

The result follows from the Squeeze Theorem.


 * $$\sin 0 = 0$$:

Follows directly from the definition: $$\sin 0 = 0 - \frac {0^3} {3!} + \frac {0^5} {5!} - \cdots = 0$$.


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

From Sign of Odd Power, we have that $$\forall n \in \mathbb{N}: -\left({x^{2n+1}}\right) = \left({-x}\right)^{2n+1}$$.

The result follows from the definition.