Sine Function is Continuous

Theorem

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

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

Then:

$\sin x$ is continuous on $\R$.

Proof

Recall the definition of the sine function:

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

Thus $\sin x$ is expressible in the form of a power series.

From Sine Function is Absolutely Convergent, we have that the interval of convergence of $\sin x$ is the whole of $\R$.

From Power Series Differentiable on Interval of Convergence, it follows that $\sin x$ is continuous on the whole of $\R$.

$\blacksquare$