TITLE: How to prove periodicity of $\sin(x)$ or $\cos(x)$ starting from the Taylor series expansion?
QUESTION [22 upvotes]: With the Taylor series representation of $\sin$ or $\cos$ as a starting point (and assuming no other knowledge about those functions), how can one:
a. prove they are periodic?
b. find the value of the period?

REPLY [8 votes]: Start from the end of Beni Bogosel's first paragraph: We must find a positive $t$ such that $e^{it}=1$. Go through the usual proof that $$ \sum_{k=0}^{\infty} \frac{z^k}{k!} = \lim_{n\to\infty} \left(1+ \frac{z}{n}\right)^n$$
So $e^{it} = \displaystyle\lim_{n\to\infty} \left( 1+\frac{it}{n} \right)^n$. Since $1\leq \displaystyle\left(1+\frac{t^2}{n^2}\right)^n \leq \left(e^{t^2}\right)^{1/n}\to 1$, $|e^{it}|=1$. 
Define $\rm si(x), co(x), ta(x) $ to be the usual trigonometric functions defined by points on the unit circle. Draw a picture to convince yourself that $\rm si(x) \leq x \leq ta(x) $ for small positive $x$, and $\rm ta(x) \leq x \leq si(x)$ for small negative $x$. Argue via squeeze theorem that $\rm ta(x) \sim x $ as $x\to 0$. 
Now note $\arg(e^{it}) = \displaystyle\lim_{n\to\infty} \rm n\cdot ta^{-1}(t/n) = t$ . Thus, $e^{it}$ is the point on the unit circle whose argument is $t$, so has period equal to the circumference of the unit circle, $2\pi$. 
Note that we don't have to depart from our geometric definition of $\pi$, instead it arises naturally. As freebies, we get that $\sin x = \rm si(x)$ and $\cos x = \rm co(x)$, i.e these power series definitions agree with the historical definition with triangles.