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?
A rough sketch for (a) could be
For part (b), you have to determine the period numerically in general. Of course the answer is $2\pi$, but proving this depends on what your definition of $\pi$ is. A popular definition is that $\pi$ is simply twice the smallest positive $\theta$ such that $\cos(\theta)=0$, in which case period $2\pi$ is just a tautology.
GH Hardy sketches a proof as follows (Hardy, Pure Mathematics, Section 224)
Use the sequences, which are absolutely convergent for all real $x$ to demonstrate the addition formulae (e.g. $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$)
He then says “The property of periodicity is a little more troublesome.”
Prove from the series for cosine that it changes sign just once in the interval (0,2) [I think this is really the key insight in the proof]
Call the zero $\pi/2$ and show that $\sin(\pi/2)=1$, $\cos(\pi)=-1$, $\sin(\pi)=0$
Use the addition formulae to establish periodicity.
Hardy cites a full proof in Whittaker and Watson’s Modern Analysis, Appendix A.
Some ideas can be found in Rudin’s Real and Complex Analysis, first chapter about the exponential function.
The exponential is defined to be
$$e^t=1+t+\frac{t^2}{2!}+…$$
Using the Taylor representation for $\sin, \cos$ it is easy to see that $e^{it}=\cos t+i\sin t$, so to prove that $\cos, \sin$ are periodic it is enough to prove that $t\mapsto e^{it}$ is periodic, and for this is enough to prove that there exists $t_0>0$ with $e^{it_0}=1$. Such a $t_0$ is $2\pi$ (since $\sin 2\pi=0$ and $\cos 2\pi=1$) and we are done.
If we don’t know anything about the existence of $\pi$ and the values of $\sin,\cos$ in $2\pi$, then we can proceed as follows:
$\cos 0=1$; this is obvious from the series. $\cos 2< 1-\frac{4}{2}+\frac{16}{24}=-\frac{1}{3}$ (the inequality is easy to prove). From the series we can see that $\cos$ is a continuous function and therefore there exists a smallest $t_0>0$ with $\cos t_0=0$. Define $\pi=2t_0$. $|e^{it}|=1$ for every $t$ since $e^{-it}=\overline{e^{it}}$ and $e^{a+b}=e^a\cdot e^b$.
Then $\sin t_0 \in \{-1,1\}$ and since $\sin 't =\cos t>0$ on $(0,t_0)$ we deduce that $\sin t_0=1$. Therefore $e^{i \cdot \pi/2}=i$, and this means $e^{2\pi i}=1$.
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.