Proving properties of sine and cosine from their Taylor series definitions I would really like to know whether someone has a rigorous way to prove that the function $t \mapsto (\cos t, \sin t )$ gives a parametrization of the unit circle in the plane, taking as a starting point the definition of the circle functions by means of their Taylor series. It would be nice to also have other proofs of their most basic properties. How does one, for example, prove that these functions are periodic? How does one properly define $\pi$ in this framework? I get the feeling that in uni these facts are seen as 'elementary', but in high school we obviously never proved it. 
My own idea is to see these functions as a solution to a differential equation in $\mathbb{R}^2$. If you draw the corresponding velocity vectors in the plane, it is sort of picturewise obvious that if you start at $(1,0)$ you will go around the origin at constant speed in a circle... But I do not know how to prove these facts.
 A: Let's see what is involved in the parametrization of unit circle via circular functions. First is the obvious relation $$\cos^{2}t+\sin^{2}t=1\tag{1}$$ and further that these functions are continuous/differentiable with values $\cos 0=1,\sin 0=0$. Some idea of $\pi$ is needed so that we can prove these functions are periodic with period $2\pi$.
All of this can be easily derived using the Taylor series for these functions. From the definitions $$\cos t=1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\dots \tag{2}$$ and $$\sin t =t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\dots\tag{3}$$ we can easily see the initial values at $t=0$ and their derivatives $$(\sin t) '=\cos t, (\cos t)' =-\sin t\tag{4}$$ and if we have $g(t) =\cos^{2}t+\sin^{2}t$ then we get $g'(t) =0$ so that $g(t) =g(0)=1$ and equation $(1)$ is proved.
To introduce $\pi$ we need a bit more work. It can be shown that $\cos t$ changes sign in $[0,2]$ and hence vanishes somewhere in this interval. Moreover the derivative $—\sin t$ maintains constant sign in $(0,2)$ hence it follows that there is a unique number $\xi\in(0,2)$ such that $\cos \xi=0$ and we define $\pi=2\xi$. It is then easily proved that $\sin(\pi/2)=1$.
Proving periodicity is bit tricky and one way to do it is to establish addition formulas. This can be done by noting that both $\cos t, \sin t$ are solutions to $f''(t) +f(t) =0$ and general solution to this equation  is $f(t) = f(0)\cos t+f'(0)\sin t$. To prove the general solution consider the function $$h(t) =f(t) - f(0)\cos t-f'(0)\sin t$$ then $h''(t) +h(t) =0, h(0)=h'(0)=0$. And then we consider $\phi(t) = \{h(t)\} ^{2}+\{h'(t)\}^{2}$. Clearly $\phi'(t) =0$ and hence $\phi(t) =\phi(0)=0$ and hence $h(t) =h'(t) =0$. This gives us $f(t) =f(0)\cos t+f'(0)\sin t$. 
Since $\cos (a+t) $ also satisfies the equation $f''+f=0$, it follows from the above that $\cos (a+t) =\cos a \cos t-\sin a \sin t $ and similarly addition formula for $\sin t $ can be established. Using these addition formulas we can show that $\cos(t+2\pi)=\cos t, \sin(t+2\pi)=\sin t$. 
A: Start out by defining $\sin$ and $\cos$ to be the unique solutions to the differential equation $y'' = -y$ with the boundary conditions $y(0) = 0, y'(0) = 1$ and $y(0) = 1, y'(0) = 0$, respectively. (It's not difficult to show that such functions exist and are unique, and one can verify that the usual Taylor series satisfy the equation.) Standard uniform convergence arguments then show that $\sin$ and $\cos$ are smooth on $\mathbb{R}$. Our goal is then to show that the map $f:[0, 2\pi] \to \mathbb{R}^2$ given by $f(t) = (\cos t, \sin t)$ parametrizes the unit circle $S^1$ (as a closed curve). 
First, consider the function $g(t) = \cos^2 t + \sin^2 t$. The derivative $\varphi = \frac{d}{dt} \sin t$ also satisfies $\varphi'' = -\varphi$ with $\varphi(0) = 1$ and 
$$\varphi'(0) = -\frac{d^2}{dt^2} \sin t\big\vert_{t=0} = -\sin 0 = 0$$ 
by the definition above. Hence $\varphi(t) = \cos t$; that is, $\frac{d}{dt} \sin t = \cos t$. Applying the same argument to $\cos t$ gives $\frac{d}{dt} \cos t= - \sin t$. Hence
\begin{align*}
g'(t) = 2\cos t \sin t - 2\sin t \cos t = 0.
\end{align*}
Since $g(0) = 1$ by the definition above, we have $g\equiv 1$ everywhere.
Thus we've shown that the image of $f$ lies on $S^1$. Suppose $f(t) = f(t')$ with $t' \not = t$. Then $\sin t = \sin t'$ and $\cos t = \cos t'$, so the smooth function $\varphi(t) = \sin (x + t) - \sin (x + t')$ satisfies $\varphi'' = -\varphi$ with $\varphi(0), \varphi'(0) = 0$. The solution of that second-order equation with prescribed boundary conditions is unique, so $\varphi = 0$; that is, $\varphi$ has period $t' - t$. Such a $t' \not = t$ must exist, since $|f'| = 1$ everywhere. It follows that there exists some constant $N$ such that the map $f:[0, N] \to S^1$; that is, $f$ is onto and injective except that $f(0) = f(N)$.
The only thing left to show is that $N = 2\pi$. That's a bit trickier; it depends on what definition you want to take for $\pi$. Continuing with the calculus theme, let's define to be $\pi$ to be half the circumference of the unit circle. Then the usual formula for arc length gives $N = 2\pi$, since $f'(t) = (-\sin t, \cos t)$ has $|f'(t)| = 1$ everywhere.
(We also haven't discussed the angle sum formulas. They can be derived by recasting them in terms of the exponential function, defined to be the unique solution to the differential equation $y' = y$ with boundary condition $y(0) = 1$. Since $y(t + t_0)/y(t_0)$ is also a solution for any $t_0$, we must $y(t + t') = y(t)y(t')$ for all $t, t'$. We could also have worked from $\exp t$ at the beginning, but that doesn't seem to be what you were going for.)
A: Apostol's calculus book has (I believe) a nice approach to sine/cosine: they lay out a few "axioms" that, taken together, define sine and cosine on all points of the form $\frac{n\pi}{2^k}$, where $n$ and $k$ are integers. 
Then they show that as a function on this set, sine and cosine are continuous. 
Then they show that this set is dense in $\Bbb R$, and hence the sine and cosine functions so defined actually have a unique continuous extension to the whole real line. 
Since one of the "axioms" is the $\sin^2 t + \cos^2 t = 1$ for all $t$, it's then pretty easy to show that for every point on the circle, there's a $t$ with $(\cos t, \sin t)$ being that point, via the intermediate value theorem. 
THat's not the approach you were looking for, I know, but it IS a nice rigorous way to get sines and cosines, which may be what you wanted, and it doesn't require convergence of power series and other such things. 
