Analytic definition implies geometric definition of trigonometric functions It is well known how we can arrive to the power series definition of trigonometric functions starting from their definition in terms of the unit circle. I'm trying to do the converse, i.e. start from the definition of trigonometric functions by their power series and prove that we can parametrize the unit circle with them and that an angle of $\theta$ radians subtends an arc of length $\theta$ (in the unit circle, of course). Here is what I have done so far.
Define $\displaystyle \pi :=2\int \limits _{-1}^1\sqrt{1-x^2}\,dx$ the area of the unit circle, and define$$C(x):=\sum \limits _{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$$and$$S(x):=\sum \limits _{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}.$$The ratio test shows that this functions have infinite convergence radius, so they are in $\mathcal{C}^\infty (\mathbb{R})$, and it is clear that $C$ is an even function and $S$ is an odd function. Differentiating gives $\dfrac{d}{dx}C(x)=-S(x)$ and $\dfrac{d}{dx}S(x)=C(x)$.
Now we prove that they verify the addition formulae for $\sin$ and $\cos$, i.e. that $S(x+y)=S(x)C(y)+C(x)S(y)$ and $C(x+y)=C(x)C(y)-S(x)S(y)$ for all $x,y\in \mathbb{R}$.
Fix $y\in \mathbb{R}$ and define$$F(x):=S(x+y)-S(x)C(y)-C(x)S(y).$$Then we have that$$F'(x)=C(x+y)-C(x)C(y)+S(x)S(y).$$Now observe that $F''(x)=-F(x)$, hence$$2F'(x)[F(x)+F''(x)]=0$$and therefore$$\frac{d}{dx}\left [(F(x))^2+(F'(x))^2\right ]=0$$for all $x\in \mathbb{R}$, hence $F(x)=0$ for all $x\in \mathbb{R}$ and $F'(x)=0$ for all $x\in \mathbb{R}$, this proves that $S(x+y)=S(x)C(y)+C(x)S(y)$ and $C(x+y)=C(x)C(y)-S(x)S(y)$ for all $x,y\in \mathbb{R}$. Using that $C$ is even and $S$ is odd we find that $S(x\pm y)=S(x)C(y)\pm C(x)S(y)$ and that $C(x\pm y)=C(x)C(y)\mp S(x)S(y)$, this also gives the Pythagorean Identity$$(C(x))^2+(S(x))^2=C(x-x)=C(0)=1$$which implies that $C(x),S(x)\in [-1,1]$ for all $x\in \mathbb{R}$.
Now I want to prove that we can parametrize the unit circle using $C$ and $S$. The Pythagorean Identity implies that $(C(x),S(x))$ is in the unit circle for all $x\in \mathbb{R}$, but I don't know how to prove that every point in the unit circle has the form $(C(x_0),S(x_0))$ for some $x_0\in \mathbb{R}$.
We now prove that the perimeter of the unit circle is $2\pi$. By symmetry, it is enough to prove that the perimeter of the upper half of the unit circle is $\pi$. This curve can be parametrized as $\alpha (t):=\left (t,\sqrt{1-t^2}\right )$ for $t\in [-1,1]$, hence $\alpha '(t)=\left (1,-\dfrac{t}{\sqrt{1-t^2}}\right )$, and therefore $\|\alpha '(t)\|=\dfrac{1}{\sqrt{1-t^2}}$. To prove that the perimeter of the upper half of the unit circle is $\pi$, we have to prove that$$\int \limits _{-1}^1\frac{1}{\sqrt{1-t^2}}\,dt=2\int \limits _{-1}^1\sqrt{1-t^2}\,dt$$i.e. we have to prove that$$\int \limits _{-1}^1\frac{1}{\sqrt{1-t^2}}-2\sqrt{1-t^2}\,dt=0.$$But$$\frac{d}{dt}\left (-t\sqrt{1-t^2}\right )=\frac{1}{\sqrt{1-t^2}}-2\sqrt{1-t^2}$$and therefore$$\int \limits _{-1}^1\frac{1}{\sqrt{1-t^2}}-2\sqrt{1-t^2}\,dt=\left .-t\sqrt{1-t^2}\right |_{-1}^1=0$$as wanted.
Now assuming that we can parametrize the unit circle using $C$ and $S$, then we can show with line integration that the length of the portion of the unit circle that goes from $(1,0)$ to $(C(x),S(x))$ in counterclockwise sense is exactly $x$. In particular, this would prove that $C$ and $S$ are periodic with period $2\pi$ and that $C(x)=\cos x$ and $S(x)=\sin x$ for $x\in [0,2\pi ]$, and therefore over all of $\mathbb{R}$.
So my question here is: How can we prove that the functions $C$ and $S$ defined as the power series of (what then are going to be) the cosine and the sine functions can be used to parametrize the unit circle?
 A: Let us first show that the function $C$ vanishes at some positive number. Suppose on the contrary that $C$ is nowhere vanishing on $(0,\infty)$. By continuity of $C$ (intermediate value theorem), it must maintain the same sign on $(0,\infty)$; but now we clearly have $C(0)=1$ so by continuity there is some open interval around the origin on which $C$ is strictly positive. Hence, the sign of $C$ on $(0,\infty)$ is positive. Fix a number $a>0$. Then, for any $x\in (0,\infty)$
\begin{align}
C(x)&=C(a)+\int_a^xC'(t)\,dt\\
&=C(a)-\int_a^xS(t)\,dt
\end{align}
Now, we have $S'=C$ which is positive on $(0,\infty)$, and $S(0)=0$, which means $S$ is strictly increasing on $[0,\infty)$, and hence we get the inequality
\begin{align}
C(x)&\leq C(a)-S(a)(x-a).
\end{align}
Since $S$ is strictly increasing and $S(0)=0$ and $a>0$, it follows $S(a)>0$, so that if $x$ is large enough, the RHS of the inequality will be negative, and hence $C(x)<0$, which contradicts our assumption. Therefore, $C$ has to vanish at some point of $(0,\infty)$. Let $\beta\in (0,\infty)$ be the smallest positive number such that $C(\beta)=0$ (why does such a smallest number exist). The number $\beta$ has the following properties:

*

*$S'=C>0$ on $(0,\beta)$, so $S$ is strictly increasing here, and $S(0)=0$, so $S(\beta)>0$. Since $C(\beta)=0$ and $C^2+S^2=1$, it follows $S(\beta)=1$. Thus, $S$ increases strictly on $[0,\beta]$ from $0$ to $1$.

*Similarly $C$ decreases strictly on $[0,\beta]$ from $1$ to $0$.

So, for any $(a,b)\in \Bbb{R}^2$ with $a^2+b^2=1$ and $a,b\geq 0$ (i.e the first quadrant of the unit circle), we can find (since $C:[0,\beta]\to [0,1]$ is bijective) some $x_0\in [0,\beta]$ such that $a=C(x_0)$. Then, $b=\sqrt{1-a^2}=\sqrt{1-C(x_0)^2}=S(x_0)$ (since $S\geq 0$ on the interval $[0,\beta]$).
Lastly, by the addition formulae you can verify $(-b,a)=(C(x_0+\beta),S(x_0+\beta))$, and $(-a,-b)=(C(x_0+2\beta),S(x_0+2\beta))$ and $(b,-a)=(C(x_0+3\beta),S(x_0+3\beta))$. This takes into account what happens in the other quadrants.

To really finish off, we can prove that $\beta$ as defined here (smallest positive root of $C$) equals your $\frac{\pi}{2}$. To do this, we have the restricted mapping $S:[0,\beta]\to [0,1]$ is bijective and strictly increasing so
\begin{align}
\beta&=\int_0^{\beta}dx\\
&=\int_{S^{-1}(0)}^{S^{-1}(1)}\,dx\\
&=\int_0^1(S^{-1})'(t)\,dt\tag{change of variables}\\
&=\int_0^1\frac{1}{S'(S^{-1}(t))}\,dt\tag{IFT}\\
&=\int_0^1\frac{1}{C(S^{-1}(t))}\,dt\\
&=\int_0^1\frac{1}{\sqrt{1-t^2}}\,dt,
\end{align}
where the last step used the fact that $C\geq 0$ on $[0,\beta]$ and $C^2+S^2=1$. So, with what you established, this equals $\frac{\pi}{2}$.
A: One way to see this is the following: The power series formulas imply that $(x(t),y(t)) = (\cos t, \sin t)$ satisfies the equation
$$
(x',y') = (-y,x)\text{ and }(x(0),y(0)) = (1,0).
$$
Since $x^2 + y^2 = (x')^2 + (y')^2 = 1$, this shows that the map $t \mapsto (x(t),y(t))$ is a unit speed parameterization of the unit circle.
Here is a explanation of how to avoid using power series to define the sine and cosine functions.
A: Summarizing the useful identities proven in your question:

*

*$C(0) = 1$

*$S(0) = 0$

*$C$ is even: $C(-x) = C(x)$

*$S$ is odd: $S(-x) = -S(x)$

*$C'(x) = S(x)$

*$S'(x) = -C(x)$

*$C(x+y) = C(x)C(y) - S(x)S(y)$

*$S(x+y) = C(x)S(y) + S(x)C(y)$

*$C(x)^2 + S(x)^2 = 1$
(Quarter-)Periodicity of $C$ and $S$
Let $\eta$ be the smallest positive number with $C(\eta) = 0$.  (Solving numerically gives $\eta \approx 1.57$.)
Then from $C(x)^2 + S(x)^2 = 1$, we get $S(x) = \pm 1$.  But $S'(x) = C(x)$, which is positive on $[0, \eta)$.  (Because $C(0) = 1 > 0$, and  $C$ can't change sign on that interval without violating the definition of $\eta$ as the smallest positive solution to $C(\eta) = 0$.)  Since $S$ starts at 0 and is increasing, then we must have $S(\eta) = 1$, not $-1$.
From the above equations for $C(x+y)$ and $S(x+y)$,

*

*$C(x+\eta) = C(x)C(\eta) - S(x)S(\eta) = C(x)(0) - S(x)(1) = -S(x)$

*$S(x+\eta) = C(x)S(\eta) + S(x)C(\eta) = C(x)(1) + S(x)(0) = C(x)$

*$C(x+2\eta) = -S(x+\eta) = -C(x)$

*$S(x+2\eta) = C(x+\eta) = -S(x)$

*$C(x+3\eta) = -S(x+2\eta) = S(x)$

*$S(x+3\eta) = C(x+2\eta) = -C(x)$

*$C(x+4\eta) = -S(x+3\eta) = C(x)$

*$S(x+4\eta) = C(x+3\eta) = S(x)$
From the last two, we see that $C$ and $S$ are periodic with a period of $\tau := 4\eta \approx 6.28$.
And by plugging in $x = 0$, we get:

*

*$C(\eta) = -S(0) = 0$

*$C(2\eta) = -C(0) = -1$

*$C(3\eta) = S(0) = 0$

*$C(4\eta) = C(0) = 1$

*$S(\eta) = C(0) = 1$

*$S(2\eta) = -S(0) = 0$

*$S(3\eta) = -C(0) = -1$

*$S(4\eta) = S(0) = 0$
By the Intermediate Value Theorem, all numbers in the interval $[-1, 1]$ are included in the range of $C$ and $S$.
Also, $S'(x) = C(x) > 0$ on $|x| < \eta'$, and $C'(x) = -S(x) < 0$ on $[0, 2\eta]$, so the functions are strictly monotonic and thus one-to-one on these intervals.
So we can let $C^{-1}: [-1, 1] \rightarrow [0, 2\eta]$ and $S^{-1}: [-1, 1] \rightarrow [-\eta, \eta]$ be functions satisfying $C(C^{-1}(x)) = x$ and $S(S^{-1}(x)) = x$.
Parametrizing the circle
If we define $x = C(t)$ and $y = S(t)$, then $x^2 + y^2 = C(t)^2 + S(t)^2 = 1$, which conveniently happens to be equation of the unit circle.
Now, we need to show that all points on the curve $x^2 + y^2 = 1$ can be mapped to a $t$.  Let's consider four cases:

*

*$t \in [0, \eta] \rightarrow x \ge 0, y \ge 0$ (Quadrant 1)

*

*Let $t = C^{-1}(x) = S^{-1}(y)$.



*$t \in [\eta, 2\eta] \rightarrow x \le 0, y \ge 0$ (Quadrant 2)

*

*Let $t' = t - \eta \rightarrow (x', y') = (y, -x)$



*$t \in [2\eta, 3\eta] \rightarrow x \le 0, y \le 0$ (Quadrant 3)

*

*Let $t' = t - 2\eta \rightarrow (x', y') = (-x, -y)$



*$t \in [3\eta, 4\eta] \rightarrow x \ge 0, y \le 0$ (Quadrant 4)

*

*Let $t' = t - 2\eta \rightarrow (x', y') = (-y, x)$
Quadrant 1 points are in the intersection of $C^{-1}$ and $S^{-1}$'s domains and can be handled directly, while points in the other three quadrants can be easily mapped to “reference” points in Quadrant 1.  The entire circle can thus be covered by $t \in [0, 4\eta]$.
Arc length
A curve parametrized as $(x(t), y(t))$ has a length of
$$L = \int_{\min(t)}^{\max(t)} \sqrt{x'(t)^2 + y'(t)^2} dt$$
On our unit circle with $t \in [0, \theta]$.
$$L = \int_0^\theta \sqrt{C'(t)^2 + S'(t)^2} dt = \int_0^\theta \sqrt{(-S(t))^2 + C(t)^2} dt = \int_0^\theta 1 dt = \theta$$
Area
As stated in the equation, the area of the unit circle is:
$$A = 2 \int_{-1}^1 \sqrt{1 - x^2} dx$$
Let $t = C^{-1}(x)$.  Then $x = C(t)$ and $dx = C'(t) = -S(t) dt$, so:
$$A = 2 \int_{2\eta}^{0} \sqrt{1 - C(t)^2} (-S(t)) dt = 2 \int_{0}^{2\eta} S(t)^2 dt$$
From the summation formula for $C$, we get $C(2t) = C(t)^2 - S(t)^2 = 1 - 2S(t)^2$, from which $S(t)^2 = \frac{1 - C(2t)}{2}$.  So,
$$A = 2 \int_{0}^{2\eta} \frac{1 - C(2t)}{2} dt = \int_{0}^{2\eta} (1 - C(2t)) dt = (t - \frac{1}{2}S(2t))|_{t=0}^{t=2\eta} = 2\eta$$
Thus, the number that you call $\pi$ is what I've been calling $2\eta$ or $\frac{\tau}{2}$.
