I'm trying to construct the trigonometric functions in a analytic-geometrically way and prove all the properties about angles. I don't want to use concepts of integral or derivatives as I would like to explain it in a formal way for a student who sees Calculus for the first time (using just supremum, limits and continuity).
I started defining the arclength measure of the function $\gamma:[-1,1]\rightarrow\mathbb{R}$ such that $\gamma(t)=\sqrt{1-t^2}$. Then, I prove the function $L_{(\cdot)}^1(\gamma)=\sup\left\{L_{(\cdot)}^1(\gamma,P)\middle|P\in\mathcal{P}([\cdot,1])\right\}$ is well-defined (the set is upper-bounded for all $x\in[-1,1)$) and increasing on $[-1,1)$. I defined $\arccos:[-1,1]\rightarrow[0,\pi]$ where $\pi:=L_{-1}^{1}(\gamma)$ like this:
$$\arccos(x)=\left\{\begin{array}{l} L_x^1(\gamma) & \text{if $-1\leq x<1$.} \\ 0 & \text{if $x=1$.}\end{array}\right.$$
After that, I prove that this function is continuous, $\arccos\left(0\right)=\dfrac{\pi}{2}$ and by the Intermediate Values Theorem, we have $\text{im}[\arccos]=[0,\pi]$.
Therefore, it makes sense to consider $\left.\cos\right|_{[0,\pi]}:[0,\pi]\rightarrow[-1,1]$ such that $\left.\cos\right|_{[0,\pi]}=\arccos^{-1}$. We can expand the definition of cosine to the interval $[0,2\cdot\pi)$ like this:
$$\begin{array}{lccc} \left.\cos\right|_{[0,2\cdot\pi)}: & [0,2\cdot\pi) & \rightarrow & [-1,1] \\ & \theta & \mapsto & \left.\cos\right|_{[0,2\cdot\pi)} \end{array},$$
where:
$$\left.\cos\right|_{[0,2\cdot\pi)}(\theta)=\left\{\begin{array}{ll} \left.\cos\right|_{[0,\pi]}(\theta) & \text{if $0\leq\theta\leq\pi$.} \\ \left.\cos\right|_{[0,\pi]}(2\cdot\pi-\theta) & \text{if $\pi<\theta<2\cdot\pi$.} \\ \end{array}\right.$$
It's very easy to prove (using the theorem of existence and uniqueness of the integer part of number) that for any $x\in\mathbb{R}$, there exists $\theta\in[0,2\cdot\pi)$ and $k\in\mathbb{Z}$ such that $x=\theta+2\cdot k\cdot\pi\in[2\cdot k\cdot\pi,2\cdot (k+1)\cdot\pi)$.
This last result allows us to extend the definition of the cosine function to all the real numbers:
$$\begin{array}{lccc} \cos: & \mathbb{R} & \rightarrow & [-1,1] \\ & \theta & \mapsto & \cos(\theta)=\left.\cos\right|_{[0,2\cdot\pi)}(\theta-2\cdot k\cdot\pi) \end{array},$$
where $k\in\mathbb{Z}$ is the unique integer such that $\theta\in[2\cdot k\cdot\pi,2\cdot(k+1)\cdot\pi)$.
The definition of the sine function goes like this:
$$\begin{array}{lccc} \sin: & \mathbb{R} & \rightarrow & [-1,1] \\ & \theta & \mapsto & \sin(\theta) \end{array},$$
where:
$$\sin(\theta)=\left\{\begin{array}{lccc} \sqrt{1-\cos^2(\theta)} & \text{if $\theta\in[2\cdot k\cdot\pi,(2\cdot k+1)\cdot\pi]$.} \\ -\sqrt{1-\cos^2(\theta)} & \text{if $\theta\in((2\cdot k+1)\cdot\pi,2\cdot (k+1)\cdot\pi]$.} \end{array},k\in\mathbb{Z}.\right.$$
Using the monotony of the arccosine function is very easy to calculate the monotony of the cosine and sine function. By definition, it's trivial to deduce that these functions are even and odd respectively, and periodic with period: $2\cdot\pi$.
These are the basis I'm using in order to prove the trigonometric functions' properties. Once I prove that $\cos(\alpha+\beta)=\cos(\alpha)\cdot\cos(\beta)\mp\sin(\alpha)\cdot\sin(\beta)$ the rest is very easy.
The proof I like the most and doesn't depend almost at all of the drawing is the one I'm presenting here (I'll bold the part which it's not proved with the tools I showed before):
First, we prove that $(x,y)\in S^1$ if and only if there exists $\alpha\in\mathbb{R}$ (which it'll be unique if we force to be $\alpha\in[0,2\cdot\pi)$) such that $x=\cos(\alpha)$ and $y=\sin(\alpha)$. This part is easy and I'll skip it.
Let be $P=(\cos(\alpha),$ $\sin(\alpha)),$ $Q=(1,0),$ $R=(\cos(-\beta),\sin(-\beta)),$ $S=(\cos(\alpha+\beta),\sin(\alpha+\beta))\in S^1$. Then:
$$d^2(P,R)=2-2\cdot\cos(\alpha)\cdot\cos(-\beta)-2\cdot\sin(\alpha)\cdot\sin(-\beta)=2-2\cdot\cos(\alpha)\cdot\cos(\beta)+2\cdot\sin(\alpha)\cdot\sin(\beta).$$
$$d^2(Q,S)=2-2\cdot\cos(\alpha+\beta).$$
We claim that $d(P,R)=d(Q,S)$. Making the two quantities equal, we get the desired result.
I know that looking on a graph is "obvious" that those distances are equal as the points shared the same arclength on the unit-circunference, but I need to prove it without that and using the definitions I provided it before. How could you do it?
Thank all of you in advance.
Suggestion Let be $P=(x,y),Q=(x',y')\in S^1$ with $x\leq x'$. We define the measure of the angle formed by $P$, $O:=(0,0)$ and $Q$ like this:
$$\angle POQ=\left\{\begin{array}{ll} 0 & \text{if $P=Q$.} \\ L_x^{x'}(\gamma) & \text{if $P\neq Q,y\cdot y'\geq 0$.} \\ \min\{L_x^1(\gamma)+L_{x'}^1(\gamma),2\cdot\pi-L_x^1(\gamma)-L_{x'}^1(\gamma)\} & \text{if $y\cdot y'<0$.} \end{array}\right.$$
We know that for any $P=(x,y),Q=(x',y')\in S^1$ with $x\leq x'$, there're unique $\alpha,\beta\in[0,2\cdot\pi)$ such that $P=(\cos(\alpha),\sin(\alpha))$ and $Q=(\cos(\beta),\sin(\beta))$. We can prove first that:
$$\angle POQ=\left\{\begin{array}{ll} 0 & \text{if $P=Q$.} \\ \alpha-\beta & \text{if $\alpha,\beta\in[0,\pi]$.} \\ \beta-\alpha & \text{if $\alpha,\beta\in(\pi,2\cdot\pi)$.} \\ \min\{\alpha+2\cdot\pi-\beta,\beta-\alpha\} & \text{if $\alpha\in[0,\pi],\beta\in(\pi,2\cdot\pi)$.} \\ \min\{\beta+2\cdot\pi-\alpha,\alpha-\beta\} & \text{if $\alpha\in(\pi,2\cdot\pi),\beta\in[0,\pi]$.} \\ \end{array}\right.$$ Finally, we should demonstrate that if $P,Q,R,S\in S^1$ are such that $\angle POQ=\angle ROS$, then $d(P,Q)=d(R,S)$.
