I'm trying to construct the trigonometric functions in a analytic-geometrically way without using the concepts of integral and derivative. So I came up with the idea of using the arclength measure, but without mentioning it.
I started introducing the equation of the unit circumference:
$$x^2+y^2=1,$$
and the function $\gamma:[-1,1]\rightarrow\mathbb{R}$ defined as $\gamma(t)=\sqrt{1-t^2}$. Then, I introduce the concept of a partition of an interval and defining the length of $\left.\gamma\right|_{[a,b]}$ where $[a,b]\subset[-1,1]$ is a non-degenerate interval with respect to $P\in\mathcal{P}([a,b])$.
After that, I proved that for any $x\in[-1,1)$, the following set is upper-bounded:
$$\left\{L_a^b(\gamma,P)\middle|P\in\mathcal{P}([a,b])\right\}.$$
Let denote $L_a^b(\gamma)=\sup\left\{L_a^b(\gamma,P)\middle|P\in\mathcal{P}([a,b])\right\}$ and $\pi=L_{-1}^{1}(\gamma)$.
Therefore, let define $\arccos:[-1,1]\rightarrow[0,\pi]$ such that:
$$\arccos(x)=\left\{\begin{array}{ll} L_x^1(\gamma) & \text{if $-1\leq x<1$.} \\ 0 & \text{if $x=1$.} \end{array}\right.$$
Notice that the function is well-defined as I've proved before that $L_a^b(\gamma)=L_a^c(\gamma)+L_c^b(\gamma)$ for any $c\in(a,b)$ which allowed us to demonstrate that the function $L_{(\cdot)}^1(\gamma)$ is decreasing. Because of that, I could say that the arccosine function is injective and therefore, I can invert it. However, I cannot define the inverse of the arccosine as the cosine because I don't know yet that the image of the arccosine function is the entire interval $[0,\pi]$ for sure. How could I do it?
Solution. We'll prove that the arccosine function is continuous. Given $a\in[0,1)$ and $b\in[-1,0)$, let be $x\in(a,1)$ and $y\in(b,0)$. Let consider $P=\left\{t_i\right\}_{i=0}^{n}\in\mathcal{P}\left(\left[a,x\right]\right)$ and $Q=\left\{s_j\right\}_{j=0}^{m}\in\mathcal{P}\left(\left[b,y\right]\right)$. As $\gamma$ in continuous on $[-1,1]$, increasing on $[-1,0]$ and decreasing on $[0,1]$, then:
$$\begin{array}{l} L_a^x(\gamma,P)\leq\sum_{i=1}^{n}{(|t_{i-1}-t_i|+|\gamma(t_{i-1})-\gamma(t_i)|)}=\sum_{i=1}^{n}{(t_{i}-t_{i-1})}+\sum_{i=1}^{n}{[\gamma(t_{i-1})-\gamma(t_i)]}=x-a+\gamma(a)-\gamma(x)=|x-a|+|\gamma(x)-\gamma(a)|. \end{array}$$
$$\begin{array}{l} L_b^y(\gamma,Q)\leq\sum_{j=1}^{m}{(|s_{j-1}-s_j|+|\gamma(s_{j-1})-\gamma(s_j)|)}=\sum_{j=1}^{m}{(s_{j}-s_{j-1})}+\sum_{j=1}^{m}{[\gamma(s_{j})-\gamma(s_{j-1})]}=y-b+\gamma(y)-\gamma(b)=|y-b|+|\gamma(y)-\gamma(b)|. \end{array}$$
Therefore:
$$|\arccos(x)-\arccos(a)|=L_a^x(\gamma)\leq |x-a|+|\gamma(x)-\gamma(a)|\stackrel{x\to a^+}{\rightarrow} 0+0=0.$$ $$|\arccos(y)-\arccos(b)|=L_b^y(\gamma)\leq |y-b|+|\gamma(y)-\gamma(b)|\stackrel{y\to b^+}{\rightarrow} 0+0=0.$$
By Sandwich's theorem, the arccosine function is continuous to the right on $[-1,1)$. Analogously, it can be proved that it's continuous to the left on $(-1,1]$. In conclusion, the arccosine function is continuous on $[-1,1]$.
