arccos equals area of unit circle sector Let $C$ be the unit circle around $0$ in $\mathbb{R}^2$.
Let $x \in [-1;1]$.
I want to prove that the area of the circular sector of $K$ formed by $(0,0)$ and two intersection points $a,b \in\{(x,y)|y \in \mathbb{R}\}$ equals $arccos(x)$.

I know that I have to show that I have to show that the area of the sector is determined and that it contains $ \{(z,0)| z \in [0;1]\} $ but how would I find it?
 A: Hint: What is the angle of that circle sector?
A: For the proof we will compute the area of circular sector via formulae:
$$A = \frac{\theta r^2}{2}$$
where $\theta$ is the angle between lines $p$ and $q$.
For line $p$, we have two points:
$$[0, 0] \text{ and } [a, \sqrt{1-a^2}]$$
The latter is derived from $x^2+y^2=1$, where $x = a$ and $a \in [-1, 1]$.
For line $q$, we have two points:
$$[0, 0] \text{ and } [a, -\sqrt{1-a^2}]$$
For simplicity, we will compute half of the angle between $p$ and $q$, that is, between $p$ and $x=0$.
Following formulae for angle between vectors:
$$\text{cos}(\gamma)=\frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| \cdot ||\vec{v}||}$$
where in our case
$$\vec{u} = (a, \sqrt{1-a^2})$$
$$\vec{v} = (1, 0)$$
Plugging the $\vec{u}$ and $\vec{v}$ into formulae, we get
$$\text{cos}(\gamma)=\frac{a}{\sqrt{a^2+|1-a^2|}}$$
We know that $a \in [-1, 1]$, therefore
$$\text{cos}(\gamma)=\frac{a}{\sqrt{a^2+1-a^2}} \implies \text{cos}(\gamma)=a$$
$$\text{cos}(\gamma)=a \implies \gamma = \text{arccos}(a)$$
We received the angle between $p$ and $x = 0$. We also know that $p$ and $q$ are axial symmetric with respect to $x$-axis, therefore we multiply the angle by 2:
$$\theta = 2 * \text{arccos}(a)$$
for $\theta$ is the angle between $p$ and $q$.
To finish the proof, plug the angle back into first formulae:
$$A = \frac{2 *\text{arccos}(a) * 1^2}{2}$$
$$A = \text{arccos}(a)$$
