Prove $ \forall x\in\mathbb{R} : |\sin{x}|<|x|$ geometrically

Question

For $x\in(-\frac{\pi}{2}, -\frac{\pi}{2})\setminus \{0\}$ we can say that since the area of sector of circle is greater than the triangle (ABC) area I have $2|\sin{x}|<2|x|$. But what's about x in the 2nd and 3rd quadrant? Look at the figure to easily understand what I am saying (Note: I have drawn a circle with a radius $>1$ to have a larger circle, but I am referring to a unit circle)

Answer 1

Area of the triangle is $$S_\triangle = \frac{1\cdot 1\cdot \sin x}{2}$$ Area of the sector is $$S_\text{sec} = \frac{1^2\cdot x}{2}$$ Since it is vivid from the picture below that $S_\text{sec} > S_\triangle$, we have $$\sin x < x$$ As has already been noted in the comment, you just need to verify only for $x \in [-1, 1] \subset (-\pi/2, \pi/2)$. The above argument can easily be extended to $\forall x \in [-1, 1]$ (note: equality occurs when $x=0$).

---

$\quad\quad\quad\quad\quad\quad\quad~$

Answer 2

Hints: Arclength (with units of distance) is the product of the radius $1$ and the angle $x$. Show how this arclength is longer than the altitude by showing how on a Euclidean plane, a line segment is the shortest path between two points.

Answer 3

Hint: The inequality is trivial when $\lvert x\rvert>1$. (It is also not true when $x=0$.)