[Math] the values of the sine function coming from

Question

On high school, I was taught that I could obtain any sine value with some basic arithmetic on the values of the following image:

But I never really understood where these values where coming from, some days ago I started to explore it but I couldn't discover it. After reading for a while, I remembered that the sine function is:

$$\sin=\frac{\text{opposite}}{\text{hypotenuse}}$$

Then I thought that I just needed to calculate $\frac{1}{x}$ where $0 \leq x \leq 1$ but it gave me no good results, then I thought that perhaps I could express not as a proportion of the opposite and hypotenuse, I thought I could express it as the ratio between slices of the circumference, for example: circumference $=\pi$, then divided it by $4$ (to obtain the slice from $0$ to $90$ degrees) then I came with: $x/ \frac{\pi}{4}$ where $0\leq x \leq \frac{\pi}{4} $ but it also didn't work, the best guess I could make was $\sqrt{x/ \frac{\pi}{4}}$, the result is in the following plot:

The last guess I made seems to be (at least visually) very similar to the original sine function, it seems it needs only to be rotated but from here, I'm out of ideas. Can you help me?

Answer 1

All of the values in your picture can be deduced from two theorems:

1. The Pythagorean theorem: If a right triangle has sides $a,b,c$ where $c$ is the hypotenuse, then $a^2+b^2=c^2$
2. If a right triangle has an angle of $\frac{\pi}{6} = 30^\circ$, then the length of the side opposite to that angle is half the length of the hypotenuse.

Both can be proven with elementary high school geometry.

Let's see how this works for Quadrant I (angles between $0$ and $90^\circ$), as the rest follows from identities.

1. $\sin 0 = 0$; this is clear from the definition.
2. $\sin 90^\circ = 1$; less intuitive because it breaks the triangle, but $\sin 90^\circ =\cos0$, and $\cos 0 = 1$ because the adjacent side and the hypotenuse coincide when the angle is $0$.
3. $\sin 45^\circ = \frac{1}{\sqrt{2}}$; this corresponds to an isosceles triangle, and if we set the sides to be $1$, then by the Pythagorean theorem, the hypotenuse is $\sqrt{2}$.
4. $\sin30^\circ = \frac{1}{2}$; this follows immediately from theorem 2 above.
5. $\sin60^\circ = \frac{\sqrt{3}}{2}$; if we take a $30^\circ-60^\circ-90^\circ$ triangle, and set the side opposite to the $30^\circ$ angle to be $1$, then the hypotenuse is $2$ and the side opposite to the $60^\circ$ angle satisfies $x^2 + 1 = 2^2$, so its length is $\sqrt{3}$ - and thus $\sin 60^\circ = \frac{\sqrt{3}}{2}$.