Chris from Saint John Payne School sent in clear diagrams to explain the first part.
solution picture 1

Diana thought about a general result. Here's what she sent us.

In general, suppose that we've placed points

A

B

and

C

in such a way that

∠ A O B = ∠ B O C = 2 θ

. I'm going to show that

\sin (2 θ) = 2 \sin θ \cos θ

. This is called a double angle formula.

From triangle $O A M$ , we know that $A M = \sin (2 θ)$ (as the circle has radius $1$ ).

From triangle $O A B$ , we know that $A B = 2 \sin θ$ (the blue line bisects the angle at $O$ and since triangle $A O B$ is isosceles, the blue line meets $A B$ at a right angle, so we can think about two right-angled triangles, each with angle $θ$ at $O$ ).

Since $∠ A B O = 90^{\circ} - θ$ (from the isosceles triangle $A O B$ ), we know that $∠ B A M = θ$ , and so from triangle $A B M$ we see that $A M = A B \cos θ$ .

Putting together the last two paragraphs, we get $A M = 2 \sin θ \cos θ$ . But also $A M = \sin (2 θ)$ , so $\sin (2 θ) = 2 \sin θ \cos θ$ .