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'Sine and Cosine for Connected Angles' printed from https://nrich.maths.org/
Chris from Saint John Payne School sent in
clear diagrams to explain the first part.
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 $\angle A O B=\angle B O C=2\theta$. I'm going to
show that $\sin(2\theta)=2\sin\theta\cos\theta$. This is called a
double angle formula.\par From triangle $O A M$, we know that $A
M=\sin(2\theta)$ (as the circle has radius $1$).
From triangle $O A B$, we know that $A B=2\sin\theta$ (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 $\theta$ at
$O$).
Since $\angle A B O=90^{\circ}-\theta$ (from the isosceles triangle
$A O B$), we know that $\angle B A M=\theta$, and so from triangle
$A B M$ we see that $A M=A B\cos\theta$.
Putting together the last two paragraphs, we get $A
M=2\sin\theta\cos\theta$. But also $A M=\sin(2\theta)$, so
$\sin(2\theta)=2\sin\theta\cos\theta$.