<s>[INST] Geometry in a circle Can somebody explain to me why the line $DB$ is equal to $\sin(\alpha + \beta)$. I haven't been able to figure it out. My image (which I would post if I could) is located here: 
For example,
if I try to use the law of cosines then,
$$(\cos \alpha)^2 + (\cos \beta)^2 - 2(\cos \alpha) (\cos \beta) (\cos(\alpha + \beta)) = DB$$
which does not obviously equal $\sin(\alpha + \beta)$.
 [/INST] (I use $a$ and $b$ instead of $\alpha$ and $\beta$ because I am lazy.)
\begin{align}
& (\cos a)^2 + (\cos b)^2 - 2(\cos a) (\cos b) (\cos(a + b)) \\
= {} & (\cos a)^2 + (\cos b)^2 - 2(\cos a)(\cos b)(\cos a \cos b - \sin a \sin b) \\
=  {} & (\cos a)^2 + (\cos b)^2 - 2(\cos a)^2(\cos b)^2
+ 2 (\cos a)(\cos b)(\sin a)(\sin b) \\
= {} & (\cos a)^2(1-(\cos b)^2) + (\cos b)^2(1-(\cos a)^2)
+ 2(\cos a)(\cos b)(\sin a \sin b) \\
= {} & (\cos a)^2 (\sin b)^2 + (\cos b)^2 (\sin a)^2
+ 2(\cos a)(\cos b)(\sin a \sin b) \\
= {} & (\cos a \sin b + \cos b \sin a)^2 \\
= {} & (\sin(a+b))^2.
\end{align}
So I think the expression you have
is for $DB^2$, not $DB$, but, other than that,
your expression was correct - just not simplified.
</s>