Q:

$2\sin(\pi/6)-1$ is not exactly 0?

So while doing some basic trig problems, I had a problem with $2\sin(\pi/6)-1$. $2\sin(\pi/6)$ is exactly $1$, but somehow $2\sin(\pi/6)-1$ is $-1.11\times10^{-16}$. 
I had previously thought that $\sin(\pi/6)$ is defined to be exactly $0.5$, especially because the ratio of a $30$-$60$-$90$ triangle is exactly $1$-$\sqrt{3}$ - $2$.
So why does this happen? Is $\sin(\pi/6)$ not actually exactly 0.5?
 

A:

It's an all too common rounding error. This claimed nonzero value is $-2^{-53}$, a common small negative value for a double due to $2^{-53}$ being the smallest possible nonzero significand error.

