Basic trigonometry - $\sin{(-a)} = -\sin{(a)}$ If $\sin (-a) = -\sin (a)$,
why isn't $\sin (-210°) = - \sin (180° + 30°) = - \sin (30°) = -0.5$ ?
The answer is positive $0.5$ instead. 
 A: \begin{align}
\sin (-210°) &= -\sin(210°) \tag{$\sin(a) = -\sin(a)$} \\
&= - \sin (180° + 30°) \\
&= - (-\sin (30°)) \tag{$\sin(180° + a) = -\sin(a)$} \\
&= 0.5
\end{align}
A: When you have such doubts, always think to the geometrical meaning of $\sin \theta$ and in which quadrant $\theta$ is.
Notably $\sin \theta$ represent the $y$-coordinate of the point on the trigonometric circle, since $\theta=-210°=-210°+360°=150°$ is in the second quadrant, $\sin -210°$ must be positive.
Of course, to determine the value you need more precise evaluation by trigonometric identities but you need to have the correct idea and feeling on the sign.
A: For a very simple and important reason, surprisingly not enough emphatized in the previous two answers, in my opinion.
\begin{align}
\sin(-210°)&=-\sin(210°)=\\
&=-\sin(180°+30°)\color{red}{\ne}\\
&\color{red}{\ne}-(\sin(180°)+\sin(30°)).
\end{align}
The first equality holds because, as you know,
$$\sin(-a)=-\sin(a)$$
for all $a\in\mathbb R$. But while $f(-a)=-f(a)$ is a property of linear functions, is important to say that

$\sin(a)$ is not a linear function of $a$

and, for example,
$$\sin(a+b)=\sin(a)+\sin(b)$$
does not hold for all $a,b\in\mathbb R$. In particular, it holds for $a=180°$ and $b=0°$ but it does not hold for $a=180°$ and $b=30°$. On the other hand we know that
$$\sin(180°+b)=-\sin(b)$$
holds for all $b\in\mathbb R$.
Hence:
\begin{align}
\sin(-210°)&=-\sin(210°)=\\
&=-\sin(180°+30°)=\\
&=-(-\sin(30°))=\\
&=\sin(30°)=\frac{1}{2}>0.
\end{align}
