Basic trigonometry - $\sin{(-a)} = -\sin{(a)}$

Question

If $\sin (-a) = -\sin (a)$, why isn't $\sin (-210°) = - \sin (180° + 30°) = - \sin (30°) = -0.5$ ?

The answer is positive $0.5$ instead.

Answer 1

\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}

Answer 2

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.

Answer 3

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}