Why do we consider the third quadrant to decide the sign of sin$(180^° + \theta)$? sin$(300^°)$ $=$ sin$(180^°+120^°)$ $=$ – sin$(120^°)$

In the above example, we have given a negative sign as it is of the form sin$(180^° + \theta)$ and it lies in the third quadrant where sin is negative. So we write -sin$(\theta)$. But the angle 180° + 120° lies in the fourth quadrant. So why are we considering the third quadrant to decide the sign? Shouldn't we consider the fourth quadrant?
 A: One of the identities that holds for the sine function is
$$\sin(180° + x) = -\sin(x)$$
which does not require to consider quadrants or anything like that to be used!
A: By definition $\forall \theta$
$$\sin(\pi + \theta) = -\sin(\theta)=\sin(-\theta)$$
as we can check drawing the unitary circle to see that $\sin (\pi + \theta)$ has the same magnitude of $\sin \theta$ but with the reversed sign.
This is a straightforward check for $\theta$ in the first quadrant. For $\theta$ in the second quadrant just consider
$$\alpha=\theta -\frac \pi 2  \implies \theta = \frac \pi 2+\alpha$$
then
$$\sin(\pi + \theta)=\sin\left(\pi +  \frac \pi 2+\alpha\right)=\sin\left( \frac {3\pi} 2+\alpha\right)=-\cos (\alpha)=-\cos\left( \theta -\frac \pi 2\right)=$$
$$=\cos\left( \frac \pi 2-\theta \right)=\sin(\theta)$$
and similarly for all other cases, just using symmetry.
A: For any real $x$ we have $$\sin(\pi+x)=\sin\pi\cos{x}+\cos\pi\sin{x}=-\sin{x}.$$
Id est, $x$ may be ended in any quadrant.
A: Although I would express the idea somewhat differently, I completely agree with
the OP's analysis.
That is, it is geometrically clear that $\sin(300^{\circ}) = \pm \sin(120^{\circ}).$
Further, $120^{\circ}$ is in the second quadrant while $300^{\circ}$ is in the fourth quadrant.
Therefore, $\sin(300^{\circ})$ must be $< 0$, while
$\sin(120^{\circ})$ must be $> 0.$
It could be that reference to the third quadrant was an unnoticed error, re all angles in the third quadrant (also) have a negative value for their sine.
