How to calculate sin ( - 150 ° ) ?

Given solution: ${\displaystyle \sin (-{150}^{\circ })}$
${\displaystyle =-\sin \left({150}^{\circ }\right)}$
${\displaystyle =-\sin ({180}^{\circ }-{30}^{\circ })}$
${\displaystyle =-\sin \left({30}^{\circ }\right) (\because \sin ({180}^{\circ }-\theta )=\sin \theta )}$
${\displaystyle =-\frac{1}{2}}$

Recall the identification of the negative angle.
${\displaystyle \sin (-\theta )=-\sin \left(\theta \right)}$
With this in mind, we can rewrite #sin(-150)# as #-sin(150)#
${\displaystyle {150}^{\circ }}$ has a reference angle of ${\displaystyle {30}^{\circ }}$, which means it will have the same trig values as ${\displaystyle {30}^{\circ }}$
On the Unit Circle, we know the coordinates for ${\displaystyle {30}^{\circ }}$ are ${\displaystyle (\frac{\sqrt{3}}{2},\frac{1}{2})}$, where the #y#-coordinate is the ${\displaystyle \sin }$ value.
This implies ${\displaystyle \sin (-150)=-\frac{1}{2}}$