## Precalculus (6th Edition) Blitzer

$\dfrac{\sqrt 2}{2}$
The reference angle of an angle $0 \leq \theta \lt 2\pi$ based on its position can be computed by using the following steps: a) Quadrant- I: $\theta$ b) Quadrant- II: $180^{\circ}-\theta$ c) Quadrant -III: $\theta - 180^o$ d) Quadrant -IV: $360^{\circ}-\theta$ Reference angle of $225^{\circ}$ is equal to $=225^{\circ}-180^{\circ}=40^{\circ}$ Since, $\sin 45^{\circ}=\dfrac{\sqrt 2}{2}$ So, $\sin (-225)^{\circ}= \dfrac{\sqrt 2}{2}$; Because $\theta$ lies in Quadrant-III.