Answer
$ -\dfrac{\sqrt 2}{2}$
Work Step by Step
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: $(\pi-\theta)$
c) Quadrant- III: $(\theta - \pi)$
d) Quadrant - IV: $(2\pi - \theta)$
Thus, the reference angle of $\dfrac{8 \pi}{4}-\dfrac{7\pi}{4}=\dfrac{ \pi}{4}$
So, $ \sin \dfrac{ \pi}{4}=\dfrac{\sqrt 2}{2}$
Thus, we have $ \sin \dfrac{\pi}{4}= -\dfrac{\sqrt 2}{2}$; Because $\theta $ lies in Quadrant-IV.
