## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Test - Page 647: 9

#### 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.

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