## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 54

#### Answer

$\dfrac{\pi}{4}$

#### Work Step by Step

$\dfrac{11\pi}{4}$ is close to $\dfrac{12\pi}{4}=3\pi$. This means that this angle is between $2.5\pi$ and $3\pi$. RECALL: An angle $\theta$, where $2.5\pi \lt \theta \lt 3\pi$, is coterminal with: $\theta - 2\pi$ Thus, the given angle is coterminal with: $=\dfrac{11\pi}{4} - 2\pi=\dfrac{11\pi}{4} - \dfrac{8\pi}{4} = \dfrac{3\pi}{4}$ $\dfrac{11\pi}{4}$ is coterminal with $\dfrac{3\pi}{4}$. $\dfrac{3\pi}{4}$ is in Quadrant II so $\dfrac{11\pi}{4}$ is also in Quadrant II. RECALL: The following are the means on how to find the reference angle of an angle $0 \leq \theta \lt 2\pi$ based on its position: (1) Quadrant I: $\theta$ (itself) (2) Quadrant II: $\pi-\theta$ (3) Quadrant III: $\theta - \pi$ (4) Quadrant IV: $2\pi - \theta$ Use formula (2) above to obtain: reference angle of $\dfrac{11\pi}{4}$ = reference angle of $\dfrac{3\pi}{4}$, which is $\\=\pi - \dfrac{3\pi}{4} \\=\dfrac{4\pi}{4} - \dfrac{3\pi}{4} \\=\dfrac{\pi}{4}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.