Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 403: 64

Answer

1. $sin(-\frac{13\pi}{6})= -\frac{1}{2}$. 2. $cos(-\frac{13\pi}{6})= \frac{\sqrt 3}{2}$. 3. $tan(-\frac{13\pi}{6})= -\frac{\sqrt 3}{3}$. 4. $cot(-\frac{13\pi}{6})= -\sqrt 3$. 5. $sec(-\frac{13\pi}{6})= \frac{2\sqrt 3}{3}$. 6. $csc(-\frac{13\pi}{6})= -2$.

Work Step by Step

1. $sin(-\frac{13\pi}{6})=sin(-2\pi-\frac{\pi}{6})=-sin(\frac{\pi}{6})=-\frac{1}{2}$. 2. $cos(-\frac{13\pi}{6})=cos(-2\pi-\frac{\pi}{6})=cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$. 3. $tan(-\frac{13\pi}{6})=tan(-2\pi-\frac{\pi}{6})=-tan(\frac{\pi}{6})=-\frac{\sqrt 3}{3}$. 4. $cot(-\frac{13\pi}{6})=cot(-2\pi-\frac{\pi}{6})=-cot(\frac{\pi}{6})=-\sqrt 3$. 5. $sec(-\frac{13\pi}{6})=\frac{1}{cos(-\frac{13\pi}{6})}=\frac{2\sqrt 3}{3}$. 6. $csc(-\frac{13\pi}{6})=\frac{1}{sin(-\frac{13\pi}{6})}=-2$.
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