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: 63

Answer

1. $sin(-\frac{14\pi}{3})= -\frac{\sqrt 3}{2}$. 2. $cos(-\frac{14\pi}{3})= -\frac{1}{2}$. 3. $tan(-\frac{14\pi}{3})= \sqrt 3$. 4. $cot(-\frac{14\pi}{3})= \frac{\sqrt 3}{3}$. 5. $sec(-\frac{14\pi}{3})= -2$. 6. $csc(-\frac{14\pi}{3})= -\frac{2\sqrt 3}{3}$.

Work Step by Step

1. $sin(-\frac{14\pi}{3})=sin(-5\pi+\frac{\pi}{3})=-sin\frac{\pi}{3}=-\frac{\sqrt 3}{2}$. 2. $cos(-\frac{14\pi}{3})=cos(-5\pi+\frac{\pi}{3})=-cos\frac{\pi}{3}=-\frac{1}{2}$. 3. $tan(-\frac{14\pi}{3})=tan(-5\pi+\frac{\pi}{3})=tan\frac{\pi}{3}=\sqrt 3$. 4. $cot(-\frac{14\pi}{3})=cot(-5\pi+\frac{\pi}{3})=cot\frac{\pi}{3}=\frac{\sqrt 3}{3}$. 5. $sec(-\frac{14\pi}{3})=\frac{1}{cos(-\frac{14\pi}{3})}=-2$. 6. $csc(-\frac{14\pi}{3})=\frac{1}{sin(-\frac{14\pi}{3})}=-\frac{2\sqrt 3}{3}$.
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