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: 47

Answer

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

Work Step by Step

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