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: 48

Answer

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

Work Step by Step

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