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: 54

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. $cot\frac{13\pi}{6}=\frac{1}{tan\frac{13\pi}{6}}=\sqrt 3$. 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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