Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.1 - Using Fundamental Identities - 7.1 Exercises - Page 513: 8

Answer

$sin~x=-\frac{6}{7}$ $cos~x=-\frac{\sqrt {13}}{7}$ $tan~x=\frac{6\sqrt {13}}{13}$ $cot~x=\frac{\sqrt {13}}{6}$ $sec~x=-\frac{7\sqrt {13}}{13}$

Work Step by Step

$csc~x=\frac{1}{sin~x}\lt0$ $tan~x=\frac{sin~x}{cos~x}\gt0$ We can conclude that: $sin~x\lt0$ $cos~x\lt0$ $cot~x=\frac{1}{tan~x}\gt0$ $sec~x=\frac{1}{cos~x}\lt0$ $sin~x=\frac{1}{csc~x}=\frac{1}{-\frac{7}{6}}=-\frac{6}{7}$ $cos^2x+sin^2x=1$ $cos^2x=1-\frac{36}{49}=\frac{13}{49}$ $cos~x=-\frac{\sqrt {13}}{7}$ $tan~x=\frac{sin~x}{cos~x}=\frac{-\frac{6}{7}}{-\frac{\sqrt {13}}{7}}=\frac{6}{\sqrt {13}}=\frac{6\sqrt {13}}{13}$ $cot~x=\frac{cos~x}{sin~x}=\frac{-\frac{\sqrt {13}}{7}}{-\frac{6}{7}}=\frac{\sqrt {13}}{6}$ $sec~x=\frac{1}{cos~x}=-\frac{7}{\sqrt {13}}=-\frac{7\sqrt {13}}{13}$
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