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: 9

Answer

$sin~x=-\frac{3}{4}$ $cos~x=\frac{\sqrt {7}}{4}$ $tan~x=-\frac{3\sqrt {7}}{7}$ $cot~x=-\frac{\sqrt {7}}{3}$ $csc~x=-\frac{4}{3}$ $sec~x=\frac{4\sqrt {7}}{7}$

Work Step by Step

$sin~x\lt0$ $cos~x\gt0$ We can conclude that: $tan~x=\frac{sin~x}{cos~x}\lt0$ $cot~x=\frac{cosx}{~sin~x}\lt0$ $csc~x=\frac{1}{sin~x}\lt0$ $sec~x=\frac{1}{cos~x}\gt0$ $cos^2x+sin^2x=1$ $cos^2x=1-\frac{9}{16}=\frac{7}{16}$ $cos~x=\frac{\sqrt {7}}{4}$ $tan~x=\frac{sin~x}{cos~x}=\frac{-\frac{3}{4}}{\frac{\sqrt {7}}{4}}=-\frac{3}{\sqrt {7}}=-\frac{3\sqrt {7}}{7}$ $cot~x=\frac{cos~x}{sin~x}=\frac{\frac{\sqrt {7}}{4}}{-\frac{3}{4}}=-\frac{\sqrt {7}}{3}$ $csc~x=\frac{1}{sin~x}=-\frac{4}{3}$ $sec~x=\frac{1}{cos~x}=\frac{4}{\sqrt {7}}=\frac{4\sqrt {7}}{7}$
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