Algebra and Trigonometry 10th Edition

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

Chapter 7 - Review Exercises - Page 552: 5

Answer

$sin~x=-\frac{\sqrt {21}}{5}$ $cos~x=-\frac{2}{5}$ $tan~x=\frac{\sqrt {21}}{2}$ $cot~x=\frac{2\sqrt {21}}{21}$ $csc~x=-\frac{5\sqrt {21}}{21}$ $sec~x=-\frac{5}{2}$

Work Step by Step

$cos~x\lt0$ $tan~x=\frac{sin~x}{cos~x}\gt0$ We can conclude that: $sin~x\lt0$ $cot~x=\frac{1}{tan~x}\gt0$ $sec~x=\frac{1}{cos~x}\lt0$ $csc~x=\frac{1}{sin~x}\lt0$ $sin^2x+cos^2x=1$ $sin^2x=1-\frac{4}{25}=\frac{21}{25}$ $sin~x=-\frac{\sqrt {21}}{5}$ $tan~x=\frac{sin~x}{cos~x}=\frac{-\frac{\sqrt {21}}{5}}{-\frac{2}{5}}=\frac{\sqrt {21}}{2}$ $cot~x=\frac{x}{y}=\frac{-\frac{2}{5}}{-\frac{\sqrt {21}}{5}}=\frac{2}{\sqrt {21}}=\frac{2\sqrt {21}}{21}$ $csc~x=\frac{1}{sin~x}=\frac{1}{-\frac{\sqrt {21}}{5}}=-\frac{5}{\sqrt {21}}=-\frac{5\sqrt {21}}{21}$ $sec~x=\frac{1}{cos~x}=\frac{1}{-\frac{2}{5}}=-\frac{5}{2}$
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