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: 55



Work Step by Step

$1+tan^2u=sec^2u$ $1+\frac{9}{16}=sec^2u$ $\frac{25}{16}=\frac{1}{cos^2u}$ $cos^2u=\frac{16}{25}$ $cos~u=±\frac{4}{5}~~$ (But, $u$ is in Quadrant I): $cos~u=\frac{4}{5}$ $cos^2u+sin^2u=1$ $\frac{16}{25}+sin^2u=1$ $sin^2u=\frac{9}{25}$ $sin~u=±\frac{3}{5}~~$ (But, $u$ is in Quadrant I): $sin~u=\frac{3}{5}$ $cos^2v+sin^2v=1$ $\frac{16}{25}+sin^2v=1$ $sin^2v=\frac{9}{25}$ $sin~v=±\frac{3}{5}~~$ (But, $v$ is in Quadrant III): $sin~v=-\frac{3}{5}$ $cos(u-v)=cos~u~cos~v+sin~u~sin~v=\frac{4}{5}(-\frac{4}{5})+\frac{3}{5}(-\frac{3}{5})=-\frac{16}{25}-\frac{9}{25}=-\frac{25}{25}=-1$
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