Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 248: 28


$sin~(x+y) = \frac{44}{125}$ $cos~(x-y) = \frac{3}{5}$ $tan~(x+y) = \frac{44}{117}$ $x+y~~$ is in quadrant I

Work Step by Step

$sin~x = \frac{3}{5}$ Since $x$ is in quadrant I, $cos~x$ is also positive. We can find $cos~x$: $sin^2~x+cos^2~x = 1$ $cos~x = \sqrt{1-sin^2~x}$ $cos~x = \sqrt{1-(\frac{3}{5})^2}$ $cos~x = \sqrt{1-\frac{9}{25}}$ $cos~x = \frac{4}{5}$ $cos~y = \frac{24}{25}$ Since $y$ is in quadrant IV, $sin~y$ is negative. We can find $sin~y$: $sin^2~y+cos^2~y = 1$ $sin~y = -\sqrt{1-cos^2~y}$ $sin~y = -\sqrt{1-(\frac{24}{25})^2}$ $sin~y = -\sqrt{1-\frac{576}{625}}$ $sin~y = -\frac{7}{25}$ We can find $sin(x+y)$: $sin~(x+y) = sin~x~cos~y+cos~x~sin~y$ $sin~(x+y) = (\frac{3}{5})(\frac{24}{25})+(\frac{4}{5})(-\frac{7}{25})$ $sin~(x+y) = \frac{72}{125}-\frac{28}{125}$ $sin~(x+y) = \frac{44}{125}$ We can find $cos(x-y)$: $cos~(x-y) = cos~x~cos~(-y)-sin~x~sin~(-y)$ $cos~(x-y) = cos~x~cos~y-sin~x~(-sin~y)$ $cos~(x-y) = (\frac{4}{5})(\frac{24}{25})-(\frac{3}{5})(-\frac{-7}{25})$ $cos~(x-y) = \frac{96}{125}-\frac{21}{125}$ $cos~(x-y) = \frac{75}{125}$ $cos~(x-y) = \frac{3}{5}$ We can find $cos(x+y)$: $cos~(x+y) = cos~x~cos~y-sin~x~sin~y$ $cos~(x+y) = (\frac{4}{5})(\frac{24}{25})-(\frac{3}{5})(\frac{-7}{25})$ $cos~(x+y) = \frac{96}{125}+\frac{21}{125}$ $cos~(x+y) = \frac{117}{125}$ We can find $tan~(x+y)$: $tan~(x+y) = \frac{sin~(x+y)}{cos~(x+y)}$ $tan~(x+y) = \frac{ \frac{44}{125}}{\frac{117}{125}}$ $tan~(x+y) = \frac{44}{117}$ Since $sin~(x+y)$ and $tan~(x+y)$ are both positive, then $x+y$ must be in quadrant I.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.