## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Review Exercises - Page 248: 28

#### Answer

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

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