Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Review Exercises - Page 242: 31

Answer

$sin~(x+y) = \frac{4-9\sqrt{11}}{50}$ $cos~(x-y) = \frac{12\sqrt{11}-3}{50}$ $tan~(x+y) = \frac{4-9\sqrt{11}}{12\sqrt{11}+3}$ $x+y~~$ is in quadrant IV.

Work Step by Step

$sin~x = \frac{1}{10}$ Since $x$ is in quadrant I, $cos~x$ is also positive. We can find $cos~x$: $sin^2~x+cos^2~x = 1$ $cos~x = \sqrt{1-sin^2~x}$ $cos~x = \sqrt{1-(\frac{1}{10})^2}$ $cos~x = \sqrt{1-\frac{1}{100}}$ $cos~x = \frac{3\sqrt{11}}{10}$ $cos~y = \frac{4}{5}$ Since $y$ is in quadrant IV, $sin~y$ is negative. We can find $sin~y$: $sin^2~y+cos^2~y = 1$ $sin~y = -\sqrt{1-cos^2~y}$ $sin~y = -\sqrt{1-(\frac{4}{5})^2}$ $sin~y = -\sqrt{1-\frac{16}{25}}$ $sin~y = -\frac{3}{5}$ We can find $sin(x+y)$: $sin~(x+y) = sin~x~cos~y+cos~x~sin~y$ $sin~(x+y) = (\frac{1}{10})(\frac{4}{5})+(\frac{3\sqrt{11}}{10})(-\frac{3}{5})$ $sin~(x+y) = \frac{2}{25}-\frac{9\sqrt{11}}{50}$ $sin~(x+y) = \frac{4-9\sqrt{11}}{50}$ We can find $cos(x-y)$: $cos~(x-y) = cos~x~cos~(-y)-sin~x~sin~(-y)$ $cos~(x-y) = cos~x~cos~y-sin~x~(-sin~y)$ $cos~(x-y) = (\frac{3\sqrt{11}}{10})(\frac{4}{5})-(\frac{1}{10})(-\frac{-3}{5})$ $cos~(x-y) = \frac{12\sqrt{11}}{50}-\frac{3}{50}$ $cos~(x-y) = \frac{12\sqrt{11}-3}{50}$ We can find $cos(x+y)$: $cos~(x+y) = cos~x~cos~y-sin~x~sin~y$ $cos~(x+y) = (\frac{3\sqrt{11}}{10})(\frac{4}{5})-(\frac{1}{10})(-\frac{3}{5})$ $cos~(x+y) = \frac{12\sqrt{11}}{50}+\frac{3}{50}$ $cos~(x+y) = \frac{12\sqrt{11}+3}{50}$ We can find $tan~(x+y)$: $tan~(x+y) = \frac{sin~(x+y)}{cos~(x+y)}$ $tan~(x+y) = \frac{\frac{4-9\sqrt{11}}{50}}{\frac{12\sqrt{11}+3}{50}}$ $tan~(x+y) = \frac{4-9\sqrt{11}}{12\sqrt{11}+3}$ Since $sin~(x+y)$ and $tan~(x+y)$ are both negative, then $x+y$ must be in quadrant IV.
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