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 249: 36


$sin~2y = -\frac{24}{25}$ $cos~2y = -\frac{7}{25}$

Work Step by Step

Since $sec~y$ is negative and $sin~y$ is positive, the angle $y$ lies in quadrant II. $sec~y = \frac{hyp}{adj} = -\frac{5}{3} = \frac{5}{-3}$ If the hypotenuse is 5 and the adjacent side is -3, then the adjacent side is $\sqrt{(5)^2-(-3)^2} = 4$ $sin~y = \frac{opp}{hyp} = \frac{4}{5}$ $cos~y = \frac{adj}{hyp} = \frac{-3}{5}$ We can find $sin~2y$: $sin~2y = 2~sin~y~cos~y$ $sin~2y = (2)(\frac{4}{5})(\frac{-3}{5})$ $sin~2y = -\frac{24}{25}$ We can find $cos~2y$: $cos~2y = cos^2~y-sin^2~y$ $cos~2y = (\frac{-3}{5})^2-(\frac{4}{5})^2$ $cos~2y = \frac{9}{25}-\frac{16}{25}$ $cos~2y = -\frac{7}{25}$
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.