Trigonometry (11th Edition) Clone

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

Chapter 3 - Review Exercises - Page 136: 41

Answer

$\tan{1} \gt \tan{2}$

Work Step by Step

Recall: (1) The tangent function's period is $\pi$ or around $3.14$; (2) The tangent function is increasing in the intervals $(-1.57, 1.57), (1.57, 4.71), (4.71, 7.85), ...$ (3) The value of the tangent function is $0$ when $x=0, \pi, 2\pi, 3\pi, ...$ or, in approximate decimal form, $x=0, 3.14, 6.28, 9.42, ...$. (4) The value of the tangent function is negative in the intervals $(-1.57, 0), (1.57, 3.14), (4.71, 6.28), ....$ (5) The value of the tangent function is positive in the intervals $(0, 1.57), (3.14, 4.71), (6.28, 7.85), ....$ Thus, the value of $\tan{1}$ is positive, and the value of $\tan{2}$ is negative. Therefore, $\tan{1} \gt \tan{2}$.
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.