Trigonometry (10th Edition)

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

Chapter 3 - Review Exercises - Page 130: 39

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